from __future__ import annotations import math import warnings from collections import namedtuple import numpy as np from numpy import (isscalar, r_, log, around, unique, asarray, zeros, arange, sort, amin, amax, atleast_1d, sqrt, array, compress, pi, exp, ravel, count_nonzero, sin, cos, arctan2, hypot) from scipy import optimize, special, interpolate, stats from scipy._lib._bunch import _make_tuple_bunch from scipy._lib._util import _rename_parameter, _contains_nan, _get_nan from . import _statlib from . import _stats_py from ._fit import FitResult from ._stats_py import find_repeats, _normtest_finish, SignificanceResult from .contingency import chi2_contingency from . import distributions from ._distn_infrastructure import rv_generic from ._hypotests import _get_wilcoxon_distr from ._axis_nan_policy import _axis_nan_policy_factory from .._lib.deprecation import _deprecated __all__ = ['mvsdist', 'bayes_mvs', 'kstat', 'kstatvar', 'probplot', 'ppcc_max', 'ppcc_plot', 'boxcox_llf', 'boxcox', 'boxcox_normmax', 'boxcox_normplot', 'shapiro', 'anderson', 'ansari', 'bartlett', 'levene', 'binom_test', 'fligner', 'mood', 'wilcoxon', 'median_test', 'circmean', 'circvar', 'circstd', 'anderson_ksamp', 'yeojohnson_llf', 'yeojohnson', 'yeojohnson_normmax', 'yeojohnson_normplot', 'directional_stats', 'false_discovery_control' ] Mean = namedtuple('Mean', ('statistic', 'minmax')) Variance = namedtuple('Variance', ('statistic', 'minmax')) Std_dev = namedtuple('Std_dev', ('statistic', 'minmax')) def bayes_mvs(data, alpha=0.90): r""" Bayesian confidence intervals for the mean, var, and std. Parameters ---------- data : array_like Input data, if multi-dimensional it is flattened to 1-D by `bayes_mvs`. Requires 2 or more data points. alpha : float, optional Probability that the returned confidence interval contains the true parameter. Returns ------- mean_cntr, var_cntr, std_cntr : tuple The three results are for the mean, variance and standard deviation, respectively. Each result is a tuple of the form:: (center, (lower, upper)) with `center` the mean of the conditional pdf of the value given the data, and `(lower, upper)` a confidence interval, centered on the median, containing the estimate to a probability ``alpha``. See Also -------- mvsdist Notes ----- Each tuple of mean, variance, and standard deviation estimates represent the (center, (lower, upper)) with center the mean of the conditional pdf of the value given the data and (lower, upper) is a confidence interval centered on the median, containing the estimate to a probability ``alpha``. Converts data to 1-D and assumes all data has the same mean and variance. Uses Jeffrey's prior for variance and std. Equivalent to ``tuple((x.mean(), x.interval(alpha)) for x in mvsdist(dat))`` References ---------- T.E. Oliphant, "A Bayesian perspective on estimating mean, variance, and standard-deviation from data", https://scholarsarchive.byu.edu/facpub/278, 2006. Examples -------- First a basic example to demonstrate the outputs: >>> from scipy import stats >>> data = [6, 9, 12, 7, 8, 8, 13] >>> mean, var, std = stats.bayes_mvs(data) >>> mean Mean(statistic=9.0, minmax=(7.103650222612533, 10.896349777387467)) >>> var Variance(statistic=10.0, minmax=(3.176724206..., 24.45910382...)) >>> std Std_dev(statistic=2.9724954732045084, minmax=(1.7823367265645143, 4.945614605014631)) Now we generate some normally distributed random data, and get estimates of mean and standard deviation with 95% confidence intervals for those estimates: >>> n_samples = 100000 >>> data = stats.norm.rvs(size=n_samples) >>> res_mean, res_var, res_std = stats.bayes_mvs(data, alpha=0.95) >>> import matplotlib.pyplot as plt >>> fig = plt.figure() >>> ax = fig.add_subplot(111) >>> ax.hist(data, bins=100, density=True, label='Histogram of data') >>> ax.vlines(res_mean.statistic, 0, 0.5, colors='r', label='Estimated mean') >>> ax.axvspan(res_mean.minmax[0],res_mean.minmax[1], facecolor='r', ... alpha=0.2, label=r'Estimated mean (95% limits)') >>> ax.vlines(res_std.statistic, 0, 0.5, colors='g', label='Estimated scale') >>> ax.axvspan(res_std.minmax[0],res_std.minmax[1], facecolor='g', alpha=0.2, ... label=r'Estimated scale (95% limits)') >>> ax.legend(fontsize=10) >>> ax.set_xlim([-4, 4]) >>> ax.set_ylim([0, 0.5]) >>> plt.show() """ m, v, s = mvsdist(data) if alpha >= 1 or alpha <= 0: raise ValueError("0 < alpha < 1 is required, but alpha=%s was given." % alpha) m_res = Mean(m.mean(), m.interval(alpha)) v_res = Variance(v.mean(), v.interval(alpha)) s_res = Std_dev(s.mean(), s.interval(alpha)) return m_res, v_res, s_res def mvsdist(data): """ 'Frozen' distributions for mean, variance, and standard deviation of data. Parameters ---------- data : array_like Input array. Converted to 1-D using ravel. Requires 2 or more data-points. Returns ------- mdist : "frozen" distribution object Distribution object representing the mean of the data. vdist : "frozen" distribution object Distribution object representing the variance of the data. sdist : "frozen" distribution object Distribution object representing the standard deviation of the data. See Also -------- bayes_mvs Notes ----- The return values from ``bayes_mvs(data)`` is equivalent to ``tuple((x.mean(), x.interval(0.90)) for x in mvsdist(data))``. In other words, calling ``.mean()`` and ``.interval(0.90)`` on the three distribution objects returned from this function will give the same results that are returned from `bayes_mvs`. References ---------- T.E. Oliphant, "A Bayesian perspective on estimating mean, variance, and standard-deviation from data", https://scholarsarchive.byu.edu/facpub/278, 2006. Examples -------- >>> from scipy import stats >>> data = [6, 9, 12, 7, 8, 8, 13] >>> mean, var, std = stats.mvsdist(data) We now have frozen distribution objects "mean", "var" and "std" that we can examine: >>> mean.mean() 9.0 >>> mean.interval(0.95) (6.6120585482655692, 11.387941451734431) >>> mean.std() 1.1952286093343936 """ x = ravel(data) n = len(x) if n < 2: raise ValueError("Need at least 2 data-points.") xbar = x.mean() C = x.var() if n > 1000: # gaussian approximations for large n mdist = distributions.norm(loc=xbar, scale=math.sqrt(C / n)) sdist = distributions.norm(loc=math.sqrt(C), scale=math.sqrt(C / (2. * n))) vdist = distributions.norm(loc=C, scale=math.sqrt(2.0 / n) * C) else: nm1 = n - 1 fac = n * C / 2. val = nm1 / 2. mdist = distributions.t(nm1, loc=xbar, scale=math.sqrt(C / nm1)) sdist = distributions.gengamma(val, -2, scale=math.sqrt(fac)) vdist = distributions.invgamma(val, scale=fac) return mdist, vdist, sdist @_axis_nan_policy_factory( lambda x: x, result_to_tuple=lambda x: (x,), n_outputs=1, default_axis=None ) def kstat(data, n=2): r""" Return the nth k-statistic (1<=n<=4 so far). The nth k-statistic k_n is the unique symmetric unbiased estimator of the nth cumulant kappa_n. Parameters ---------- data : array_like Input array. Note that n-D input gets flattened. n : int, {1, 2, 3, 4}, optional Default is equal to 2. Returns ------- kstat : float The nth k-statistic. See Also -------- kstatvar : Returns an unbiased estimator of the variance of the k-statistic moment : Returns the n-th central moment about the mean for a sample. Notes ----- For a sample size n, the first few k-statistics are given by: .. math:: k_{1} = \mu k_{2} = \frac{n}{n-1} m_{2} k_{3} = \frac{ n^{2} } {(n-1) (n-2)} m_{3} k_{4} = \frac{ n^{2} [(n + 1)m_{4} - 3(n - 1) m^2_{2}]} {(n-1) (n-2) (n-3)} where :math:`\mu` is the sample mean, :math:`m_2` is the sample variance, and :math:`m_i` is the i-th sample central moment. References ---------- http://mathworld.wolfram.com/k-Statistic.html http://mathworld.wolfram.com/Cumulant.html Examples -------- >>> from scipy import stats >>> from numpy.random import default_rng >>> rng = default_rng() As sample size increases, n-th moment and n-th k-statistic converge to the same number (although they aren't identical). In the case of the normal distribution, they converge to zero. >>> for n in [2, 3, 4, 5, 6, 7]: ... x = rng.normal(size=10**n) ... m, k = stats.moment(x, 3), stats.kstat(x, 3) ... print("%.3g %.3g %.3g" % (m, k, m-k)) -0.631 -0.651 0.0194 # random 0.0282 0.0283 -8.49e-05 -0.0454 -0.0454 1.36e-05 7.53e-05 7.53e-05 -2.26e-09 0.00166 0.00166 -4.99e-09 -2.88e-06 -2.88e-06 8.63e-13 """ if n > 4 or n < 1: raise ValueError("k-statistics only supported for 1<=n<=4") n = int(n) S = np.zeros(n + 1, np.float64) data = ravel(data) N = data.size # raise ValueError on empty input if N == 0: raise ValueError("Data input must not be empty") # on nan input, return nan without warning if np.isnan(np.sum(data)): return np.nan for k in range(1, n + 1): S[k] = np.sum(data**k, axis=0) if n == 1: return S[1] * 1.0/N elif n == 2: return (N*S[2] - S[1]**2.0) / (N*(N - 1.0)) elif n == 3: return (2*S[1]**3 - 3*N*S[1]*S[2] + N*N*S[3]) / (N*(N - 1.0)*(N - 2.0)) elif n == 4: return ((-6*S[1]**4 + 12*N*S[1]**2 * S[2] - 3*N*(N-1.0)*S[2]**2 - 4*N*(N+1)*S[1]*S[3] + N*N*(N+1)*S[4]) / (N*(N-1.0)*(N-2.0)*(N-3.0))) else: raise ValueError("Should not be here.") @_axis_nan_policy_factory( lambda x: x, result_to_tuple=lambda x: (x,), n_outputs=1, default_axis=None ) def kstatvar(data, n=2): r"""Return an unbiased estimator of the variance of the k-statistic. See `kstat` for more details of the k-statistic. Parameters ---------- data : array_like Input array. Note that n-D input gets flattened. n : int, {1, 2}, optional Default is equal to 2. Returns ------- kstatvar : float The nth k-statistic variance. See Also -------- kstat : Returns the n-th k-statistic. moment : Returns the n-th central moment about the mean for a sample. Notes ----- The variances of the first few k-statistics are given by: .. math:: var(k_{1}) = \frac{\kappa^2}{n} var(k_{2}) = \frac{\kappa^4}{n} + \frac{2\kappa^2_{2}}{n - 1} var(k_{3}) = \frac{\kappa^6}{n} + \frac{9 \kappa_2 \kappa_4}{n - 1} + \frac{9 \kappa^2_{3}}{n - 1} + \frac{6 n \kappa^3_{2}}{(n-1) (n-2)} var(k_{4}) = \frac{\kappa^8}{n} + \frac{16 \kappa_2 \kappa_6}{n - 1} + \frac{48 \kappa_{3} \kappa_5}{n - 1} + \frac{34 \kappa^2_{4}}{n-1} + \frac{72 n \kappa^2_{2} \kappa_4}{(n - 1) (n - 2)} + \frac{144 n \kappa_{2} \kappa^2_{3}}{(n - 1) (n - 2)} + \frac{24 (n + 1) n \kappa^4_{2}}{(n - 1) (n - 2) (n - 3)} """ data = ravel(data) N = len(data) if n == 1: return kstat(data, n=2) * 1.0/N elif n == 2: k2 = kstat(data, n=2) k4 = kstat(data, n=4) return (2*N*k2**2 + (N-1)*k4) / (N*(N+1)) else: raise ValueError("Only n=1 or n=2 supported.") def _calc_uniform_order_statistic_medians(n): """Approximations of uniform order statistic medians. Parameters ---------- n : int Sample size. Returns ------- v : 1d float array Approximations of the order statistic medians. References ---------- .. [1] James J. Filliben, "The Probability Plot Correlation Coefficient Test for Normality", Technometrics, Vol. 17, pp. 111-117, 1975. Examples -------- Order statistics of the uniform distribution on the unit interval are marginally distributed according to beta distributions. The expectations of these order statistic are evenly spaced across the interval, but the distributions are skewed in a way that pushes the medians slightly towards the endpoints of the unit interval: >>> import numpy as np >>> n = 4 >>> k = np.arange(1, n+1) >>> from scipy.stats import beta >>> a = k >>> b = n-k+1 >>> beta.mean(a, b) array([0.2, 0.4, 0.6, 0.8]) >>> beta.median(a, b) array([0.15910358, 0.38572757, 0.61427243, 0.84089642]) The Filliben approximation uses the exact medians of the smallest and greatest order statistics, and the remaining medians are approximated by points spread evenly across a sub-interval of the unit interval: >>> from scipy.stats._morestats import _calc_uniform_order_statistic_medians >>> _calc_uniform_order_statistic_medians(n) array([0.15910358, 0.38545246, 0.61454754, 0.84089642]) This plot shows the skewed distributions of the order statistics of a sample of size four from a uniform distribution on the unit interval: >>> import matplotlib.pyplot as plt >>> x = np.linspace(0.0, 1.0, num=50, endpoint=True) >>> pdfs = [beta.pdf(x, a[i], b[i]) for i in range(n)] >>> plt.figure() >>> plt.plot(x, pdfs[0], x, pdfs[1], x, pdfs[2], x, pdfs[3]) """ v = np.empty(n, dtype=np.float64) v[-1] = 0.5**(1.0 / n) v[0] = 1 - v[-1] i = np.arange(2, n) v[1:-1] = (i - 0.3175) / (n + 0.365) return v def _parse_dist_kw(dist, enforce_subclass=True): """Parse `dist` keyword. Parameters ---------- dist : str or stats.distributions instance. Several functions take `dist` as a keyword, hence this utility function. enforce_subclass : bool, optional If True (default), `dist` needs to be a `_distn_infrastructure.rv_generic` instance. It can sometimes be useful to set this keyword to False, if a function wants to accept objects that just look somewhat like such an instance (for example, they have a ``ppf`` method). """ if isinstance(dist, rv_generic): pass elif isinstance(dist, str): try: dist = getattr(distributions, dist) except AttributeError as e: raise ValueError("%s is not a valid distribution name" % dist) from e elif enforce_subclass: msg = ("`dist` should be a stats.distributions instance or a string " "with the name of such a distribution.") raise ValueError(msg) return dist def _add_axis_labels_title(plot, xlabel, ylabel, title): """Helper function to add axes labels and a title to stats plots.""" try: if hasattr(plot, 'set_title'): # Matplotlib Axes instance or something that looks like it plot.set_title(title) plot.set_xlabel(xlabel) plot.set_ylabel(ylabel) else: # matplotlib.pyplot module plot.title(title) plot.xlabel(xlabel) plot.ylabel(ylabel) except Exception: # Not an MPL object or something that looks (enough) like it. # Don't crash on adding labels or title pass def probplot(x, sparams=(), dist='norm', fit=True, plot=None, rvalue=False): """ Calculate quantiles for a probability plot, and optionally show the plot. Generates a probability plot of sample data against the quantiles of a specified theoretical distribution (the normal distribution by default). `probplot` optionally calculates a best-fit line for the data and plots the results using Matplotlib or a given plot function. Parameters ---------- x : array_like Sample/response data from which `probplot` creates the plot. sparams : tuple, optional Distribution-specific shape parameters (shape parameters plus location and scale). dist : str or stats.distributions instance, optional Distribution or distribution function name. The default is 'norm' for a normal probability plot. Objects that look enough like a stats.distributions instance (i.e. they have a ``ppf`` method) are also accepted. fit : bool, optional Fit a least-squares regression (best-fit) line to the sample data if True (default). plot : object, optional If given, plots the quantiles. If given and `fit` is True, also plots the least squares fit. `plot` is an object that has to have methods "plot" and "text". The `matplotlib.pyplot` module or a Matplotlib Axes object can be used, or a custom object with the same methods. Default is None, which means that no plot is created. rvalue : bool, optional If `plot` is provided and `fit` is True, setting `rvalue` to True includes the coefficient of determination on the plot. Default is False. Returns ------- (osm, osr) : tuple of ndarrays Tuple of theoretical quantiles (osm, or order statistic medians) and ordered responses (osr). `osr` is simply sorted input `x`. For details on how `osm` is calculated see the Notes section. (slope, intercept, r) : tuple of floats, optional Tuple containing the result of the least-squares fit, if that is performed by `probplot`. `r` is the square root of the coefficient of determination. If ``fit=False`` and ``plot=None``, this tuple is not returned. Notes ----- Even if `plot` is given, the figure is not shown or saved by `probplot`; ``plt.show()`` or ``plt.savefig('figname.png')`` should be used after calling `probplot`. `probplot` generates a probability plot, which should not be confused with a Q-Q or a P-P plot. Statsmodels has more extensive functionality of this type, see ``statsmodels.api.ProbPlot``. The formula used for the theoretical quantiles (horizontal axis of the probability plot) is Filliben's estimate:: quantiles = dist.ppf(val), for 0.5**(1/n), for i = n val = (i - 0.3175) / (n + 0.365), for i = 2, ..., n-1 1 - 0.5**(1/n), for i = 1 where ``i`` indicates the i-th ordered value and ``n`` is the total number of values. Examples -------- >>> import numpy as np >>> from scipy import stats >>> import matplotlib.pyplot as plt >>> nsample = 100 >>> rng = np.random.default_rng() A t distribution with small degrees of freedom: >>> ax1 = plt.subplot(221) >>> x = stats.t.rvs(3, size=nsample, random_state=rng) >>> res = stats.probplot(x, plot=plt) A t distribution with larger degrees of freedom: >>> ax2 = plt.subplot(222) >>> x = stats.t.rvs(25, size=nsample, random_state=rng) >>> res = stats.probplot(x, plot=plt) A mixture of two normal distributions with broadcasting: >>> ax3 = plt.subplot(223) >>> x = stats.norm.rvs(loc=[0,5], scale=[1,1.5], ... size=(nsample//2,2), random_state=rng).ravel() >>> res = stats.probplot(x, plot=plt) A standard normal distribution: >>> ax4 = plt.subplot(224) >>> x = stats.norm.rvs(loc=0, scale=1, size=nsample, random_state=rng) >>> res = stats.probplot(x, plot=plt) Produce a new figure with a loggamma distribution, using the ``dist`` and ``sparams`` keywords: >>> fig = plt.figure() >>> ax = fig.add_subplot(111) >>> x = stats.loggamma.rvs(c=2.5, size=500, random_state=rng) >>> res = stats.probplot(x, dist=stats.loggamma, sparams=(2.5,), plot=ax) >>> ax.set_title("Probplot for loggamma dist with shape parameter 2.5") Show the results with Matplotlib: >>> plt.show() """ x = np.asarray(x) if x.size == 0: if fit: return (x, x), (np.nan, np.nan, 0.0) else: return x, x osm_uniform = _calc_uniform_order_statistic_medians(len(x)) dist = _parse_dist_kw(dist, enforce_subclass=False) if sparams is None: sparams = () if isscalar(sparams): sparams = (sparams,) if not isinstance(sparams, tuple): sparams = tuple(sparams) osm = dist.ppf(osm_uniform, *sparams) osr = sort(x) if fit: # perform a linear least squares fit. slope, intercept, r, prob, _ = _stats_py.linregress(osm, osr) if plot is not None: plot.plot(osm, osr, 'bo') if fit: plot.plot(osm, slope*osm + intercept, 'r-') _add_axis_labels_title(plot, xlabel='Theoretical quantiles', ylabel='Ordered Values', title='Probability Plot') # Add R^2 value to the plot as text if fit and rvalue: xmin = amin(osm) xmax = amax(osm) ymin = amin(x) ymax = amax(x) posx = xmin + 0.70 * (xmax - xmin) posy = ymin + 0.01 * (ymax - ymin) plot.text(posx, posy, "$R^2=%1.4f$" % r**2) if fit: return (osm, osr), (slope, intercept, r) else: return osm, osr def ppcc_max(x, brack=(0.0, 1.0), dist='tukeylambda'): """Calculate the shape parameter that maximizes the PPCC. The probability plot correlation coefficient (PPCC) plot can be used to determine the optimal shape parameter for a one-parameter family of distributions. ``ppcc_max`` returns the shape parameter that would maximize the probability plot correlation coefficient for the given data to a one-parameter family of distributions. Parameters ---------- x : array_like Input array. brack : tuple, optional Triple (a,b,c) where (a>> import numpy as np >>> from scipy import stats >>> import matplotlib.pyplot as plt >>> rng = np.random.default_rng() >>> c = 2.5 >>> x = stats.weibull_min.rvs(c, scale=4, size=2000, random_state=rng) Generate the PPCC plot for this data with the Weibull distribution. >>> fig, ax = plt.subplots(figsize=(8, 6)) >>> res = stats.ppcc_plot(x, c/2, 2*c, dist='weibull_min', plot=ax) We calculate the value where the shape should reach its maximum and a red line is drawn there. The line should coincide with the highest point in the PPCC graph. >>> cmax = stats.ppcc_max(x, brack=(c/2, 2*c), dist='weibull_min') >>> ax.axvline(cmax, color='r') >>> plt.show() """ dist = _parse_dist_kw(dist) osm_uniform = _calc_uniform_order_statistic_medians(len(x)) osr = sort(x) # this function computes the x-axis values of the probability plot # and computes a linear regression (including the correlation) # and returns 1-r so that a minimization function maximizes the # correlation def tempfunc(shape, mi, yvals, func): xvals = func(mi, shape) r, prob = _stats_py.pearsonr(xvals, yvals) return 1 - r return optimize.brent(tempfunc, brack=brack, args=(osm_uniform, osr, dist.ppf)) def ppcc_plot(x, a, b, dist='tukeylambda', plot=None, N=80): """Calculate and optionally plot probability plot correlation coefficient. The probability plot correlation coefficient (PPCC) plot can be used to determine the optimal shape parameter for a one-parameter family of distributions. It cannot be used for distributions without shape parameters (like the normal distribution) or with multiple shape parameters. By default a Tukey-Lambda distribution (`stats.tukeylambda`) is used. A Tukey-Lambda PPCC plot interpolates from long-tailed to short-tailed distributions via an approximately normal one, and is therefore particularly useful in practice. Parameters ---------- x : array_like Input array. a, b : scalar Lower and upper bounds of the shape parameter to use. dist : str or stats.distributions instance, optional Distribution or distribution function name. Objects that look enough like a stats.distributions instance (i.e. they have a ``ppf`` method) are also accepted. The default is ``'tukeylambda'``. plot : object, optional If given, plots PPCC against the shape parameter. `plot` is an object that has to have methods "plot" and "text". The `matplotlib.pyplot` module or a Matplotlib Axes object can be used, or a custom object with the same methods. Default is None, which means that no plot is created. N : int, optional Number of points on the horizontal axis (equally distributed from `a` to `b`). Returns ------- svals : ndarray The shape values for which `ppcc` was calculated. ppcc : ndarray The calculated probability plot correlation coefficient values. See Also -------- ppcc_max, probplot, boxcox_normplot, tukeylambda References ---------- J.J. Filliben, "The Probability Plot Correlation Coefficient Test for Normality", Technometrics, Vol. 17, pp. 111-117, 1975. Examples -------- First we generate some random data from a Weibull distribution with shape parameter 2.5, and plot the histogram of the data: >>> import numpy as np >>> from scipy import stats >>> import matplotlib.pyplot as plt >>> rng = np.random.default_rng() >>> c = 2.5 >>> x = stats.weibull_min.rvs(c, scale=4, size=2000, random_state=rng) Take a look at the histogram of the data. >>> fig1, ax = plt.subplots(figsize=(9, 4)) >>> ax.hist(x, bins=50) >>> ax.set_title('Histogram of x') >>> plt.show() Now we explore this data with a PPCC plot as well as the related probability plot and Box-Cox normplot. A red line is drawn where we expect the PPCC value to be maximal (at the shape parameter ``c`` used above): >>> fig2 = plt.figure(figsize=(12, 4)) >>> ax1 = fig2.add_subplot(1, 3, 1) >>> ax2 = fig2.add_subplot(1, 3, 2) >>> ax3 = fig2.add_subplot(1, 3, 3) >>> res = stats.probplot(x, plot=ax1) >>> res = stats.boxcox_normplot(x, -4, 4, plot=ax2) >>> res = stats.ppcc_plot(x, c/2, 2*c, dist='weibull_min', plot=ax3) >>> ax3.axvline(c, color='r') >>> plt.show() """ if b <= a: raise ValueError("`b` has to be larger than `a`.") svals = np.linspace(a, b, num=N) ppcc = np.empty_like(svals) for k, sval in enumerate(svals): _, r2 = probplot(x, sval, dist=dist, fit=True) ppcc[k] = r2[-1] if plot is not None: plot.plot(svals, ppcc, 'x') _add_axis_labels_title(plot, xlabel='Shape Values', ylabel='Prob Plot Corr. Coef.', title='(%s) PPCC Plot' % dist) return svals, ppcc def boxcox_llf(lmb, data): r"""The boxcox log-likelihood function. Parameters ---------- lmb : scalar Parameter for Box-Cox transformation. See `boxcox` for details. data : array_like Data to calculate Box-Cox log-likelihood for. If `data` is multi-dimensional, the log-likelihood is calculated along the first axis. Returns ------- llf : float or ndarray Box-Cox log-likelihood of `data` given `lmb`. A float for 1-D `data`, an array otherwise. See Also -------- boxcox, probplot, boxcox_normplot, boxcox_normmax Notes ----- The Box-Cox log-likelihood function is defined here as .. math:: llf = (\lambda - 1) \sum_i(\log(x_i)) - N/2 \log(\sum_i (y_i - \bar{y})^2 / N), where ``y`` is the Box-Cox transformed input data ``x``. Examples -------- >>> import numpy as np >>> from scipy import stats >>> import matplotlib.pyplot as plt >>> from mpl_toolkits.axes_grid1.inset_locator import inset_axes Generate some random variates and calculate Box-Cox log-likelihood values for them for a range of ``lmbda`` values: >>> rng = np.random.default_rng() >>> x = stats.loggamma.rvs(5, loc=10, size=1000, random_state=rng) >>> lmbdas = np.linspace(-2, 10) >>> llf = np.zeros(lmbdas.shape, dtype=float) >>> for ii, lmbda in enumerate(lmbdas): ... llf[ii] = stats.boxcox_llf(lmbda, x) Also find the optimal lmbda value with `boxcox`: >>> x_most_normal, lmbda_optimal = stats.boxcox(x) Plot the log-likelihood as function of lmbda. Add the optimal lmbda as a horizontal line to check that that's really the optimum: >>> fig = plt.figure() >>> ax = fig.add_subplot(111) >>> ax.plot(lmbdas, llf, 'b.-') >>> ax.axhline(stats.boxcox_llf(lmbda_optimal, x), color='r') >>> ax.set_xlabel('lmbda parameter') >>> ax.set_ylabel('Box-Cox log-likelihood') Now add some probability plots to show that where the log-likelihood is maximized the data transformed with `boxcox` looks closest to normal: >>> locs = [3, 10, 4] # 'lower left', 'center', 'lower right' >>> for lmbda, loc in zip([-1, lmbda_optimal, 9], locs): ... xt = stats.boxcox(x, lmbda=lmbda) ... (osm, osr), (slope, intercept, r_sq) = stats.probplot(xt) ... ax_inset = inset_axes(ax, width="20%", height="20%", loc=loc) ... ax_inset.plot(osm, osr, 'c.', osm, slope*osm + intercept, 'k-') ... ax_inset.set_xticklabels([]) ... ax_inset.set_yticklabels([]) ... ax_inset.set_title(r'$\lambda=%1.2f$' % lmbda) >>> plt.show() """ data = np.asarray(data) N = data.shape[0] if N == 0: return np.nan logdata = np.log(data) # Compute the variance of the transformed data. if lmb == 0: variance = np.var(logdata, axis=0) else: # Transform without the constant offset 1/lmb. The offset does # not effect the variance, and the subtraction of the offset can # lead to loss of precision. variance = np.var(data**lmb / lmb, axis=0) return (lmb - 1) * np.sum(logdata, axis=0) - N/2 * np.log(variance) def _boxcox_conf_interval(x, lmax, alpha): # Need to find the lambda for which # f(x,lmbda) >= f(x,lmax) - 0.5*chi^2_alpha;1 fac = 0.5 * distributions.chi2.ppf(1 - alpha, 1) target = boxcox_llf(lmax, x) - fac def rootfunc(lmbda, data, target): return boxcox_llf(lmbda, data) - target # Find positive endpoint of interval in which answer is to be found newlm = lmax + 0.5 N = 0 while (rootfunc(newlm, x, target) > 0.0) and (N < 500): newlm += 0.1 N += 1 if N == 500: raise RuntimeError("Could not find endpoint.") lmplus = optimize.brentq(rootfunc, lmax, newlm, args=(x, target)) # Now find negative interval in the same way newlm = lmax - 0.5 N = 0 while (rootfunc(newlm, x, target) > 0.0) and (N < 500): newlm -= 0.1 N += 1 if N == 500: raise RuntimeError("Could not find endpoint.") lmminus = optimize.brentq(rootfunc, newlm, lmax, args=(x, target)) return lmminus, lmplus def boxcox(x, lmbda=None, alpha=None, optimizer=None): r"""Return a dataset transformed by a Box-Cox power transformation. Parameters ---------- x : ndarray Input array to be transformed. If `lmbda` is not None, this is an alias of `scipy.special.boxcox`. Returns nan if ``x < 0``; returns -inf if ``x == 0 and lmbda < 0``. If `lmbda` is None, array must be positive, 1-dimensional, and non-constant. lmbda : scalar, optional If `lmbda` is None (default), find the value of `lmbda` that maximizes the log-likelihood function and return it as the second output argument. If `lmbda` is not None, do the transformation for that value. alpha : float, optional If `lmbda` is None and `alpha` is not None (default), return the ``100 * (1-alpha)%`` confidence interval for `lmbda` as the third output argument. Must be between 0.0 and 1.0. If `lmbda` is not None, `alpha` is ignored. optimizer : callable, optional If `lmbda` is None, `optimizer` is the scalar optimizer used to find the value of `lmbda` that minimizes the negative log-likelihood function. `optimizer` is a callable that accepts one argument: fun : callable The objective function, which evaluates the negative log-likelihood function at a provided value of `lmbda` and returns an object, such as an instance of `scipy.optimize.OptimizeResult`, which holds the optimal value of `lmbda` in an attribute `x`. See the example in `boxcox_normmax` or the documentation of `scipy.optimize.minimize_scalar` for more information. If `lmbda` is not None, `optimizer` is ignored. Returns ------- boxcox : ndarray Box-Cox power transformed array. maxlog : float, optional If the `lmbda` parameter is None, the second returned argument is the `lmbda` that maximizes the log-likelihood function. (min_ci, max_ci) : tuple of float, optional If `lmbda` parameter is None and `alpha` is not None, this returned tuple of floats represents the minimum and maximum confidence limits given `alpha`. See Also -------- probplot, boxcox_normplot, boxcox_normmax, boxcox_llf Notes ----- The Box-Cox transform is given by:: y = (x**lmbda - 1) / lmbda, for lmbda != 0 log(x), for lmbda = 0 `boxcox` requires the input data to be positive. Sometimes a Box-Cox transformation provides a shift parameter to achieve this; `boxcox` does not. Such a shift parameter is equivalent to adding a positive constant to `x` before calling `boxcox`. The confidence limits returned when `alpha` is provided give the interval where: .. math:: llf(\hat{\lambda}) - llf(\lambda) < \frac{1}{2}\chi^2(1 - \alpha, 1), with ``llf`` the log-likelihood function and :math:`\chi^2` the chi-squared function. References ---------- G.E.P. Box and D.R. Cox, "An Analysis of Transformations", Journal of the Royal Statistical Society B, 26, 211-252 (1964). Examples -------- >>> from scipy import stats >>> import matplotlib.pyplot as plt We generate some random variates from a non-normal distribution and make a probability plot for it, to show it is non-normal in the tails: >>> fig = plt.figure() >>> ax1 = fig.add_subplot(211) >>> x = stats.loggamma.rvs(5, size=500) + 5 >>> prob = stats.probplot(x, dist=stats.norm, plot=ax1) >>> ax1.set_xlabel('') >>> ax1.set_title('Probplot against normal distribution') We now use `boxcox` to transform the data so it's closest to normal: >>> ax2 = fig.add_subplot(212) >>> xt, _ = stats.boxcox(x) >>> prob = stats.probplot(xt, dist=stats.norm, plot=ax2) >>> ax2.set_title('Probplot after Box-Cox transformation') >>> plt.show() """ x = np.asarray(x) if lmbda is not None: # single transformation return special.boxcox(x, lmbda) if x.ndim != 1: raise ValueError("Data must be 1-dimensional.") if x.size == 0: return x if np.all(x == x[0]): raise ValueError("Data must not be constant.") if np.any(x <= 0): raise ValueError("Data must be positive.") # If lmbda=None, find the lmbda that maximizes the log-likelihood function. lmax = boxcox_normmax(x, method='mle', optimizer=optimizer) y = boxcox(x, lmax) if alpha is None: return y, lmax else: # Find confidence interval interval = _boxcox_conf_interval(x, lmax, alpha) return y, lmax, interval def boxcox_normmax(x, brack=None, method='pearsonr', optimizer=None): """Compute optimal Box-Cox transform parameter for input data. Parameters ---------- x : array_like Input array. brack : 2-tuple, optional, default (-2.0, 2.0) The starting interval for a downhill bracket search for the default `optimize.brent` solver. Note that this is in most cases not critical; the final result is allowed to be outside this bracket. If `optimizer` is passed, `brack` must be None. method : str, optional The method to determine the optimal transform parameter (`boxcox` ``lmbda`` parameter). Options are: 'pearsonr' (default) Maximizes the Pearson correlation coefficient between ``y = boxcox(x)`` and the expected values for ``y`` if `x` would be normally-distributed. 'mle' Minimizes the log-likelihood `boxcox_llf`. This is the method used in `boxcox`. 'all' Use all optimization methods available, and return all results. Useful to compare different methods. optimizer : callable, optional `optimizer` is a callable that accepts one argument: fun : callable The objective function to be optimized. `fun` accepts one argument, the Box-Cox transform parameter `lmbda`, and returns the negative log-likelihood function at the provided value. The job of `optimizer` is to find the value of `lmbda` that minimizes `fun`. and returns an object, such as an instance of `scipy.optimize.OptimizeResult`, which holds the optimal value of `lmbda` in an attribute `x`. See the example below or the documentation of `scipy.optimize.minimize_scalar` for more information. Returns ------- maxlog : float or ndarray The optimal transform parameter found. An array instead of a scalar for ``method='all'``. See Also -------- boxcox, boxcox_llf, boxcox_normplot, scipy.optimize.minimize_scalar Examples -------- >>> import numpy as np >>> from scipy import stats >>> import matplotlib.pyplot as plt We can generate some data and determine the optimal ``lmbda`` in various ways: >>> rng = np.random.default_rng() >>> x = stats.loggamma.rvs(5, size=30, random_state=rng) + 5 >>> y, lmax_mle = stats.boxcox(x) >>> lmax_pearsonr = stats.boxcox_normmax(x) >>> lmax_mle 2.217563431465757 >>> lmax_pearsonr 2.238318660200961 >>> stats.boxcox_normmax(x, method='all') array([2.23831866, 2.21756343]) >>> fig = plt.figure() >>> ax = fig.add_subplot(111) >>> prob = stats.boxcox_normplot(x, -10, 10, plot=ax) >>> ax.axvline(lmax_mle, color='r') >>> ax.axvline(lmax_pearsonr, color='g', ls='--') >>> plt.show() Alternatively, we can define our own `optimizer` function. Suppose we are only interested in values of `lmbda` on the interval [6, 7], we want to use `scipy.optimize.minimize_scalar` with ``method='bounded'``, and we want to use tighter tolerances when optimizing the log-likelihood function. To do this, we define a function that accepts positional argument `fun` and uses `scipy.optimize.minimize_scalar` to minimize `fun` subject to the provided bounds and tolerances: >>> from scipy import optimize >>> options = {'xatol': 1e-12} # absolute tolerance on `x` >>> def optimizer(fun): ... return optimize.minimize_scalar(fun, bounds=(6, 7), ... method="bounded", options=options) >>> stats.boxcox_normmax(x, optimizer=optimizer) 6.000... """ # If optimizer is not given, define default 'brent' optimizer. if optimizer is None: # Set default value for `brack`. if brack is None: brack = (-2.0, 2.0) def _optimizer(func, args): return optimize.brent(func, args=args, brack=brack) # Otherwise check optimizer. else: if not callable(optimizer): raise ValueError("`optimizer` must be a callable") if brack is not None: raise ValueError("`brack` must be None if `optimizer` is given") # `optimizer` is expected to return a `OptimizeResult` object, we here # get the solution to the optimization problem. def _optimizer(func, args): def func_wrapped(x): return func(x, *args) return getattr(optimizer(func_wrapped), 'x', None) def _pearsonr(x): osm_uniform = _calc_uniform_order_statistic_medians(len(x)) xvals = distributions.norm.ppf(osm_uniform) def _eval_pearsonr(lmbda, xvals, samps): # This function computes the x-axis values of the probability plot # and computes a linear regression (including the correlation) and # returns ``1 - r`` so that a minimization function maximizes the # correlation. y = boxcox(samps, lmbda) yvals = np.sort(y) r, prob = _stats_py.pearsonr(xvals, yvals) return 1 - r return _optimizer(_eval_pearsonr, args=(xvals, x)) def _mle(x): def _eval_mle(lmb, data): # function to minimize return -boxcox_llf(lmb, data) return _optimizer(_eval_mle, args=(x,)) def _all(x): maxlog = np.empty(2, dtype=float) maxlog[0] = _pearsonr(x) maxlog[1] = _mle(x) return maxlog methods = {'pearsonr': _pearsonr, 'mle': _mle, 'all': _all} if method not in methods.keys(): raise ValueError("Method %s not recognized." % method) optimfunc = methods[method] res = optimfunc(x) if res is None: message = ("`optimizer` must return an object containing the optimal " "`lmbda` in attribute `x`") raise ValueError(message) return res def _normplot(method, x, la, lb, plot=None, N=80): """Compute parameters for a Box-Cox or Yeo-Johnson normality plot, optionally show it. See `boxcox_normplot` or `yeojohnson_normplot` for details. """ if method == 'boxcox': title = 'Box-Cox Normality Plot' transform_func = boxcox else: title = 'Yeo-Johnson Normality Plot' transform_func = yeojohnson x = np.asarray(x) if x.size == 0: return x if lb <= la: raise ValueError("`lb` has to be larger than `la`.") if method == 'boxcox' and np.any(x <= 0): raise ValueError("Data must be positive.") lmbdas = np.linspace(la, lb, num=N) ppcc = lmbdas * 0.0 for i, val in enumerate(lmbdas): # Determine for each lmbda the square root of correlation coefficient # of transformed x z = transform_func(x, lmbda=val) _, (_, _, r) = probplot(z, dist='norm', fit=True) ppcc[i] = r if plot is not None: plot.plot(lmbdas, ppcc, 'x') _add_axis_labels_title(plot, xlabel='$\\lambda$', ylabel='Prob Plot Corr. Coef.', title=title) return lmbdas, ppcc def boxcox_normplot(x, la, lb, plot=None, N=80): """Compute parameters for a Box-Cox normality plot, optionally show it. A Box-Cox normality plot shows graphically what the best transformation parameter is to use in `boxcox` to obtain a distribution that is close to normal. Parameters ---------- x : array_like Input array. la, lb : scalar The lower and upper bounds for the ``lmbda`` values to pass to `boxcox` for Box-Cox transformations. These are also the limits of the horizontal axis of the plot if that is generated. plot : object, optional If given, plots the quantiles and least squares fit. `plot` is an object that has to have methods "plot" and "text". The `matplotlib.pyplot` module or a Matplotlib Axes object can be used, or a custom object with the same methods. Default is None, which means that no plot is created. N : int, optional Number of points on the horizontal axis (equally distributed from `la` to `lb`). Returns ------- lmbdas : ndarray The ``lmbda`` values for which a Box-Cox transform was done. ppcc : ndarray Probability Plot Correlelation Coefficient, as obtained from `probplot` when fitting the Box-Cox transformed input `x` against a normal distribution. See Also -------- probplot, boxcox, boxcox_normmax, boxcox_llf, ppcc_max Notes ----- Even if `plot` is given, the figure is not shown or saved by `boxcox_normplot`; ``plt.show()`` or ``plt.savefig('figname.png')`` should be used after calling `probplot`. Examples -------- >>> from scipy import stats >>> import matplotlib.pyplot as plt Generate some non-normally distributed data, and create a Box-Cox plot: >>> x = stats.loggamma.rvs(5, size=500) + 5 >>> fig = plt.figure() >>> ax = fig.add_subplot(111) >>> prob = stats.boxcox_normplot(x, -20, 20, plot=ax) Determine and plot the optimal ``lmbda`` to transform ``x`` and plot it in the same plot: >>> _, maxlog = stats.boxcox(x) >>> ax.axvline(maxlog, color='r') >>> plt.show() """ return _normplot('boxcox', x, la, lb, plot, N) def yeojohnson(x, lmbda=None): r"""Return a dataset transformed by a Yeo-Johnson power transformation. Parameters ---------- x : ndarray Input array. Should be 1-dimensional. lmbda : float, optional If ``lmbda`` is ``None``, find the lambda that maximizes the log-likelihood function and return it as the second output argument. Otherwise the transformation is done for the given value. Returns ------- yeojohnson: ndarray Yeo-Johnson power transformed array. maxlog : float, optional If the `lmbda` parameter is None, the second returned argument is the lambda that maximizes the log-likelihood function. See Also -------- probplot, yeojohnson_normplot, yeojohnson_normmax, yeojohnson_llf, boxcox Notes ----- The Yeo-Johnson transform is given by:: y = ((x + 1)**lmbda - 1) / lmbda, for x >= 0, lmbda != 0 log(x + 1), for x >= 0, lmbda = 0 -((-x + 1)**(2 - lmbda) - 1) / (2 - lmbda), for x < 0, lmbda != 2 -log(-x + 1), for x < 0, lmbda = 2 Unlike `boxcox`, `yeojohnson` does not require the input data to be positive. .. versionadded:: 1.2.0 References ---------- I. Yeo and R.A. Johnson, "A New Family of Power Transformations to Improve Normality or Symmetry", Biometrika 87.4 (2000): Examples -------- >>> from scipy import stats >>> import matplotlib.pyplot as plt We generate some random variates from a non-normal distribution and make a probability plot for it, to show it is non-normal in the tails: >>> fig = plt.figure() >>> ax1 = fig.add_subplot(211) >>> x = stats.loggamma.rvs(5, size=500) + 5 >>> prob = stats.probplot(x, dist=stats.norm, plot=ax1) >>> ax1.set_xlabel('') >>> ax1.set_title('Probplot against normal distribution') We now use `yeojohnson` to transform the data so it's closest to normal: >>> ax2 = fig.add_subplot(212) >>> xt, lmbda = stats.yeojohnson(x) >>> prob = stats.probplot(xt, dist=stats.norm, plot=ax2) >>> ax2.set_title('Probplot after Yeo-Johnson transformation') >>> plt.show() """ x = np.asarray(x) if x.size == 0: return x if np.issubdtype(x.dtype, np.complexfloating): raise ValueError('Yeo-Johnson transformation is not defined for ' 'complex numbers.') if np.issubdtype(x.dtype, np.integer): x = x.astype(np.float64, copy=False) if lmbda is not None: return _yeojohnson_transform(x, lmbda) # if lmbda=None, find the lmbda that maximizes the log-likelihood function. lmax = yeojohnson_normmax(x) y = _yeojohnson_transform(x, lmax) return y, lmax def _yeojohnson_transform(x, lmbda): """Returns `x` transformed by the Yeo-Johnson power transform with given parameter `lmbda`. """ out = np.zeros_like(x) pos = x >= 0 # binary mask # when x >= 0 if abs(lmbda) < np.spacing(1.): out[pos] = np.log1p(x[pos]) else: # lmbda != 0 out[pos] = (np.power(x[pos] + 1, lmbda) - 1) / lmbda # when x < 0 if abs(lmbda - 2) > np.spacing(1.): out[~pos] = -(np.power(-x[~pos] + 1, 2 - lmbda) - 1) / (2 - lmbda) else: # lmbda == 2 out[~pos] = -np.log1p(-x[~pos]) return out def yeojohnson_llf(lmb, data): r"""The yeojohnson log-likelihood function. Parameters ---------- lmb : scalar Parameter for Yeo-Johnson transformation. See `yeojohnson` for details. data : array_like Data to calculate Yeo-Johnson log-likelihood for. If `data` is multi-dimensional, the log-likelihood is calculated along the first axis. Returns ------- llf : float Yeo-Johnson log-likelihood of `data` given `lmb`. See Also -------- yeojohnson, probplot, yeojohnson_normplot, yeojohnson_normmax Notes ----- The Yeo-Johnson log-likelihood function is defined here as .. math:: llf = -N/2 \log(\hat{\sigma}^2) + (\lambda - 1) \sum_i \text{ sign }(x_i)\log(|x_i| + 1) where :math:`\hat{\sigma}^2` is estimated variance of the Yeo-Johnson transformed input data ``x``. .. versionadded:: 1.2.0 Examples -------- >>> import numpy as np >>> from scipy import stats >>> import matplotlib.pyplot as plt >>> from mpl_toolkits.axes_grid1.inset_locator import inset_axes Generate some random variates and calculate Yeo-Johnson log-likelihood values for them for a range of ``lmbda`` values: >>> x = stats.loggamma.rvs(5, loc=10, size=1000) >>> lmbdas = np.linspace(-2, 10) >>> llf = np.zeros(lmbdas.shape, dtype=float) >>> for ii, lmbda in enumerate(lmbdas): ... llf[ii] = stats.yeojohnson_llf(lmbda, x) Also find the optimal lmbda value with `yeojohnson`: >>> x_most_normal, lmbda_optimal = stats.yeojohnson(x) Plot the log-likelihood as function of lmbda. Add the optimal lmbda as a horizontal line to check that that's really the optimum: >>> fig = plt.figure() >>> ax = fig.add_subplot(111) >>> ax.plot(lmbdas, llf, 'b.-') >>> ax.axhline(stats.yeojohnson_llf(lmbda_optimal, x), color='r') >>> ax.set_xlabel('lmbda parameter') >>> ax.set_ylabel('Yeo-Johnson log-likelihood') Now add some probability plots to show that where the log-likelihood is maximized the data transformed with `yeojohnson` looks closest to normal: >>> locs = [3, 10, 4] # 'lower left', 'center', 'lower right' >>> for lmbda, loc in zip([-1, lmbda_optimal, 9], locs): ... xt = stats.yeojohnson(x, lmbda=lmbda) ... (osm, osr), (slope, intercept, r_sq) = stats.probplot(xt) ... ax_inset = inset_axes(ax, width="20%", height="20%", loc=loc) ... ax_inset.plot(osm, osr, 'c.', osm, slope*osm + intercept, 'k-') ... ax_inset.set_xticklabels([]) ... ax_inset.set_yticklabels([]) ... ax_inset.set_title(r'$\lambda=%1.2f$' % lmbda) >>> plt.show() """ data = np.asarray(data) n_samples = data.shape[0] if n_samples == 0: return np.nan trans = _yeojohnson_transform(data, lmb) trans_var = trans.var(axis=0) loglike = np.empty_like(trans_var) # Avoid RuntimeWarning raised by np.log when the variance is too low tiny_variance = trans_var < np.finfo(trans_var.dtype).tiny loglike[tiny_variance] = np.inf loglike[~tiny_variance] = ( -n_samples / 2 * np.log(trans_var[~tiny_variance])) loglike[~tiny_variance] += ( (lmb - 1) * (np.sign(data) * np.log(np.abs(data) + 1)).sum(axis=0)) return loglike def yeojohnson_normmax(x, brack=(-2, 2)): """Compute optimal Yeo-Johnson transform parameter. Compute optimal Yeo-Johnson transform parameter for input data, using maximum likelihood estimation. Parameters ---------- x : array_like Input array. brack : 2-tuple, optional The starting interval for a downhill bracket search with `optimize.brent`. Note that this is in most cases not critical; the final result is allowed to be outside this bracket. Returns ------- maxlog : float The optimal transform parameter found. See Also -------- yeojohnson, yeojohnson_llf, yeojohnson_normplot Notes ----- .. versionadded:: 1.2.0 Examples -------- >>> import numpy as np >>> from scipy import stats >>> import matplotlib.pyplot as plt Generate some data and determine optimal ``lmbda`` >>> rng = np.random.default_rng() >>> x = stats.loggamma.rvs(5, size=30, random_state=rng) + 5 >>> lmax = stats.yeojohnson_normmax(x) >>> fig = plt.figure() >>> ax = fig.add_subplot(111) >>> prob = stats.yeojohnson_normplot(x, -10, 10, plot=ax) >>> ax.axvline(lmax, color='r') >>> plt.show() """ def _neg_llf(lmbda, data): llf = yeojohnson_llf(lmbda, data) # reject likelihoods that are inf which are likely due to small # variance in the transformed space llf[np.isinf(llf)] = -np.inf return -llf with np.errstate(invalid='ignore'): return optimize.brent(_neg_llf, brack=brack, args=(x,)) def yeojohnson_normplot(x, la, lb, plot=None, N=80): """Compute parameters for a Yeo-Johnson normality plot, optionally show it. A Yeo-Johnson normality plot shows graphically what the best transformation parameter is to use in `yeojohnson` to obtain a distribution that is close to normal. Parameters ---------- x : array_like Input array. la, lb : scalar The lower and upper bounds for the ``lmbda`` values to pass to `yeojohnson` for Yeo-Johnson transformations. These are also the limits of the horizontal axis of the plot if that is generated. plot : object, optional If given, plots the quantiles and least squares fit. `plot` is an object that has to have methods "plot" and "text". The `matplotlib.pyplot` module or a Matplotlib Axes object can be used, or a custom object with the same methods. Default is None, which means that no plot is created. N : int, optional Number of points on the horizontal axis (equally distributed from `la` to `lb`). Returns ------- lmbdas : ndarray The ``lmbda`` values for which a Yeo-Johnson transform was done. ppcc : ndarray Probability Plot Correlelation Coefficient, as obtained from `probplot` when fitting the Box-Cox transformed input `x` against a normal distribution. See Also -------- probplot, yeojohnson, yeojohnson_normmax, yeojohnson_llf, ppcc_max Notes ----- Even if `plot` is given, the figure is not shown or saved by `boxcox_normplot`; ``plt.show()`` or ``plt.savefig('figname.png')`` should be used after calling `probplot`. .. versionadded:: 1.2.0 Examples -------- >>> from scipy import stats >>> import matplotlib.pyplot as plt Generate some non-normally distributed data, and create a Yeo-Johnson plot: >>> x = stats.loggamma.rvs(5, size=500) + 5 >>> fig = plt.figure() >>> ax = fig.add_subplot(111) >>> prob = stats.yeojohnson_normplot(x, -20, 20, plot=ax) Determine and plot the optimal ``lmbda`` to transform ``x`` and plot it in the same plot: >>> _, maxlog = stats.yeojohnson(x) >>> ax.axvline(maxlog, color='r') >>> plt.show() """ return _normplot('yeojohnson', x, la, lb, plot, N) ShapiroResult = namedtuple('ShapiroResult', ('statistic', 'pvalue')) def shapiro(x): r"""Perform the Shapiro-Wilk test for normality. The Shapiro-Wilk test tests the null hypothesis that the data was drawn from a normal distribution. Parameters ---------- x : array_like Array of sample data. Returns ------- statistic : float The test statistic. p-value : float The p-value for the hypothesis test. See Also -------- anderson : The Anderson-Darling test for normality kstest : The Kolmogorov-Smirnov test for goodness of fit. Notes ----- The algorithm used is described in [4]_ but censoring parameters as described are not implemented. For N > 5000 the W test statistic is accurate, but the p-value may not be. References ---------- .. [1] https://www.itl.nist.gov/div898/handbook/prc/section2/prc213.htm .. [2] Shapiro, S. S. & Wilk, M.B (1965). An analysis of variance test for normality (complete samples), Biometrika, Vol. 52, pp. 591-611. .. [3] Razali, N. M. & Wah, Y. B. (2011) Power comparisons of Shapiro-Wilk, Kolmogorov-Smirnov, Lilliefors and Anderson-Darling tests, Journal of Statistical Modeling and Analytics, Vol. 2, pp. 21-33. .. [4] ALGORITHM AS R94 APPL. STATIST. (1995) VOL. 44, NO. 4. .. [5] B. Phipson and G. K. Smyth. "Permutation P-values Should Never Be Zero: Calculating Exact P-values When Permutations Are Randomly Drawn." Statistical Applications in Genetics and Molecular Biology 9.1 (2010). .. [6] Panagiotakos, D. B. (2008). The value of p-value in biomedical research. The open cardiovascular medicine journal, 2, 97. Examples -------- Suppose we wish to infer from measurements whether the weights of adult human males in a medical study are not normally distributed [2]_. The weights (lbs) are recorded in the array ``x`` below. >>> import numpy as np >>> x = np.array([148, 154, 158, 160, 161, 162, 166, 170, 182, 195, 236]) The normality test of [1]_ and [2]_ begins by computing a statistic based on the relationship between the observations and the expected order statistics of a normal distribution. >>> from scipy import stats >>> res = stats.shapiro(x) >>> res.statistic 0.7888147830963135 The value of this statistic tends to be high (close to 1) for samples drawn from a normal distribution. The test is performed by comparing the observed value of the statistic against the null distribution: the distribution of statistic values formed under the null hypothesis that the weights were drawn from a normal distribution. For this normality test, the null distribution is not easy to calculate exactly, so it is usually approximated by Monte Carlo methods, that is, drawing many samples of the same size as ``x`` from a normal distribution and computing the values of the statistic for each. >>> def statistic(x): ... # Get only the `shapiro` statistic; ignore its p-value ... return stats.shapiro(x).statistic >>> ref = stats.monte_carlo_test(x, stats.norm.rvs, statistic, ... alternative='less') >>> import matplotlib.pyplot as plt >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> bins = np.linspace(0.65, 1, 50) >>> def plot(ax): # we'll re-use this ... ax.hist(ref.null_distribution, density=True, bins=bins) ... ax.set_title("Shapiro-Wilk Test Null Distribution \n" ... "(Monte Carlo Approximation, 11 Observations)") ... ax.set_xlabel("statistic") ... ax.set_ylabel("probability density") >>> plot(ax) >>> plt.show() The comparison is quantified by the p-value: the proportion of values in the null distribution less than or equal to the observed value of the statistic. >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> plot(ax) >>> annotation = (f'p-value={res.pvalue:.6f}\n(highlighted area)') >>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8) >>> _ = ax.annotate(annotation, (0.75, 0.1), (0.68, 0.7), arrowprops=props) >>> i_extreme = np.where(bins <= res.statistic)[0] >>> for i in i_extreme: ... ax.patches[i].set_color('C1') >>> plt.xlim(0.65, 0.9) >>> plt.ylim(0, 4) >>> plt.show >>> res.pvalue 0.006703833118081093 If the p-value is "small" - that is, if there is a low probability of sampling data from a normally distributed population that produces such an extreme value of the statistic - this may be taken as evidence against the null hypothesis in favor of the alternative: the weights were not drawn from a normal distribution. Note that: - The inverse is not true; that is, the test is not used to provide evidence *for* the null hypothesis. - The threshold for values that will be considered "small" is a choice that should be made before the data is analyzed [5]_ with consideration of the risks of both false positives (incorrectly rejecting the null hypothesis) and false negatives (failure to reject a false null hypothesis). """ x = np.ravel(x) N = len(x) if N < 3: raise ValueError("Data must be at least length 3.") x = x - np.median(x) a = zeros(N, 'f') init = 0 y = sort(x) a, w, pw, ifault = _statlib.swilk(y, a[:N//2], init) if ifault not in [0, 2]: warnings.warn("Input data for shapiro has range zero. The results " "may not be accurate.") if N > 5000: warnings.warn("p-value may not be accurate for N > 5000.") # `swilk` can return negative p-values for N==3; see gh-18322. if N == 3: # Potential improvement: precision for small p-values pw = 1 - 6/np.pi*np.arccos(np.sqrt(w)) return ShapiroResult(w, pw) # Values from Stephens, M A, "EDF Statistics for Goodness of Fit and # Some Comparisons", Journal of the American Statistical # Association, Vol. 69, Issue 347, Sept. 1974, pp 730-737 _Avals_norm = array([0.576, 0.656, 0.787, 0.918, 1.092]) _Avals_expon = array([0.922, 1.078, 1.341, 1.606, 1.957]) # From Stephens, M A, "Goodness of Fit for the Extreme Value Distribution", # Biometrika, Vol. 64, Issue 3, Dec. 1977, pp 583-588. _Avals_gumbel = array([0.474, 0.637, 0.757, 0.877, 1.038]) # From Stephens, M A, "Tests of Fit for the Logistic Distribution Based # on the Empirical Distribution Function.", Biometrika, # Vol. 66, Issue 3, Dec. 1979, pp 591-595. _Avals_logistic = array([0.426, 0.563, 0.660, 0.769, 0.906, 1.010]) # From Richard A. Lockhart and Michael A. Stephens "Estimation and Tests of # Fit for the Three-Parameter Weibull Distribution" # Journal of the Royal Statistical Society.Series B(Methodological) # Vol. 56, No. 3 (1994), pp. 491-500, table 1. Keys are c*100 _Avals_weibull = [[0.292, 0.395, 0.467, 0.522, 0.617, 0.711, 0.836, 0.931], [0.295, 0.399, 0.471, 0.527, 0.623, 0.719, 0.845, 0.941], [0.298, 0.403, 0.476, 0.534, 0.631, 0.728, 0.856, 0.954], [0.301, 0.408, 0.483, 0.541, 0.640, 0.738, 0.869, 0.969], [0.305, 0.414, 0.490, 0.549, 0.650, 0.751, 0.885, 0.986], [0.309, 0.421, 0.498, 0.559, 0.662, 0.765, 0.902, 1.007], [0.314, 0.429, 0.508, 0.570, 0.676, 0.782, 0.923, 1.030], [0.320, 0.438, 0.519, 0.583, 0.692, 0.802, 0.947, 1.057], [0.327, 0.448, 0.532, 0.598, 0.711, 0.824, 0.974, 1.089], [0.334, 0.469, 0.547, 0.615, 0.732, 0.850, 1.006, 1.125], [0.342, 0.472, 0.563, 0.636, 0.757, 0.879, 1.043, 1.167]] _Avals_weibull = np.array(_Avals_weibull) _cvals_weibull = np.linspace(0, 0.5, 11) _get_As_weibull = interpolate.interp1d(_cvals_weibull, _Avals_weibull.T, kind='linear', bounds_error=False, fill_value=_Avals_weibull[-1]) def _weibull_fit_check(params, x): # Refine the fit returned by `weibull_min.fit` to ensure that the first # order necessary conditions are satisfied. If not, raise an error. # Here, use `m` for the shape parameter to be consistent with [7] # and avoid confusion with `c` as defined in [7]. n = len(x) m, u, s = params def dnllf_dm(m, u): # Partial w.r.t. shape w/ optimal scale. See [7] Equation 5. xu = x-u return (1/m - (xu**m*np.log(xu)).sum()/(xu**m).sum() + np.log(xu).sum()/n) def dnllf_du(m, u): # Partial w.r.t. loc w/ optimal scale. See [7] Equation 6. xu = x-u return (m-1)/m*(xu**-1).sum() - n*(xu**(m-1)).sum()/(xu**m).sum() def get_scale(m, u): # Partial w.r.t. scale solved in terms of shape and location. # See [7] Equation 7. return ((x-u)**m/n).sum()**(1/m) def dnllf(params): # Partial derivatives of the NLLF w.r.t. parameters, i.e. # first order necessary conditions for MLE fit. return [dnllf_dm(*params), dnllf_du(*params)] suggestion = ("Maximum likelihood estimation is known to be challenging " "for the three-parameter Weibull distribution. Consider " "performing a custom goodness-of-fit test using " "`scipy.stats.monte_carlo_test`.") if np.allclose(u, np.min(x)) or m < 1: # The critical values provided by [7] don't seem to control the # Type I error rate in this case. Error out. message = ("Maximum likelihood estimation has converged to " "a solution in which the location is equal to the minimum " "of the data, the shape parameter is less than 2, or both. " "The table of critical values in [7] does not " "include this case. " + suggestion) raise ValueError(message) try: # Refine the MLE / verify that first-order necessary conditions are # satisfied. If so, the critical values provided in [7] seem reliable. with np.errstate(over='raise', invalid='raise'): res = optimize.root(dnllf, params[:-1]) message = ("Solution of MLE first-order conditions failed: " f"{res.message}. `anderson` cannot continue. " + suggestion) if not res.success: raise ValueError(message) except (FloatingPointError, ValueError) as e: message = ("An error occurred while fitting the Weibull distribution " "to the data, so `anderson` cannot continue. " + suggestion) raise ValueError(message) from e m, u = res.x s = get_scale(m, u) return m, u, s AndersonResult = _make_tuple_bunch('AndersonResult', ['statistic', 'critical_values', 'significance_level'], ['fit_result']) def anderson(x, dist='norm'): """Anderson-Darling test for data coming from a particular distribution. The Anderson-Darling test tests the null hypothesis that a sample is drawn from a population that follows a particular distribution. For the Anderson-Darling test, the critical values depend on which distribution is being tested against. This function works for normal, exponential, logistic, weibull_min, or Gumbel (Extreme Value Type I) distributions. Parameters ---------- x : array_like Array of sample data. dist : {'norm', 'expon', 'logistic', 'gumbel', 'gumbel_l', 'gumbel_r', 'extreme1', 'weibull_min'}, optional The type of distribution to test against. The default is 'norm'. The names 'extreme1', 'gumbel_l' and 'gumbel' are synonyms for the same distribution. Returns ------- result : AndersonResult An object with the following attributes: statistic : float The Anderson-Darling test statistic. critical_values : list The critical values for this distribution. significance_level : list The significance levels for the corresponding critical values in percents. The function returns critical values for a differing set of significance levels depending on the distribution that is being tested against. fit_result : `~scipy.stats._result_classes.FitResult` An object containing the results of fitting the distribution to the data. See Also -------- kstest : The Kolmogorov-Smirnov test for goodness-of-fit. Notes ----- Critical values provided are for the following significance levels: normal/exponential 15%, 10%, 5%, 2.5%, 1% logistic 25%, 10%, 5%, 2.5%, 1%, 0.5% gumbel_l / gumbel_r 25%, 10%, 5%, 2.5%, 1% weibull_min 50%, 25%, 15%, 10%, 5%, 2.5%, 1%, 0.5% If the returned statistic is larger than these critical values then for the corresponding significance level, the null hypothesis that the data come from the chosen distribution can be rejected. The returned statistic is referred to as 'A2' in the references. For `weibull_min`, maximum likelihood estimation is known to be challenging. If the test returns successfully, then the first order conditions for a maximum likehood estimate have been verified and the critical values correspond relatively well to the significance levels, provided that the sample is sufficiently large (>10 observations [7]). However, for some data - especially data with no left tail - `anderson` is likely to result in an error message. In this case, consider performing a custom goodness of fit test using `scipy.stats.monte_carlo_test`. References ---------- .. [1] https://www.itl.nist.gov/div898/handbook/prc/section2/prc213.htm .. [2] Stephens, M. A. (1974). EDF Statistics for Goodness of Fit and Some Comparisons, Journal of the American Statistical Association, Vol. 69, pp. 730-737. .. [3] Stephens, M. A. (1976). Asymptotic Results for Goodness-of-Fit Statistics with Unknown Parameters, Annals of Statistics, Vol. 4, pp. 357-369. .. [4] Stephens, M. A. (1977). Goodness of Fit for the Extreme Value Distribution, Biometrika, Vol. 64, pp. 583-588. .. [5] Stephens, M. A. (1977). Goodness of Fit with Special Reference to Tests for Exponentiality , Technical Report No. 262, Department of Statistics, Stanford University, Stanford, CA. .. [6] Stephens, M. A. (1979). Tests of Fit for the Logistic Distribution Based on the Empirical Distribution Function, Biometrika, Vol. 66, pp. 591-595. .. [7] Richard A. Lockhart and Michael A. Stephens "Estimation and Tests of Fit for the Three-Parameter Weibull Distribution" Journal of the Royal Statistical Society.Series B(Methodological) Vol. 56, No. 3 (1994), pp. 491-500, Table 0. Examples -------- Test the null hypothesis that a random sample was drawn from a normal distribution (with unspecified mean and standard deviation). >>> import numpy as np >>> from scipy.stats import anderson >>> rng = np.random.default_rng() >>> data = rng.random(size=35) >>> res = anderson(data) >>> res.statistic 0.8398018749744764 >>> res.critical_values array([0.527, 0.6 , 0.719, 0.839, 0.998]) >>> res.significance_level array([15. , 10. , 5. , 2.5, 1. ]) The value of the statistic (barely) exceeds the critical value associated with a significance level of 2.5%, so the null hypothesis may be rejected at a significance level of 2.5%, but not at a significance level of 1%. """ # noqa dist = dist.lower() if dist in {'extreme1', 'gumbel'}: dist = 'gumbel_l' dists = {'norm', 'expon', 'gumbel_l', 'gumbel_r', 'logistic', 'weibull_min'} if dist not in dists: raise ValueError(f"Invalid distribution; dist must be in {dists}.") y = sort(x) xbar = np.mean(x, axis=0) N = len(y) if dist == 'norm': s = np.std(x, ddof=1, axis=0) w = (y - xbar) / s fit_params = xbar, s logcdf = distributions.norm.logcdf(w) logsf = distributions.norm.logsf(w) sig = array([15, 10, 5, 2.5, 1]) critical = around(_Avals_norm / (1.0 + 4.0/N - 25.0/N/N), 3) elif dist == 'expon': w = y / xbar fit_params = 0, xbar logcdf = distributions.expon.logcdf(w) logsf = distributions.expon.logsf(w) sig = array([15, 10, 5, 2.5, 1]) critical = around(_Avals_expon / (1.0 + 0.6/N), 3) elif dist == 'logistic': def rootfunc(ab, xj, N): a, b = ab tmp = (xj - a) / b tmp2 = exp(tmp) val = [np.sum(1.0/(1+tmp2), axis=0) - 0.5*N, np.sum(tmp*(1.0-tmp2)/(1+tmp2), axis=0) + N] return array(val) sol0 = array([xbar, np.std(x, ddof=1, axis=0)]) sol = optimize.fsolve(rootfunc, sol0, args=(x, N), xtol=1e-5) w = (y - sol[0]) / sol[1] fit_params = sol logcdf = distributions.logistic.logcdf(w) logsf = distributions.logistic.logsf(w) sig = array([25, 10, 5, 2.5, 1, 0.5]) critical = around(_Avals_logistic / (1.0 + 0.25/N), 3) elif dist == 'gumbel_r': xbar, s = distributions.gumbel_r.fit(x) w = (y - xbar) / s fit_params = xbar, s logcdf = distributions.gumbel_r.logcdf(w) logsf = distributions.gumbel_r.logsf(w) sig = array([25, 10, 5, 2.5, 1]) critical = around(_Avals_gumbel / (1.0 + 0.2/sqrt(N)), 3) elif dist == 'gumbel_l': xbar, s = distributions.gumbel_l.fit(x) w = (y - xbar) / s fit_params = xbar, s logcdf = distributions.gumbel_l.logcdf(w) logsf = distributions.gumbel_l.logsf(w) sig = array([25, 10, 5, 2.5, 1]) critical = around(_Avals_gumbel / (1.0 + 0.2/sqrt(N)), 3) elif dist == 'weibull_min': message = ("Critical values of the test statistic are given for the " "asymptotic distribution. These may not be accurate for " "samples with fewer than 10 observations. Consider using " "`scipy.stats.monte_carlo_test`.") if N < 10: warnings.warn(message, stacklevel=2) # [7] writes our 'c' as 'm', and they write `c = 1/m`. Use their names. m, loc, scale = distributions.weibull_min.fit(y) m, loc, scale = _weibull_fit_check((m, loc, scale), y) fit_params = m, loc, scale logcdf = stats.weibull_min(*fit_params).logcdf(y) logsf = stats.weibull_min(*fit_params).logsf(y) c = 1 / m # m and c are as used in [7] sig = array([0.5, 0.75, 0.85, 0.9, 0.95, 0.975, 0.99, 0.995]) critical = _get_As_weibull(c) # Goodness-of-fit tests should only be used to provide evidence # _against_ the null hypothesis. Be conservative and round up. critical = np.round(critical + 0.0005, decimals=3) i = arange(1, N + 1) A2 = -N - np.sum((2*i - 1.0) / N * (logcdf + logsf[::-1]), axis=0) # FitResult initializer expects an optimize result, so let's work with it message = '`anderson` successfully fit the distribution to the data.' res = optimize.OptimizeResult(success=True, message=message) res.x = np.array(fit_params) fit_result = FitResult(getattr(distributions, dist), y, discrete=False, res=res) return AndersonResult(A2, critical, sig, fit_result=fit_result) def _anderson_ksamp_midrank(samples, Z, Zstar, k, n, N): """Compute A2akN equation 7 of Scholz and Stephens. Parameters ---------- samples : sequence of 1-D array_like Array of sample arrays. Z : array_like Sorted array of all observations. Zstar : array_like Sorted array of unique observations. k : int Number of samples. n : array_like Number of observations in each sample. N : int Total number of observations. Returns ------- A2aKN : float The A2aKN statistics of Scholz and Stephens 1987. """ A2akN = 0. Z_ssorted_left = Z.searchsorted(Zstar, 'left') if N == Zstar.size: lj = 1. else: lj = Z.searchsorted(Zstar, 'right') - Z_ssorted_left Bj = Z_ssorted_left + lj / 2. for i in arange(0, k): s = np.sort(samples[i]) s_ssorted_right = s.searchsorted(Zstar, side='right') Mij = s_ssorted_right.astype(float) fij = s_ssorted_right - s.searchsorted(Zstar, 'left') Mij -= fij / 2. inner = lj / float(N) * (N*Mij - Bj*n[i])**2 / (Bj*(N - Bj) - N*lj/4.) A2akN += inner.sum() / n[i] A2akN *= (N - 1.) / N return A2akN def _anderson_ksamp_right(samples, Z, Zstar, k, n, N): """Compute A2akN equation 6 of Scholz & Stephens. Parameters ---------- samples : sequence of 1-D array_like Array of sample arrays. Z : array_like Sorted array of all observations. Zstar : array_like Sorted array of unique observations. k : int Number of samples. n : array_like Number of observations in each sample. N : int Total number of observations. Returns ------- A2KN : float The A2KN statistics of Scholz and Stephens 1987. """ A2kN = 0. lj = Z.searchsorted(Zstar[:-1], 'right') - Z.searchsorted(Zstar[:-1], 'left') Bj = lj.cumsum() for i in arange(0, k): s = np.sort(samples[i]) Mij = s.searchsorted(Zstar[:-1], side='right') inner = lj / float(N) * (N * Mij - Bj * n[i])**2 / (Bj * (N - Bj)) A2kN += inner.sum() / n[i] return A2kN Anderson_ksampResult = _make_tuple_bunch( 'Anderson_ksampResult', ['statistic', 'critical_values', 'pvalue'], [] ) def anderson_ksamp(samples, midrank=True): """The Anderson-Darling test for k-samples. The k-sample Anderson-Darling test is a modification of the one-sample Anderson-Darling test. It tests the null hypothesis that k-samples are drawn from the same population without having to specify the distribution function of that population. The critical values depend on the number of samples. Parameters ---------- samples : sequence of 1-D array_like Array of sample data in arrays. midrank : bool, optional Type of Anderson-Darling test which is computed. Default (True) is the midrank test applicable to continuous and discrete populations. If False, the right side empirical distribution is used. Returns ------- res : Anderson_ksampResult An object containing attributes: statistic : float Normalized k-sample Anderson-Darling test statistic. critical_values : array The critical values for significance levels 25%, 10%, 5%, 2.5%, 1%, 0.5%, 0.1%. pvalue : float The approximate p-value of the test. The value is floored / capped at 0.1% / 25%. Raises ------ ValueError If less than 2 samples are provided, a sample is empty, or no distinct observations are in the samples. See Also -------- ks_2samp : 2 sample Kolmogorov-Smirnov test anderson : 1 sample Anderson-Darling test Notes ----- [1]_ defines three versions of the k-sample Anderson-Darling test: one for continuous distributions and two for discrete distributions, in which ties between samples may occur. The default of this routine is to compute the version based on the midrank empirical distribution function. This test is applicable to continuous and discrete data. If midrank is set to False, the right side empirical distribution is used for a test for discrete data. According to [1]_, the two discrete test statistics differ only slightly if a few collisions due to round-off errors occur in the test not adjusted for ties between samples. The critical values corresponding to the significance levels from 0.01 to 0.25 are taken from [1]_. p-values are floored / capped at 0.1% / 25%. Since the range of critical values might be extended in future releases, it is recommended not to test ``p == 0.25``, but rather ``p >= 0.25`` (analogously for the lower bound). .. versionadded:: 0.14.0 References ---------- .. [1] Scholz, F. W and Stephens, M. A. (1987), K-Sample Anderson-Darling Tests, Journal of the American Statistical Association, Vol. 82, pp. 918-924. Examples -------- >>> import numpy as np >>> from scipy import stats >>> rng = np.random.default_rng() >>> res = stats.anderson_ksamp([rng.normal(size=50), ... rng.normal(loc=0.5, size=30)]) >>> res.statistic, res.pvalue (1.974403288713695, 0.04991293614572478) >>> res.critical_values array([0.325, 1.226, 1.961, 2.718, 3.752, 4.592, 6.546]) The null hypothesis that the two random samples come from the same distribution can be rejected at the 5% level because the returned test value is greater than the critical value for 5% (1.961) but not at the 2.5% level. The interpolation gives an approximate p-value of 4.99%. >>> res = stats.anderson_ksamp([rng.normal(size=50), ... rng.normal(size=30), rng.normal(size=20)]) >>> res.statistic, res.pvalue (-0.29103725200789504, 0.25) >>> res.critical_values array([ 0.44925884, 1.3052767 , 1.9434184 , 2.57696569, 3.41634856, 4.07210043, 5.56419101]) The null hypothesis cannot be rejected for three samples from an identical distribution. The reported p-value (25%) has been capped and may not be very accurate (since it corresponds to the value 0.449 whereas the statistic is -0.291). """ k = len(samples) if (k < 2): raise ValueError("anderson_ksamp needs at least two samples") samples = list(map(np.asarray, samples)) Z = np.sort(np.hstack(samples)) N = Z.size Zstar = np.unique(Z) if Zstar.size < 2: raise ValueError("anderson_ksamp needs more than one distinct " "observation") n = np.array([sample.size for sample in samples]) if np.any(n == 0): raise ValueError("anderson_ksamp encountered sample without " "observations") if midrank: A2kN = _anderson_ksamp_midrank(samples, Z, Zstar, k, n, N) else: A2kN = _anderson_ksamp_right(samples, Z, Zstar, k, n, N) H = (1. / n).sum() hs_cs = (1. / arange(N - 1, 1, -1)).cumsum() h = hs_cs[-1] + 1 g = (hs_cs / arange(2, N)).sum() a = (4*g - 6) * (k - 1) + (10 - 6*g)*H b = (2*g - 4)*k**2 + 8*h*k + (2*g - 14*h - 4)*H - 8*h + 4*g - 6 c = (6*h + 2*g - 2)*k**2 + (4*h - 4*g + 6)*k + (2*h - 6)*H + 4*h d = (2*h + 6)*k**2 - 4*h*k sigmasq = (a*N**3 + b*N**2 + c*N + d) / ((N - 1.) * (N - 2.) * (N - 3.)) m = k - 1 A2 = (A2kN - m) / math.sqrt(sigmasq) # The b_i values are the interpolation coefficients from Table 2 # of Scholz and Stephens 1987 b0 = np.array([0.675, 1.281, 1.645, 1.96, 2.326, 2.573, 3.085]) b1 = np.array([-0.245, 0.25, 0.678, 1.149, 1.822, 2.364, 3.615]) b2 = np.array([-0.105, -0.305, -0.362, -0.391, -0.396, -0.345, -0.154]) critical = b0 + b1 / math.sqrt(m) + b2 / m sig = np.array([0.25, 0.1, 0.05, 0.025, 0.01, 0.005, 0.001]) if A2 < critical.min(): p = sig.max() warnings.warn("p-value capped: true value larger than {}".format(p), stacklevel=2) elif A2 > critical.max(): p = sig.min() warnings.warn("p-value floored: true value smaller than {}".format(p), stacklevel=2) else: # interpolation of probit of significance level pf = np.polyfit(critical, log(sig), 2) p = math.exp(np.polyval(pf, A2)) # create result object with alias for backward compatibility res = Anderson_ksampResult(A2, critical, p) res.significance_level = p return res AnsariResult = namedtuple('AnsariResult', ('statistic', 'pvalue')) class _ABW: """Distribution of Ansari-Bradley W-statistic under the null hypothesis.""" # TODO: calculate exact distribution considering ties # We could avoid summing over more than half the frequencies, # but inititally it doesn't seem worth the extra complexity def __init__(self): """Minimal initializer.""" self.m = None self.n = None self.astart = None self.total = None self.freqs = None def _recalc(self, n, m): """When necessary, recalculate exact distribution.""" if n != self.n or m != self.m: self.n, self.m = n, m # distribution is NOT symmetric when m + n is odd # n is len(x), m is len(y), and ratio of scales is defined x/y astart, a1, _ = _statlib.gscale(n, m) self.astart = astart # minimum value of statistic # Exact distribution of test statistic under null hypothesis # expressed as frequencies/counts/integers to maintain precision. # Stored as floats to avoid overflow of sums. self.freqs = a1.astype(np.float64) self.total = self.freqs.sum() # could calculate from m and n # probability mass is self.freqs / self.total; def pmf(self, k, n, m): """Probability mass function.""" self._recalc(n, m) # The convention here is that PMF at k = 12.5 is the same as at k = 12, # -> use `floor` in case of ties. ind = np.floor(k - self.astart).astype(int) return self.freqs[ind] / self.total def cdf(self, k, n, m): """Cumulative distribution function.""" self._recalc(n, m) # Null distribution derived without considering ties is # approximate. Round down to avoid Type I error. ind = np.ceil(k - self.astart).astype(int) return self.freqs[:ind+1].sum() / self.total def sf(self, k, n, m): """Survival function.""" self._recalc(n, m) # Null distribution derived without considering ties is # approximate. Round down to avoid Type I error. ind = np.floor(k - self.astart).astype(int) return self.freqs[ind:].sum() / self.total # Maintain state for faster repeat calls to ansari w/ method='exact' _abw_state = _ABW() def ansari(x, y, alternative='two-sided'): """Perform the Ansari-Bradley test for equal scale parameters. The Ansari-Bradley test ([1]_, [2]_) is a non-parametric test for the equality of the scale parameter of the distributions from which two samples were drawn. The null hypothesis states that the ratio of the scale of the distribution underlying `x` to the scale of the distribution underlying `y` is 1. Parameters ---------- x, y : array_like Arrays of sample data. alternative : {'two-sided', 'less', 'greater'}, optional Defines the alternative hypothesis. Default is 'two-sided'. The following options are available: * 'two-sided': the ratio of scales is not equal to 1. * 'less': the ratio of scales is less than 1. * 'greater': the ratio of scales is greater than 1. .. versionadded:: 1.7.0 Returns ------- statistic : float The Ansari-Bradley test statistic. pvalue : float The p-value of the hypothesis test. See Also -------- fligner : A non-parametric test for the equality of k variances mood : A non-parametric test for the equality of two scale parameters Notes ----- The p-value given is exact when the sample sizes are both less than 55 and there are no ties, otherwise a normal approximation for the p-value is used. References ---------- .. [1] Ansari, A. R. and Bradley, R. A. (1960) Rank-sum tests for dispersions, Annals of Mathematical Statistics, 31, 1174-1189. .. [2] Sprent, Peter and N.C. Smeeton. Applied nonparametric statistical methods. 3rd ed. Chapman and Hall/CRC. 2001. Section 5.8.2. .. [3] Nathaniel E. Helwig "Nonparametric Dispersion and Equality Tests" at http://users.stat.umn.edu/~helwig/notes/npde-Notes.pdf Examples -------- >>> import numpy as np >>> from scipy.stats import ansari >>> rng = np.random.default_rng() For these examples, we'll create three random data sets. The first two, with sizes 35 and 25, are drawn from a normal distribution with mean 0 and standard deviation 2. The third data set has size 25 and is drawn from a normal distribution with standard deviation 1.25. >>> x1 = rng.normal(loc=0, scale=2, size=35) >>> x2 = rng.normal(loc=0, scale=2, size=25) >>> x3 = rng.normal(loc=0, scale=1.25, size=25) First we apply `ansari` to `x1` and `x2`. These samples are drawn from the same distribution, so we expect the Ansari-Bradley test should not lead us to conclude that the scales of the distributions are different. >>> ansari(x1, x2) AnsariResult(statistic=541.0, pvalue=0.9762532927399098) With a p-value close to 1, we cannot conclude that there is a significant difference in the scales (as expected). Now apply the test to `x1` and `x3`: >>> ansari(x1, x3) AnsariResult(statistic=425.0, pvalue=0.0003087020407974518) The probability of observing such an extreme value of the statistic under the null hypothesis of equal scales is only 0.03087%. We take this as evidence against the null hypothesis in favor of the alternative: the scales of the distributions from which the samples were drawn are not equal. We can use the `alternative` parameter to perform a one-tailed test. In the above example, the scale of `x1` is greater than `x3` and so the ratio of scales of `x1` and `x3` is greater than 1. This means that the p-value when ``alternative='greater'`` should be near 0 and hence we should be able to reject the null hypothesis: >>> ansari(x1, x3, alternative='greater') AnsariResult(statistic=425.0, pvalue=0.0001543510203987259) As we can see, the p-value is indeed quite low. Use of ``alternative='less'`` should thus yield a large p-value: >>> ansari(x1, x3, alternative='less') AnsariResult(statistic=425.0, pvalue=0.9998643258449039) """ if alternative not in {'two-sided', 'greater', 'less'}: raise ValueError("'alternative' must be 'two-sided'," " 'greater', or 'less'.") x, y = asarray(x), asarray(y) n = len(x) m = len(y) if m < 1: raise ValueError("Not enough other observations.") if n < 1: raise ValueError("Not enough test observations.") N = m + n xy = r_[x, y] # combine rank = _stats_py.rankdata(xy) symrank = amin(array((rank, N - rank + 1)), 0) AB = np.sum(symrank[:n], axis=0) uxy = unique(xy) repeats = (len(uxy) != len(xy)) exact = ((m < 55) and (n < 55) and not repeats) if repeats and (m < 55 or n < 55): warnings.warn("Ties preclude use of exact statistic.") if exact: if alternative == 'two-sided': pval = 2.0 * np.minimum(_abw_state.cdf(AB, n, m), _abw_state.sf(AB, n, m)) elif alternative == 'greater': # AB statistic is _smaller_ when ratio of scales is larger, # so this is the opposite of the usual calculation pval = _abw_state.cdf(AB, n, m) else: pval = _abw_state.sf(AB, n, m) return AnsariResult(AB, min(1.0, pval)) # otherwise compute normal approximation if N % 2: # N odd mnAB = n * (N+1.0)**2 / 4.0 / N varAB = n * m * (N+1.0) * (3+N**2) / (48.0 * N**2) else: mnAB = n * (N+2.0) / 4.0 varAB = m * n * (N+2) * (N-2.0) / 48 / (N-1.0) if repeats: # adjust variance estimates # compute np.sum(tj * rj**2,axis=0) fac = np.sum(symrank**2, axis=0) if N % 2: # N odd varAB = m * n * (16*N*fac - (N+1)**4) / (16.0 * N**2 * (N-1)) else: # N even varAB = m * n * (16*fac - N*(N+2)**2) / (16.0 * N * (N-1)) # Small values of AB indicate larger dispersion for the x sample. # Large values of AB indicate larger dispersion for the y sample. # This is opposite to the way we define the ratio of scales. see [1]_. z = (mnAB - AB) / sqrt(varAB) z, pval = _normtest_finish(z, alternative) return AnsariResult(AB, pval) BartlettResult = namedtuple('BartlettResult', ('statistic', 'pvalue')) def bartlett(*samples): r"""Perform Bartlett's test for equal variances. Bartlett's test tests the null hypothesis that all input samples are from populations with equal variances. For samples from significantly non-normal populations, Levene's test `levene` is more robust. Parameters ---------- sample1, sample2, ... : array_like arrays of sample data. Only 1d arrays are accepted, they may have different lengths. Returns ------- statistic : float The test statistic. pvalue : float The p-value of the test. See Also -------- fligner : A non-parametric test for the equality of k variances levene : A robust parametric test for equality of k variances Notes ----- Conover et al. (1981) examine many of the existing parametric and nonparametric tests by extensive simulations and they conclude that the tests proposed by Fligner and Killeen (1976) and Levene (1960) appear to be superior in terms of robustness of departures from normality and power ([3]_). References ---------- .. [1] https://www.itl.nist.gov/div898/handbook/eda/section3/eda357.htm .. [2] Snedecor, George W. and Cochran, William G. (1989), Statistical Methods, Eighth Edition, Iowa State University Press. .. [3] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and Hypothesis Testing based on Quadratic Inference Function. Technical Report #99-03, Center for Likelihood Studies, Pennsylvania State University. .. [4] Bartlett, M. S. (1937). Properties of Sufficiency and Statistical Tests. Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, Vol. 160, No.901, pp. 268-282. .. [5] C.I. BLISS (1952), The Statistics of Bioassay: With Special Reference to the Vitamins, pp 499-503, :doi:`10.1016/C2013-0-12584-6`. .. [6] B. Phipson and G. K. Smyth. "Permutation P-values Should Never Be Zero: Calculating Exact P-values When Permutations Are Randomly Drawn." Statistical Applications in Genetics and Molecular Biology 9.1 (2010). .. [7] Ludbrook, J., & Dudley, H. (1998). Why permutation tests are superior to t and F tests in biomedical research. The American Statistician, 52(2), 127-132. Examples -------- In [5]_, the influence of vitamin C on the tooth growth of guinea pigs was investigated. In a control study, 60 subjects were divided into small dose, medium dose, and large dose groups that received daily doses of 0.5, 1.0 and 2.0 mg of vitamin C, respectively. After 42 days, the tooth growth was measured. The ``small_dose``, ``medium_dose``, and ``large_dose`` arrays below record tooth growth measurements of the three groups in microns. >>> import numpy as np >>> small_dose = np.array([ ... 4.2, 11.5, 7.3, 5.8, 6.4, 10, 11.2, 11.2, 5.2, 7, ... 15.2, 21.5, 17.6, 9.7, 14.5, 10, 8.2, 9.4, 16.5, 9.7 ... ]) >>> medium_dose = np.array([ ... 16.5, 16.5, 15.2, 17.3, 22.5, 17.3, 13.6, 14.5, 18.8, 15.5, ... 19.7, 23.3, 23.6, 26.4, 20, 25.2, 25.8, 21.2, 14.5, 27.3 ... ]) >>> large_dose = np.array([ ... 23.6, 18.5, 33.9, 25.5, 26.4, 32.5, 26.7, 21.5, 23.3, 29.5, ... 25.5, 26.4, 22.4, 24.5, 24.8, 30.9, 26.4, 27.3, 29.4, 23 ... ]) The `bartlett` statistic is sensitive to differences in variances between the samples. >>> from scipy import stats >>> res = stats.bartlett(small_dose, medium_dose, large_dose) >>> res.statistic 0.6654670663030519 The value of the statistic tends to be high when there is a large difference in variances. We can test for inequality of variance among the groups by comparing the observed value of the statistic against the null distribution: the distribution of statistic values derived under the null hypothesis that the population variances of the three groups are equal. For this test, the null distribution follows the chi-square distribution as shown below. >>> import matplotlib.pyplot as plt >>> k = 3 # number of samples >>> dist = stats.chi2(df=k-1) >>> val = np.linspace(0, 5, 100) >>> pdf = dist.pdf(val) >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> def plot(ax): # we'll re-use this ... ax.plot(val, pdf, color='C0') ... ax.set_title("Bartlett Test Null Distribution") ... ax.set_xlabel("statistic") ... ax.set_ylabel("probability density") ... ax.set_xlim(0, 5) ... ax.set_ylim(0, 1) >>> plot(ax) >>> plt.show() The comparison is quantified by the p-value: the proportion of values in the null distribution greater than or equal to the observed value of the statistic. >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> plot(ax) >>> pvalue = dist.sf(res.statistic) >>> annotation = (f'p-value={pvalue:.3f}\n(shaded area)') >>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8) >>> _ = ax.annotate(annotation, (1.5, 0.22), (2.25, 0.3), arrowprops=props) >>> i = val >= res.statistic >>> ax.fill_between(val[i], y1=0, y2=pdf[i], color='C0') >>> plt.show() >>> res.pvalue 0.71696121509966 If the p-value is "small" - that is, if there is a low probability of sampling data from distributions with identical variances that produces such an extreme value of the statistic - this may be taken as evidence against the null hypothesis in favor of the alternative: the variances of the groups are not equal. Note that: - The inverse is not true; that is, the test is not used to provide evidence for the null hypothesis. - The threshold for values that will be considered "small" is a choice that should be made before the data is analyzed [6]_ with consideration of the risks of both false positives (incorrectly rejecting the null hypothesis) and false negatives (failure to reject a false null hypothesis). - Small p-values are not evidence for a *large* effect; rather, they can only provide evidence for a "significant" effect, meaning that they are unlikely to have occurred under the null hypothesis. Note that the chi-square distribution provides the null distribution when the observations are normally distributed. For small samples drawn from non-normal populations, it may be more appropriate to perform a permutation test: Under the null hypothesis that all three samples were drawn from the same population, each of the measurements is equally likely to have been observed in any of the three samples. Therefore, we can form a randomized null distribution by calculating the statistic under many randomly-generated partitionings of the observations into the three samples. >>> def statistic(*samples): ... return stats.bartlett(*samples).statistic >>> ref = stats.permutation_test( ... (small_dose, medium_dose, large_dose), statistic, ... permutation_type='independent', alternative='greater' ... ) >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> plot(ax) >>> bins = np.linspace(0, 5, 25) >>> ax.hist( ... ref.null_distribution, bins=bins, density=True, facecolor="C1" ... ) >>> ax.legend(['aymptotic approximation\n(many observations)', ... 'randomized null distribution']) >>> plot(ax) >>> plt.show() >>> ref.pvalue # randomized test p-value 0.5387 # may vary Note that there is significant disagreement between the p-value calculated here and the asymptotic approximation returned by `bartlett` above. The statistical inferences that can be drawn rigorously from a permutation test are limited; nonetheless, they may be the preferred approach in many circumstances [7]_. Following is another generic example where the null hypothesis would be rejected. Test whether the lists `a`, `b` and `c` come from populations with equal variances. >>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99] >>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05] >>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98] >>> stat, p = stats.bartlett(a, b, c) >>> p 1.1254782518834628e-05 The very small p-value suggests that the populations do not have equal variances. This is not surprising, given that the sample variance of `b` is much larger than that of `a` and `c`: >>> [np.var(x, ddof=1) for x in [a, b, c]] [0.007054444444444413, 0.13073888888888888, 0.008890000000000002] """ # Handle empty input and input that is not 1d for sample in samples: if np.asanyarray(sample).size == 0: return BartlettResult(np.nan, np.nan) if np.asanyarray(sample).ndim > 1: raise ValueError('Samples must be one-dimensional.') k = len(samples) if k < 2: raise ValueError("Must enter at least two input sample vectors.") Ni = np.empty(k) ssq = np.empty(k, 'd') for j in range(k): Ni[j] = len(samples[j]) ssq[j] = np.var(samples[j], ddof=1) Ntot = np.sum(Ni, axis=0) spsq = np.sum((Ni - 1)*ssq, axis=0) / (1.0*(Ntot - k)) numer = (Ntot*1.0 - k) * log(spsq) - np.sum((Ni - 1.0)*log(ssq), axis=0) denom = 1.0 + 1.0/(3*(k - 1)) * ((np.sum(1.0/(Ni - 1.0), axis=0)) - 1.0/(Ntot - k)) T = numer / denom pval = distributions.chi2.sf(T, k - 1) # 1 - cdf return BartlettResult(T, pval) LeveneResult = namedtuple('LeveneResult', ('statistic', 'pvalue')) def levene(*samples, center='median', proportiontocut=0.05): r"""Perform Levene test for equal variances. The Levene test tests the null hypothesis that all input samples are from populations with equal variances. Levene's test is an alternative to Bartlett's test `bartlett` in the case where there are significant deviations from normality. Parameters ---------- sample1, sample2, ... : array_like The sample data, possibly with different lengths. Only one-dimensional samples are accepted. center : {'mean', 'median', 'trimmed'}, optional Which function of the data to use in the test. The default is 'median'. proportiontocut : float, optional When `center` is 'trimmed', this gives the proportion of data points to cut from each end. (See `scipy.stats.trim_mean`.) Default is 0.05. Returns ------- statistic : float The test statistic. pvalue : float The p-value for the test. See Also -------- fligner : A non-parametric test for the equality of k variances bartlett : A parametric test for equality of k variances in normal samples Notes ----- Three variations of Levene's test are possible. The possibilities and their recommended usages are: * 'median' : Recommended for skewed (non-normal) distributions> * 'mean' : Recommended for symmetric, moderate-tailed distributions. * 'trimmed' : Recommended for heavy-tailed distributions. The test version using the mean was proposed in the original article of Levene ([2]_) while the median and trimmed mean have been studied by Brown and Forsythe ([3]_), sometimes also referred to as Brown-Forsythe test. References ---------- .. [1] https://www.itl.nist.gov/div898/handbook/eda/section3/eda35a.htm .. [2] Levene, H. (1960). In Contributions to Probability and Statistics: Essays in Honor of Harold Hotelling, I. Olkin et al. eds., Stanford University Press, pp. 278-292. .. [3] Brown, M. B. and Forsythe, A. B. (1974), Journal of the American Statistical Association, 69, 364-367 .. [4] C.I. BLISS (1952), The Statistics of Bioassay: With Special Reference to the Vitamins, pp 499-503, :doi:`10.1016/C2013-0-12584-6`. .. [5] B. Phipson and G. K. Smyth. "Permutation P-values Should Never Be Zero: Calculating Exact P-values When Permutations Are Randomly Drawn." Statistical Applications in Genetics and Molecular Biology 9.1 (2010). .. [6] Ludbrook, J., & Dudley, H. (1998). Why permutation tests are superior to t and F tests in biomedical research. The American Statistician, 52(2), 127-132. Examples -------- In [4]_, the influence of vitamin C on the tooth growth of guinea pigs was investigated. In a control study, 60 subjects were divided into small dose, medium dose, and large dose groups that received daily doses of 0.5, 1.0 and 2.0 mg of vitamin C, respectively. After 42 days, the tooth growth was measured. The ``small_dose``, ``medium_dose``, and ``large_dose`` arrays below record tooth growth measurements of the three groups in microns. >>> import numpy as np >>> small_dose = np.array([ ... 4.2, 11.5, 7.3, 5.8, 6.4, 10, 11.2, 11.2, 5.2, 7, ... 15.2, 21.5, 17.6, 9.7, 14.5, 10, 8.2, 9.4, 16.5, 9.7 ... ]) >>> medium_dose = np.array([ ... 16.5, 16.5, 15.2, 17.3, 22.5, 17.3, 13.6, 14.5, 18.8, 15.5, ... 19.7, 23.3, 23.6, 26.4, 20, 25.2, 25.8, 21.2, 14.5, 27.3 ... ]) >>> large_dose = np.array([ ... 23.6, 18.5, 33.9, 25.5, 26.4, 32.5, 26.7, 21.5, 23.3, 29.5, ... 25.5, 26.4, 22.4, 24.5, 24.8, 30.9, 26.4, 27.3, 29.4, 23 ... ]) The `levene` statistic is sensitive to differences in variances between the samples. >>> from scipy import stats >>> res = stats.levene(small_dose, medium_dose, large_dose) >>> res.statistic 0.6457341109631506 The value of the statistic tends to be high when there is a large difference in variances. We can test for inequality of variance among the groups by comparing the observed value of the statistic against the null distribution: the distribution of statistic values derived under the null hypothesis that the population variances of the three groups are equal. For this test, the null distribution follows the F distribution as shown below. >>> import matplotlib.pyplot as plt >>> k, n = 3, 60 # number of samples, total number of observations >>> dist = stats.f(dfn=k-1, dfd=n-k) >>> val = np.linspace(0, 5, 100) >>> pdf = dist.pdf(val) >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> def plot(ax): # we'll re-use this ... ax.plot(val, pdf, color='C0') ... ax.set_title("Levene Test Null Distribution") ... ax.set_xlabel("statistic") ... ax.set_ylabel("probability density") ... ax.set_xlim(0, 5) ... ax.set_ylim(0, 1) >>> plot(ax) >>> plt.show() The comparison is quantified by the p-value: the proportion of values in the null distribution greater than or equal to the observed value of the statistic. >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> plot(ax) >>> pvalue = dist.sf(res.statistic) >>> annotation = (f'p-value={pvalue:.3f}\n(shaded area)') >>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8) >>> _ = ax.annotate(annotation, (1.5, 0.22), (2.25, 0.3), arrowprops=props) >>> i = val >= res.statistic >>> ax.fill_between(val[i], y1=0, y2=pdf[i], color='C0') >>> plt.show() >>> res.pvalue 0.5280694573759905 If the p-value is "small" - that is, if there is a low probability of sampling data from distributions with identical variances that produces such an extreme value of the statistic - this may be taken as evidence against the null hypothesis in favor of the alternative: the variances of the groups are not equal. Note that: - The inverse is not true; that is, the test is not used to provide evidence for the null hypothesis. - The threshold for values that will be considered "small" is a choice that should be made before the data is analyzed [5]_ with consideration of the risks of both false positives (incorrectly rejecting the null hypothesis) and false negatives (failure to reject a false null hypothesis). - Small p-values are not evidence for a *large* effect; rather, they can only provide evidence for a "significant" effect, meaning that they are unlikely to have occurred under the null hypothesis. Note that the F distribution provides an asymptotic approximation of the null distribution. For small samples, it may be more appropriate to perform a permutation test: Under the null hypothesis that all three samples were drawn from the same population, each of the measurements is equally likely to have been observed in any of the three samples. Therefore, we can form a randomized null distribution by calculating the statistic under many randomly-generated partitionings of the observations into the three samples. >>> def statistic(*samples): ... return stats.levene(*samples).statistic >>> ref = stats.permutation_test( ... (small_dose, medium_dose, large_dose), statistic, ... permutation_type='independent', alternative='greater' ... ) >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> plot(ax) >>> bins = np.linspace(0, 5, 25) >>> ax.hist( ... ref.null_distribution, bins=bins, density=True, facecolor="C1" ... ) >>> ax.legend(['aymptotic approximation\n(many observations)', ... 'randomized null distribution']) >>> plot(ax) >>> plt.show() >>> ref.pvalue # randomized test p-value 0.4559 # may vary Note that there is significant disagreement between the p-value calculated here and the asymptotic approximation returned by `levene` above. The statistical inferences that can be drawn rigorously from a permutation test are limited; nonetheless, they may be the preferred approach in many circumstances [6]_. Following is another generic example where the null hypothesis would be rejected. Test whether the lists `a`, `b` and `c` come from populations with equal variances. >>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99] >>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05] >>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98] >>> stat, p = stats.levene(a, b, c) >>> p 0.002431505967249681 The small p-value suggests that the populations do not have equal variances. This is not surprising, given that the sample variance of `b` is much larger than that of `a` and `c`: >>> [np.var(x, ddof=1) for x in [a, b, c]] [0.007054444444444413, 0.13073888888888888, 0.008890000000000002] """ if center not in ['mean', 'median', 'trimmed']: raise ValueError("center must be 'mean', 'median' or 'trimmed'.") k = len(samples) if k < 2: raise ValueError("Must enter at least two input sample vectors.") # check for 1d input for j in range(k): if np.asanyarray(samples[j]).ndim > 1: raise ValueError('Samples must be one-dimensional.') Ni = np.empty(k) Yci = np.empty(k, 'd') if center == 'median': def func(x): return np.median(x, axis=0) elif center == 'mean': def func(x): return np.mean(x, axis=0) else: # center == 'trimmed' samples = tuple(_stats_py.trimboth(np.sort(sample), proportiontocut) for sample in samples) def func(x): return np.mean(x, axis=0) for j in range(k): Ni[j] = len(samples[j]) Yci[j] = func(samples[j]) Ntot = np.sum(Ni, axis=0) # compute Zij's Zij = [None] * k for i in range(k): Zij[i] = abs(asarray(samples[i]) - Yci[i]) # compute Zbari Zbari = np.empty(k, 'd') Zbar = 0.0 for i in range(k): Zbari[i] = np.mean(Zij[i], axis=0) Zbar += Zbari[i] * Ni[i] Zbar /= Ntot numer = (Ntot - k) * np.sum(Ni * (Zbari - Zbar)**2, axis=0) # compute denom_variance dvar = 0.0 for i in range(k): dvar += np.sum((Zij[i] - Zbari[i])**2, axis=0) denom = (k - 1.0) * dvar W = numer / denom pval = distributions.f.sf(W, k-1, Ntot-k) # 1 - cdf return LeveneResult(W, pval) @_deprecated("'binom_test' is deprecated in favour of" " 'binomtest' from version 1.7.0 and will" " be removed in Scipy 1.12.0.") def binom_test(x, n=None, p=0.5, alternative='two-sided'): """Perform a test that the probability of success is p. This is an exact, two-sided test of the null hypothesis that the probability of success in a Bernoulli experiment is `p`. .. deprecated:: 1.10.0 `binom_test` is deprecated in favour of `binomtest` and will be removed in Scipy 1.12.0. Parameters ---------- x : int or array_like The number of successes, or if x has length 2, it is the number of successes and the number of failures. n : int The number of trials. This is ignored if x gives both the number of successes and failures. p : float, optional The hypothesized probability of success. ``0 <= p <= 1``. The default value is ``p = 0.5``. alternative : {'two-sided', 'greater', 'less'}, optional Indicates the alternative hypothesis. The default value is 'two-sided'. Returns ------- p-value : float The p-value of the hypothesis test. References ---------- .. [1] https://en.wikipedia.org/wiki/Binomial_test Examples -------- >>> from scipy import stats A car manufacturer claims that no more than 10% of their cars are unsafe. 15 cars are inspected for safety, 3 were found to be unsafe. Test the manufacturer's claim: >>> stats.binom_test(3, n=15, p=0.1, alternative='greater') 0.18406106910639114 The null hypothesis cannot be rejected at the 5% level of significance because the returned p-value is greater than the critical value of 5%. """ x = atleast_1d(x).astype(np.int_) if len(x) == 2: n = x[1] + x[0] x = x[0] elif len(x) == 1: x = x[0] if n is None or n < x: raise ValueError("n must be >= x") n = np.int_(n) else: raise ValueError("Incorrect length for x.") if (p > 1.0) or (p < 0.0): raise ValueError("p must be in range [0,1]") if alternative not in ('two-sided', 'less', 'greater'): raise ValueError("alternative not recognized\n" "should be 'two-sided', 'less' or 'greater'") if alternative == 'less': pval = distributions.binom.cdf(x, n, p) return pval if alternative == 'greater': pval = distributions.binom.sf(x-1, n, p) return pval # if alternative was neither 'less' nor 'greater', then it's 'two-sided' d = distributions.binom.pmf(x, n, p) rerr = 1 + 1e-7 if x == p * n: # special case as shortcut, would also be handled by `else` below pval = 1. elif x < p * n: i = np.arange(np.ceil(p * n), n+1) y = np.sum(distributions.binom.pmf(i, n, p) <= d*rerr, axis=0) pval = (distributions.binom.cdf(x, n, p) + distributions.binom.sf(n - y, n, p)) else: i = np.arange(np.floor(p*n) + 1) y = np.sum(distributions.binom.pmf(i, n, p) <= d*rerr, axis=0) pval = (distributions.binom.cdf(y-1, n, p) + distributions.binom.sf(x-1, n, p)) return min(1.0, pval) def _apply_func(x, g, func): # g is list of indices into x # separating x into different groups # func should be applied over the groups g = unique(r_[0, g, len(x)]) output = [func(x[g[k]:g[k+1]]) for k in range(len(g) - 1)] return asarray(output) FlignerResult = namedtuple('FlignerResult', ('statistic', 'pvalue')) def fligner(*samples, center='median', proportiontocut=0.05): r"""Perform Fligner-Killeen test for equality of variance. Fligner's test tests the null hypothesis that all input samples are from populations with equal variances. Fligner-Killeen's test is distribution free when populations are identical [2]_. Parameters ---------- sample1, sample2, ... : array_like Arrays of sample data. Need not be the same length. center : {'mean', 'median', 'trimmed'}, optional Keyword argument controlling which function of the data is used in computing the test statistic. The default is 'median'. proportiontocut : float, optional When `center` is 'trimmed', this gives the proportion of data points to cut from each end. (See `scipy.stats.trim_mean`.) Default is 0.05. Returns ------- statistic : float The test statistic. pvalue : float The p-value for the hypothesis test. See Also -------- bartlett : A parametric test for equality of k variances in normal samples levene : A robust parametric test for equality of k variances Notes ----- As with Levene's test there are three variants of Fligner's test that differ by the measure of central tendency used in the test. See `levene` for more information. Conover et al. (1981) examine many of the existing parametric and nonparametric tests by extensive simulations and they conclude that the tests proposed by Fligner and Killeen (1976) and Levene (1960) appear to be superior in terms of robustness of departures from normality and power [3]_. References ---------- .. [1] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and Hypothesis Testing based on Quadratic Inference Function. Technical Report #99-03, Center for Likelihood Studies, Pennsylvania State University. https://cecas.clemson.edu/~cspark/cv/paper/qif/draftqif2.pdf .. [2] Fligner, M.A. and Killeen, T.J. (1976). Distribution-free two-sample tests for scale. 'Journal of the American Statistical Association.' 71(353), 210-213. .. [3] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and Hypothesis Testing based on Quadratic Inference Function. Technical Report #99-03, Center for Likelihood Studies, Pennsylvania State University. .. [4] Conover, W. J., Johnson, M. E. and Johnson M. M. (1981). A comparative study of tests for homogeneity of variances, with applications to the outer continental shelf biding data. Technometrics, 23(4), 351-361. .. [5] C.I. BLISS (1952), The Statistics of Bioassay: With Special Reference to the Vitamins, pp 499-503, :doi:`10.1016/C2013-0-12584-6`. .. [6] B. Phipson and G. K. Smyth. "Permutation P-values Should Never Be Zero: Calculating Exact P-values When Permutations Are Randomly Drawn." Statistical Applications in Genetics and Molecular Biology 9.1 (2010). .. [7] Ludbrook, J., & Dudley, H. (1998). Why permutation tests are superior to t and F tests in biomedical research. The American Statistician, 52(2), 127-132. Examples -------- In [5]_, the influence of vitamin C on the tooth growth of guinea pigs was investigated. In a control study, 60 subjects were divided into small dose, medium dose, and large dose groups that received daily doses of 0.5, 1.0 and 2.0 mg of vitamin C, respectively. After 42 days, the tooth growth was measured. The ``small_dose``, ``medium_dose``, and ``large_dose`` arrays below record tooth growth measurements of the three groups in microns. >>> import numpy as np >>> small_dose = np.array([ ... 4.2, 11.5, 7.3, 5.8, 6.4, 10, 11.2, 11.2, 5.2, 7, ... 15.2, 21.5, 17.6, 9.7, 14.5, 10, 8.2, 9.4, 16.5, 9.7 ... ]) >>> medium_dose = np.array([ ... 16.5, 16.5, 15.2, 17.3, 22.5, 17.3, 13.6, 14.5, 18.8, 15.5, ... 19.7, 23.3, 23.6, 26.4, 20, 25.2, 25.8, 21.2, 14.5, 27.3 ... ]) >>> large_dose = np.array([ ... 23.6, 18.5, 33.9, 25.5, 26.4, 32.5, 26.7, 21.5, 23.3, 29.5, ... 25.5, 26.4, 22.4, 24.5, 24.8, 30.9, 26.4, 27.3, 29.4, 23 ... ]) The `fligner` statistic is sensitive to differences in variances between the samples. >>> from scipy import stats >>> res = stats.fligner(small_dose, medium_dose, large_dose) >>> res.statistic 1.3878943408857916 The value of the statistic tends to be high when there is a large difference in variances. We can test for inequality of variance among the groups by comparing the observed value of the statistic against the null distribution: the distribution of statistic values derived under the null hypothesis that the population variances of the three groups are equal. For this test, the null distribution follows the chi-square distribution as shown below. >>> import matplotlib.pyplot as plt >>> k = 3 # number of samples >>> dist = stats.chi2(df=k-1) >>> val = np.linspace(0, 8, 100) >>> pdf = dist.pdf(val) >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> def plot(ax): # we'll re-use this ... ax.plot(val, pdf, color='C0') ... ax.set_title("Fligner Test Null Distribution") ... ax.set_xlabel("statistic") ... ax.set_ylabel("probability density") ... ax.set_xlim(0, 8) ... ax.set_ylim(0, 0.5) >>> plot(ax) >>> plt.show() The comparison is quantified by the p-value: the proportion of values in the null distribution greater than or equal to the observed value of the statistic. >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> plot(ax) >>> pvalue = dist.sf(res.statistic) >>> annotation = (f'p-value={pvalue:.4f}\n(shaded area)') >>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8) >>> _ = ax.annotate(annotation, (1.5, 0.22), (2.25, 0.3), arrowprops=props) >>> i = val >= res.statistic >>> ax.fill_between(val[i], y1=0, y2=pdf[i], color='C0') >>> plt.show() >>> res.pvalue 0.49960016501182125 If the p-value is "small" - that is, if there is a low probability of sampling data from distributions with identical variances that produces such an extreme value of the statistic - this may be taken as evidence against the null hypothesis in favor of the alternative: the variances of the groups are not equal. Note that: - The inverse is not true; that is, the test is not used to provide evidence for the null hypothesis. - The threshold for values that will be considered "small" is a choice that should be made before the data is analyzed [6]_ with consideration of the risks of both false positives (incorrectly rejecting the null hypothesis) and false negatives (failure to reject a false null hypothesis). - Small p-values are not evidence for a *large* effect; rather, they can only provide evidence for a "significant" effect, meaning that they are unlikely to have occurred under the null hypothesis. Note that the chi-square distribution provides an asymptotic approximation of the null distribution. For small samples, it may be more appropriate to perform a permutation test: Under the null hypothesis that all three samples were drawn from the same population, each of the measurements is equally likely to have been observed in any of the three samples. Therefore, we can form a randomized null distribution by calculating the statistic under many randomly-generated partitionings of the observations into the three samples. >>> def statistic(*samples): ... return stats.fligner(*samples).statistic >>> ref = stats.permutation_test( ... (small_dose, medium_dose, large_dose), statistic, ... permutation_type='independent', alternative='greater' ... ) >>> fig, ax = plt.subplots(figsize=(8, 5)) >>> plot(ax) >>> bins = np.linspace(0, 8, 25) >>> ax.hist( ... ref.null_distribution, bins=bins, density=True, facecolor="C1" ... ) >>> ax.legend(['aymptotic approximation\n(many observations)', ... 'randomized null distribution']) >>> plot(ax) >>> plt.show() >>> ref.pvalue # randomized test p-value 0.4332 # may vary Note that there is significant disagreement between the p-value calculated here and the asymptotic approximation returned by `fligner` above. The statistical inferences that can be drawn rigorously from a permutation test are limited; nonetheless, they may be the preferred approach in many circumstances [7]_. Following is another generic example where the null hypothesis would be rejected. Test whether the lists `a`, `b` and `c` come from populations with equal variances. >>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99] >>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05] >>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98] >>> stat, p = stats.fligner(a, b, c) >>> p 0.00450826080004775 The small p-value suggests that the populations do not have equal variances. This is not surprising, given that the sample variance of `b` is much larger than that of `a` and `c`: >>> [np.var(x, ddof=1) for x in [a, b, c]] [0.007054444444444413, 0.13073888888888888, 0.008890000000000002] """ if center not in ['mean', 'median', 'trimmed']: raise ValueError("center must be 'mean', 'median' or 'trimmed'.") # Handle empty input for sample in samples: if np.asanyarray(sample).size == 0: return FlignerResult(np.nan, np.nan) k = len(samples) if k < 2: raise ValueError("Must enter at least two input sample vectors.") if center == 'median': def func(x): return np.median(x, axis=0) elif center == 'mean': def func(x): return np.mean(x, axis=0) else: # center == 'trimmed' samples = tuple(_stats_py.trimboth(sample, proportiontocut) for sample in samples) def func(x): return np.mean(x, axis=0) Ni = asarray([len(samples[j]) for j in range(k)]) Yci = asarray([func(samples[j]) for j in range(k)]) Ntot = np.sum(Ni, axis=0) # compute Zij's Zij = [abs(asarray(samples[i]) - Yci[i]) for i in range(k)] allZij = [] g = [0] for i in range(k): allZij.extend(list(Zij[i])) g.append(len(allZij)) ranks = _stats_py.rankdata(allZij) sample = distributions.norm.ppf(ranks / (2*(Ntot + 1.0)) + 0.5) # compute Aibar Aibar = _apply_func(sample, g, np.sum) / Ni anbar = np.mean(sample, axis=0) varsq = np.var(sample, axis=0, ddof=1) Xsq = np.sum(Ni * (asarray(Aibar) - anbar)**2.0, axis=0) / varsq pval = distributions.chi2.sf(Xsq, k - 1) # 1 - cdf return FlignerResult(Xsq, pval) @_axis_nan_policy_factory(lambda x1: (x1,), n_samples=4, n_outputs=1) def _mood_inner_lc(xy, x, diffs, sorted_xy, n, m, N) -> float: # Obtain the unique values and their frequencies from the pooled samples. # "a_j, + b_j, = t_j, for j = 1, ... k" where `k` is the number of unique # classes, and "[t]he number of values associated with the x's and y's in # the jth class will be denoted by a_j, and b_j respectively." # (Mielke, 312) # Reuse previously computed sorted array and `diff` arrays to obtain the # unique values and counts. Prepend `diffs` with a non-zero to indicate # that the first element should be marked as not matching what preceded it. diffs_prep = np.concatenate(([1], diffs)) # Unique elements are where the was a difference between elements in the # sorted array uniques = sorted_xy[diffs_prep != 0] # The count of each element is the bin size for each set of consecutive # differences where the difference is zero. Replace nonzero differences # with 1 and then use the cumulative sum to count the indices. t = np.bincount(np.cumsum(np.asarray(diffs_prep != 0, dtype=int)))[1:] k = len(uniques) js = np.arange(1, k + 1, dtype=int) # the `b` array mentioned in the paper is not used, outside of the # calculation of `t`, so we do not need to calculate it separately. Here # we calculate `a`. In plain language, `a[j]` is the number of values in # `x` that equal `uniques[j]`. sorted_xyx = np.sort(np.concatenate((xy, x))) diffs = np.diff(sorted_xyx) diffs_prep = np.concatenate(([1], diffs)) diff_is_zero = np.asarray(diffs_prep != 0, dtype=int) xyx_counts = np.bincount(np.cumsum(diff_is_zero))[1:] a = xyx_counts - t # "Define .. a_0 = b_0 = t_0 = S_0 = 0" (Mielke 312) so we shift `a` # and `t` arrays over 1 to allow a first element of 0 to accommodate this # indexing. t = np.concatenate(([0], t)) a = np.concatenate(([0], a)) # S is built from `t`, so it does not need a preceding zero added on. S = np.cumsum(t) # define a copy of `S` with a prepending zero for later use to avoid # the need for indexing. S_i_m1 = np.concatenate(([0], S[:-1])) # Psi, as defined by the 6th unnumbered equation on page 313 (Mielke). # Note that in the paper there is an error where the denominator `2` is # squared when it should be the entire equation. def psi(indicator): return (indicator - (N + 1)/2)**2 # define summation range for use in calculation of phi, as seen in sum # in the unnumbered equation on the bottom of page 312 (Mielke). s_lower = S[js - 1] + 1 s_upper = S[js] + 1 phi_J = [np.arange(s_lower[idx], s_upper[idx]) for idx in range(k)] # for every range in the above array, determine the sum of psi(I) for # every element in the range. Divide all the sums by `t`. Following the # last unnumbered equation on page 312. phis = [np.sum(psi(I_j)) for I_j in phi_J] / t[js] # `T` is equal to a[j] * phi[j], per the first unnumbered equation on # page 312. `phis` is already in the order based on `js`, so we index # into `a` with `js` as well. T = sum(phis * a[js]) # The approximate statistic E_0_T = n * (N * N - 1) / 12 varM = (m * n * (N + 1.0) * (N ** 2 - 4) / 180 - m * n / (180 * N * (N - 1)) * np.sum( t * (t**2 - 1) * (t**2 - 4 + (15 * (N - S - S_i_m1) ** 2)) )) return ((T - E_0_T) / np.sqrt(varM),) def mood(x, y, axis=0, alternative="two-sided"): """Perform Mood's test for equal scale parameters. Mood's two-sample test for scale parameters is a non-parametric test for the null hypothesis that two samples are drawn from the same distribution with the same scale parameter. Parameters ---------- x, y : array_like Arrays of sample data. axis : int, optional The axis along which the samples are tested. `x` and `y` can be of different length along `axis`. If `axis` is None, `x` and `y` are flattened and the test is done on all values in the flattened arrays. alternative : {'two-sided', 'less', 'greater'}, optional Defines the alternative hypothesis. Default is 'two-sided'. The following options are available: * 'two-sided': the scales of the distributions underlying `x` and `y` are different. * 'less': the scale of the distribution underlying `x` is less than the scale of the distribution underlying `y`. * 'greater': the scale of the distribution underlying `x` is greater than the scale of the distribution underlying `y`. .. versionadded:: 1.7.0 Returns ------- res : SignificanceResult An object containing attributes: statistic : scalar or ndarray The z-score for the hypothesis test. For 1-D inputs a scalar is returned. pvalue : scalar ndarray The p-value for the hypothesis test. See Also -------- fligner : A non-parametric test for the equality of k variances ansari : A non-parametric test for the equality of 2 variances bartlett : A parametric test for equality of k variances in normal samples levene : A parametric test for equality of k variances Notes ----- The data are assumed to be drawn from probability distributions ``f(x)`` and ``f(x/s) / s`` respectively, for some probability density function f. The null hypothesis is that ``s == 1``. For multi-dimensional arrays, if the inputs are of shapes ``(n0, n1, n2, n3)`` and ``(n0, m1, n2, n3)``, then if ``axis=1``, the resulting z and p values will have shape ``(n0, n2, n3)``. Note that ``n1`` and ``m1`` don't have to be equal, but the other dimensions do. References ---------- [1] Mielke, Paul W. "Note on Some Squared Rank Tests with Existing Ties." Technometrics, vol. 9, no. 2, 1967, pp. 312-14. JSTOR, https://doi.org/10.2307/1266427. Accessed 18 May 2022. Examples -------- >>> import numpy as np >>> from scipy import stats >>> rng = np.random.default_rng() >>> x2 = rng.standard_normal((2, 45, 6, 7)) >>> x1 = rng.standard_normal((2, 30, 6, 7)) >>> res = stats.mood(x1, x2, axis=1) >>> res.pvalue.shape (2, 6, 7) Find the number of points where the difference in scale is not significant: >>> (res.pvalue > 0.1).sum() 78 Perform the test with different scales: >>> x1 = rng.standard_normal((2, 30)) >>> x2 = rng.standard_normal((2, 35)) * 10.0 >>> stats.mood(x1, x2, axis=1) SignificanceResult(statistic=array([-5.76174136, -6.12650783]), pvalue=array([8.32505043e-09, 8.98287869e-10])) """ x = np.asarray(x, dtype=float) y = np.asarray(y, dtype=float) if axis is None: x = x.flatten() y = y.flatten() axis = 0 if axis < 0: axis = x.ndim + axis # Determine shape of the result arrays res_shape = tuple([x.shape[ax] for ax in range(len(x.shape)) if ax != axis]) if not (res_shape == tuple([y.shape[ax] for ax in range(len(y.shape)) if ax != axis])): raise ValueError("Dimensions of x and y on all axes except `axis` " "should match") n = x.shape[axis] m = y.shape[axis] N = m + n if N < 3: raise ValueError("Not enough observations.") xy = np.concatenate((x, y), axis=axis) # determine if any of the samples contain ties sorted_xy = np.sort(xy, axis=axis) diffs = np.diff(sorted_xy, axis=axis) if 0 in diffs: z = np.asarray(_mood_inner_lc(xy, x, diffs, sorted_xy, n, m, N, axis=axis)) else: if axis != 0: xy = np.moveaxis(xy, axis, 0) xy = xy.reshape(xy.shape[0], -1) # Generalized to the n-dimensional case by adding the axis argument, # and using for loops, since rankdata is not vectorized. For improving # performance consider vectorizing rankdata function. all_ranks = np.empty_like(xy) for j in range(xy.shape[1]): all_ranks[:, j] = _stats_py.rankdata(xy[:, j]) Ri = all_ranks[:n] M = np.sum((Ri - (N + 1.0) / 2) ** 2, axis=0) # Approx stat. mnM = n * (N * N - 1.0) / 12 varM = m * n * (N + 1.0) * (N + 2) * (N - 2) / 180 z = (M - mnM) / sqrt(varM) z, pval = _normtest_finish(z, alternative) if res_shape == (): # Return scalars, not 0-D arrays z = z[0] pval = pval[0] else: z.shape = res_shape pval.shape = res_shape return SignificanceResult(z, pval) WilcoxonResult = _make_tuple_bunch('WilcoxonResult', ['statistic', 'pvalue']) def wilcoxon_result_unpacker(res): if hasattr(res, 'zstatistic'): return res.statistic, res.pvalue, res.zstatistic else: return res.statistic, res.pvalue def wilcoxon_result_object(statistic, pvalue, zstatistic=None): res = WilcoxonResult(statistic, pvalue) if zstatistic is not None: res.zstatistic = zstatistic return res def wilcoxon_outputs(kwds): method = kwds.get('method', 'auto') if method == 'approx': return 3 return 2 @_rename_parameter("mode", "method") @_axis_nan_policy_factory( wilcoxon_result_object, paired=True, n_samples=lambda kwds: 2 if kwds.get('y', None) is not None else 1, result_to_tuple=wilcoxon_result_unpacker, n_outputs=wilcoxon_outputs, ) def wilcoxon(x, y=None, zero_method="wilcox", correction=False, alternative="two-sided", method='auto'): """Calculate the Wilcoxon signed-rank test. The Wilcoxon signed-rank test tests the null hypothesis that two related paired samples come from the same distribution. In particular, it tests whether the distribution of the differences ``x - y`` is symmetric about zero. It is a non-parametric version of the paired T-test. Parameters ---------- x : array_like Either the first set of measurements (in which case ``y`` is the second set of measurements), or the differences between two sets of measurements (in which case ``y`` is not to be specified.) Must be one-dimensional. y : array_like, optional Either the second set of measurements (if ``x`` is the first set of measurements), or not specified (if ``x`` is the differences between two sets of measurements.) Must be one-dimensional. .. warning:: When `y` is provided, `wilcoxon` calculates the test statistic based on the ranks of the absolute values of ``d = x - y``. Roundoff error in the subtraction can result in elements of ``d`` being assigned different ranks even when they would be tied with exact arithmetic. Rather than passing `x` and `y` separately, consider computing the difference ``x - y``, rounding as needed to ensure that only truly unique elements are numerically distinct, and passing the result as `x`, leaving `y` at the default (None). zero_method : {"wilcox", "pratt", "zsplit"}, optional There are different conventions for handling pairs of observations with equal values ("zero-differences", or "zeros"). * "wilcox": Discards all zero-differences (default); see [4]_. * "pratt": Includes zero-differences in the ranking process, but drops the ranks of the zeros (more conservative); see [3]_. In this case, the normal approximation is adjusted as in [5]_. * "zsplit": Includes zero-differences in the ranking process and splits the zero rank between positive and negative ones. correction : bool, optional If True, apply continuity correction by adjusting the Wilcoxon rank statistic by 0.5 towards the mean value when computing the z-statistic if a normal approximation is used. Default is False. alternative : {"two-sided", "greater", "less"}, optional Defines the alternative hypothesis. Default is 'two-sided'. In the following, let ``d`` represent the difference between the paired samples: ``d = x - y`` if both ``x`` and ``y`` are provided, or ``d = x`` otherwise. * 'two-sided': the distribution underlying ``d`` is not symmetric about zero. * 'less': the distribution underlying ``d`` is stochastically less than a distribution symmetric about zero. * 'greater': the distribution underlying ``d`` is stochastically greater than a distribution symmetric about zero. method : {"auto", "exact", "approx"}, optional Method to calculate the p-value, see Notes. Default is "auto". Returns ------- An object with the following attributes. statistic : array_like If `alternative` is "two-sided", the sum of the ranks of the differences above or below zero, whichever is smaller. Otherwise the sum of the ranks of the differences above zero. pvalue : array_like The p-value for the test depending on `alternative` and `method`. zstatistic : array_like When ``method = 'approx'``, this is the normalized z-statistic:: z = (T - mn - d) / se where ``T`` is `statistic` as defined above, ``mn`` is the mean of the distribution under the null hypothesis, ``d`` is a continuity correction, and ``se`` is the standard error. When ``method != 'approx'``, this attribute is not available. See Also -------- kruskal, mannwhitneyu Notes ----- In the following, let ``d`` represent the difference between the paired samples: ``d = x - y`` if both ``x`` and ``y`` are provided, or ``d = x`` otherwise. Assume that all elements of ``d`` are independent and identically distributed observations, and all are distinct and nonzero. - When ``len(d)`` is sufficiently large, the null distribution of the normalized test statistic (`zstatistic` above) is approximately normal, and ``method = 'approx'`` can be used to compute the p-value. - When ``len(d)`` is small, the normal approximation may not be accurate, and ``method='exact'`` is preferred (at the cost of additional execution time). - The default, ``method='auto'``, selects between the two: when ``len(d) <= 50``, the exact method is used; otherwise, the approximate method is used. The presence of "ties" (i.e. not all elements of ``d`` are unique) and "zeros" (i.e. elements of ``d`` are zero) changes the null distribution of the test statistic, and ``method='exact'`` no longer calculates the exact p-value. If ``method='approx'``, the z-statistic is adjusted for more accurate comparison against the standard normal, but still, for finite sample sizes, the standard normal is only an approximation of the true null distribution of the z-statistic. There is no clear consensus among references on which method most accurately approximates the p-value for small samples in the presence of zeros and/or ties. In any case, this is the behavior of `wilcoxon` when ``method='auto': ``method='exact'`` is used when ``len(d) <= 50`` *and there are no zeros*; otherwise, ``method='approx'`` is used. References ---------- .. [1] https://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test .. [2] Conover, W.J., Practical Nonparametric Statistics, 1971. .. [3] Pratt, J.W., Remarks on Zeros and Ties in the Wilcoxon Signed Rank Procedures, Journal of the American Statistical Association, Vol. 54, 1959, pp. 655-667. :doi:`10.1080/01621459.1959.10501526` .. [4] Wilcoxon, F., Individual Comparisons by Ranking Methods, Biometrics Bulletin, Vol. 1, 1945, pp. 80-83. :doi:`10.2307/3001968` .. [5] Cureton, E.E., The Normal Approximation to the Signed-Rank Sampling Distribution When Zero Differences are Present, Journal of the American Statistical Association, Vol. 62, 1967, pp. 1068-1069. :doi:`10.1080/01621459.1967.10500917` Examples -------- In [4]_, the differences in height between cross- and self-fertilized corn plants is given as follows: >>> d = [6, 8, 14, 16, 23, 24, 28, 29, 41, -48, 49, 56, 60, -67, 75] Cross-fertilized plants appear to be higher. To test the null hypothesis that there is no height difference, we can apply the two-sided test: >>> from scipy.stats import wilcoxon >>> res = wilcoxon(d) >>> res.statistic, res.pvalue (24.0, 0.041259765625) Hence, we would reject the null hypothesis at a confidence level of 5%, concluding that there is a difference in height between the groups. To confirm that the median of the differences can be assumed to be positive, we use: >>> res = wilcoxon(d, alternative='greater') >>> res.statistic, res.pvalue (96.0, 0.0206298828125) This shows that the null hypothesis that the median is negative can be rejected at a confidence level of 5% in favor of the alternative that the median is greater than zero. The p-values above are exact. Using the normal approximation gives very similar values: >>> res = wilcoxon(d, method='approx') >>> res.statistic, res.pvalue (24.0, 0.04088813291185591) Note that the statistic changed to 96 in the one-sided case (the sum of ranks of positive differences) whereas it is 24 in the two-sided case (the minimum of sum of ranks above and below zero). In the example above, the differences in height between paired plants are provided to `wilcoxon` directly. Alternatively, `wilcoxon` accepts two samples of equal length, calculates the differences between paired elements, then performs the test. Consider the samples ``x`` and ``y``: >>> import numpy as np >>> x = np.array([0.5, 0.825, 0.375, 0.5]) >>> y = np.array([0.525, 0.775, 0.325, 0.55]) >>> res = wilcoxon(x, y, alternative='greater') >>> res WilcoxonResult(statistic=5.0, pvalue=0.5625) Note that had we calculated the differences by hand, the test would have produced different results: >>> d = [-0.025, 0.05, 0.05, -0.05] >>> ref = wilcoxon(d, alternative='greater') >>> ref WilcoxonResult(statistic=6.0, pvalue=0.4375) The substantial difference is due to roundoff error in the results of ``x-y``: >>> d - (x-y) array([2.08166817e-17, 6.93889390e-17, 1.38777878e-17, 4.16333634e-17]) Even though we expected all the elements of ``(x-y)[1:]`` to have the same magnitude ``0.05``, they have slightly different magnitudes in practice, and therefore are assigned different ranks in the test. Before performing the test, consider calculating ``d`` and adjusting it as necessary to ensure that theoretically identically values are not numerically distinct. For example: >>> d2 = np.around(x - y, decimals=3) >>> wilcoxon(d2, alternative='greater') WilcoxonResult(statistic=6.0, pvalue=0.4375) """ mode = method if mode not in ["auto", "approx", "exact"]: raise ValueError("mode must be either 'auto', 'approx' or 'exact'") if zero_method not in ["wilcox", "pratt", "zsplit"]: raise ValueError("Zero method must be either 'wilcox' " "or 'pratt' or 'zsplit'") if alternative not in ["two-sided", "less", "greater"]: raise ValueError("Alternative must be either 'two-sided', " "'greater' or 'less'") if y is None: d = asarray(x) if d.ndim > 1: raise ValueError('Sample x must be one-dimensional.') else: x, y = map(asarray, (x, y)) if x.ndim > 1 or y.ndim > 1: raise ValueError('Samples x and y must be one-dimensional.') if len(x) != len(y): raise ValueError('The samples x and y must have the same length.') # Future enhancement: consider warning when elements of `d` appear to # be tied but are numerically distinct. d = x - y if len(d) == 0: NaN = _get_nan(d) res = WilcoxonResult(NaN, NaN) if method == 'approx': res.zstatistic = NaN return res if mode == "auto": if len(d) <= 50: mode = "exact" else: mode = "approx" n_zero = np.sum(d == 0) if n_zero > 0 and mode == "exact": mode = "approx" warnings.warn("Exact p-value calculation does not work if there are " "zeros. Switching to normal approximation.") if mode == "approx": if zero_method in ["wilcox", "pratt"]: if n_zero == len(d): raise ValueError("zero_method 'wilcox' and 'pratt' do not " "work if x - y is zero for all elements.") if zero_method == "wilcox": # Keep all non-zero differences d = compress(np.not_equal(d, 0), d) count = len(d) if count < 10 and mode == "approx": warnings.warn("Sample size too small for normal approximation.") r = _stats_py.rankdata(abs(d)) r_plus = np.sum((d > 0) * r) r_minus = np.sum((d < 0) * r) if zero_method == "zsplit": r_zero = np.sum((d == 0) * r) r_plus += r_zero / 2. r_minus += r_zero / 2. # return min for two-sided test, but r_plus for one-sided test # the literature is not consistent here # r_plus is more informative since r_plus + r_minus = count*(count+1)/2, # i.e. the sum of the ranks, so r_minus and the min can be inferred # (If alternative='pratt', r_plus + r_minus = count*(count+1)/2 - r_zero.) # [3] uses the r_plus for the one-sided test, keep min for two-sided test # to keep backwards compatibility if alternative == "two-sided": T = min(r_plus, r_minus) else: T = r_plus if mode == "approx": mn = count * (count + 1.) * 0.25 se = count * (count + 1.) * (2. * count + 1.) if zero_method == "pratt": r = r[d != 0] # normal approximation needs to be adjusted, see Cureton (1967) mn -= n_zero * (n_zero + 1.) * 0.25 se -= n_zero * (n_zero + 1.) * (2. * n_zero + 1.) replist, repnum = find_repeats(r) if repnum.size != 0: # Correction for repeated elements. se -= 0.5 * (repnum * (repnum * repnum - 1)).sum() se = sqrt(se / 24) # apply continuity correction if applicable d = 0 if correction: if alternative == "two-sided": d = 0.5 * np.sign(T - mn) elif alternative == "less": d = -0.5 else: d = 0.5 # compute statistic and p-value using normal approximation z = (T - mn - d) / se if alternative == "two-sided": prob = 2. * distributions.norm.sf(abs(z)) elif alternative == "greater": # large T = r_plus indicates x is greater than y; i.e. # accept alternative in that case and return small p-value (sf) prob = distributions.norm.sf(z) else: prob = distributions.norm.cdf(z) elif mode == "exact": # get pmf of the possible positive ranksums r_plus pmf = _get_wilcoxon_distr(count) # note: r_plus is int (ties not allowed), need int for slices below r_plus = int(r_plus) if alternative == "two-sided": if r_plus == (len(pmf) - 1) // 2: # r_plus is the center of the distribution. prob = 1.0 else: p_less = np.sum(pmf[:r_plus + 1]) p_greater = np.sum(pmf[r_plus:]) prob = 2*min(p_greater, p_less) elif alternative == "greater": prob = np.sum(pmf[r_plus:]) else: prob = np.sum(pmf[:r_plus + 1]) prob = np.clip(prob, 0, 1) res = WilcoxonResult(T, prob) if method == 'approx': res.zstatistic = z return res MedianTestResult = _make_tuple_bunch( 'MedianTestResult', ['statistic', 'pvalue', 'median', 'table'], [] ) def median_test(*samples, ties='below', correction=True, lambda_=1, nan_policy='propagate'): """Perform a Mood's median test. Test that two or more samples come from populations with the same median. Let ``n = len(samples)`` be the number of samples. The "grand median" of all the data is computed, and a contingency table is formed by classifying the values in each sample as being above or below the grand median. The contingency table, along with `correction` and `lambda_`, are passed to `scipy.stats.chi2_contingency` to compute the test statistic and p-value. Parameters ---------- sample1, sample2, ... : array_like The set of samples. There must be at least two samples. Each sample must be a one-dimensional sequence containing at least one value. The samples are not required to have the same length. ties : str, optional Determines how values equal to the grand median are classified in the contingency table. The string must be one of:: "below": Values equal to the grand median are counted as "below". "above": Values equal to the grand median are counted as "above". "ignore": Values equal to the grand median are not counted. The default is "below". correction : bool, optional If True, *and* there are just two samples, apply Yates' correction for continuity when computing the test statistic associated with the contingency table. Default is True. lambda_ : float or str, optional By default, the statistic computed in this test is Pearson's chi-squared statistic. `lambda_` allows a statistic from the Cressie-Read power divergence family to be used instead. See `power_divergence` for details. Default is 1 (Pearson's chi-squared statistic). nan_policy : {'propagate', 'raise', 'omit'}, optional Defines how to handle when input contains nan. 'propagate' returns nan, 'raise' throws an error, 'omit' performs the calculations ignoring nan values. Default is 'propagate'. Returns ------- res : MedianTestResult An object containing attributes: statistic : float The test statistic. The statistic that is returned is determined by `lambda_`. The default is Pearson's chi-squared statistic. pvalue : float The p-value of the test. median : float The grand median. table : ndarray The contingency table. The shape of the table is (2, n), where n is the number of samples. The first row holds the counts of the values above the grand median, and the second row holds the counts of the values below the grand median. The table allows further analysis with, for example, `scipy.stats.chi2_contingency`, or with `scipy.stats.fisher_exact` if there are two samples, without having to recompute the table. If ``nan_policy`` is "propagate" and there are nans in the input, the return value for ``table`` is ``None``. See Also -------- kruskal : Compute the Kruskal-Wallis H-test for independent samples. mannwhitneyu : Computes the Mann-Whitney rank test on samples x and y. Notes ----- .. versionadded:: 0.15.0 References ---------- .. [1] Mood, A. M., Introduction to the Theory of Statistics. McGraw-Hill (1950), pp. 394-399. .. [2] Zar, J. H., Biostatistical Analysis, 5th ed. Prentice Hall (2010). See Sections 8.12 and 10.15. Examples -------- A biologist runs an experiment in which there are three groups of plants. Group 1 has 16 plants, group 2 has 15 plants, and group 3 has 17 plants. Each plant produces a number of seeds. The seed counts for each group are:: Group 1: 10 14 14 18 20 22 24 25 31 31 32 39 43 43 48 49 Group 2: 28 30 31 33 34 35 36 40 44 55 57 61 91 92 99 Group 3: 0 3 9 22 23 25 25 33 34 34 40 45 46 48 62 67 84 The following code applies Mood's median test to these samples. >>> g1 = [10, 14, 14, 18, 20, 22, 24, 25, 31, 31, 32, 39, 43, 43, 48, 49] >>> g2 = [28, 30, 31, 33, 34, 35, 36, 40, 44, 55, 57, 61, 91, 92, 99] >>> g3 = [0, 3, 9, 22, 23, 25, 25, 33, 34, 34, 40, 45, 46, 48, 62, 67, 84] >>> from scipy.stats import median_test >>> res = median_test(g1, g2, g3) The median is >>> res.median 34.0 and the contingency table is >>> res.table array([[ 5, 10, 7], [11, 5, 10]]) `p` is too large to conclude that the medians are not the same: >>> res.pvalue 0.12609082774093244 The "G-test" can be performed by passing ``lambda_="log-likelihood"`` to `median_test`. >>> res = median_test(g1, g2, g3, lambda_="log-likelihood") >>> res.pvalue 0.12224779737117837 The median occurs several times in the data, so we'll get a different result if, for example, ``ties="above"`` is used: >>> res = median_test(g1, g2, g3, ties="above") >>> res.pvalue 0.063873276069553273 >>> res.table array([[ 5, 11, 9], [11, 4, 8]]) This example demonstrates that if the data set is not large and there are values equal to the median, the p-value can be sensitive to the choice of `ties`. """ if len(samples) < 2: raise ValueError('median_test requires two or more samples.') ties_options = ['below', 'above', 'ignore'] if ties not in ties_options: raise ValueError("invalid 'ties' option '{}'; 'ties' must be one " "of: {}".format(ties, str(ties_options)[1:-1])) data = [np.asarray(sample) for sample in samples] # Validate the sizes and shapes of the arguments. for k, d in enumerate(data): if d.size == 0: raise ValueError("Sample %d is empty. All samples must " "contain at least one value." % (k + 1)) if d.ndim != 1: raise ValueError("Sample %d has %d dimensions. All " "samples must be one-dimensional sequences." % (k + 1, d.ndim)) cdata = np.concatenate(data) contains_nan, nan_policy = _contains_nan(cdata, nan_policy) if contains_nan and nan_policy == 'propagate': return MedianTestResult(np.nan, np.nan, np.nan, None) if contains_nan: grand_median = np.median(cdata[~np.isnan(cdata)]) else: grand_median = np.median(cdata) # When the minimum version of numpy supported by scipy is 1.9.0, # the above if/else statement can be replaced by the single line: # grand_median = np.nanmedian(cdata) # Create the contingency table. table = np.zeros((2, len(data)), dtype=np.int64) for k, sample in enumerate(data): sample = sample[~np.isnan(sample)] nabove = count_nonzero(sample > grand_median) nbelow = count_nonzero(sample < grand_median) nequal = sample.size - (nabove + nbelow) table[0, k] += nabove table[1, k] += nbelow if ties == "below": table[1, k] += nequal elif ties == "above": table[0, k] += nequal # Check that no row or column of the table is all zero. # Such a table can not be given to chi2_contingency, because it would have # a zero in the table of expected frequencies. rowsums = table.sum(axis=1) if rowsums[0] == 0: raise ValueError("All values are below the grand median (%r)." % grand_median) if rowsums[1] == 0: raise ValueError("All values are above the grand median (%r)." % grand_median) if ties == "ignore": # We already checked that each sample has at least one value, but it # is possible that all those values equal the grand median. If `ties` # is "ignore", that would result in a column of zeros in `table`. We # check for that case here. zero_cols = np.nonzero((table == 0).all(axis=0))[0] if len(zero_cols) > 0: msg = ("All values in sample %d are equal to the grand " "median (%r), so they are ignored, resulting in an " "empty sample." % (zero_cols[0] + 1, grand_median)) raise ValueError(msg) stat, p, dof, expected = chi2_contingency(table, lambda_=lambda_, correction=correction) return MedianTestResult(stat, p, grand_median, table) def _circfuncs_common(samples, high, low, nan_policy='propagate'): # Ensure samples are array-like and size is not zero samples = np.asarray(samples) if samples.size == 0: return np.nan, np.asarray(np.nan), np.asarray(np.nan), None # Recast samples as radians that range between 0 and 2 pi and calculate # the sine and cosine sin_samp = sin((samples - low)*2.*pi / (high - low)) cos_samp = cos((samples - low)*2.*pi / (high - low)) # Apply the NaN policy contains_nan, nan_policy = _contains_nan(samples, nan_policy) if contains_nan and nan_policy == 'omit': mask = np.isnan(samples) # Set the sines and cosines that are NaN to zero sin_samp[mask] = 0.0 cos_samp[mask] = 0.0 else: mask = None return samples, sin_samp, cos_samp, mask def circmean(samples, high=2*pi, low=0, axis=None, nan_policy='propagate'): """Compute the circular mean for samples in a range. Parameters ---------- samples : array_like Input array. high : float or int, optional High boundary for the sample range. Default is ``2*pi``. low : float or int, optional Low boundary for the sample range. Default is 0. axis : int, optional Axis along which means are computed. The default is to compute the mean of the flattened array. nan_policy : {'propagate', 'raise', 'omit'}, optional Defines how to handle when input contains nan. 'propagate' returns nan, 'raise' throws an error, 'omit' performs the calculations ignoring nan values. Default is 'propagate'. Returns ------- circmean : float Circular mean. See Also -------- circstd : Circular standard deviation. circvar : Circular variance. Examples -------- For simplicity, all angles are printed out in degrees. >>> import numpy as np >>> from scipy.stats import circmean >>> import matplotlib.pyplot as plt >>> angles = np.deg2rad(np.array([20, 30, 330])) >>> circmean = circmean(angles) >>> np.rad2deg(circmean) 7.294976657784009 >>> mean = angles.mean() >>> np.rad2deg(mean) 126.66666666666666 Plot and compare the circular mean against the arithmetic mean. >>> plt.plot(np.cos(np.linspace(0, 2*np.pi, 500)), ... np.sin(np.linspace(0, 2*np.pi, 500)), ... c='k') >>> plt.scatter(np.cos(angles), np.sin(angles), c='k') >>> plt.scatter(np.cos(circmean), np.sin(circmean), c='b', ... label='circmean') >>> plt.scatter(np.cos(mean), np.sin(mean), c='r', label='mean') >>> plt.legend() >>> plt.axis('equal') >>> plt.show() """ samples, sin_samp, cos_samp, nmask = _circfuncs_common(samples, high, low, nan_policy=nan_policy) sin_sum = sin_samp.sum(axis=axis) cos_sum = cos_samp.sum(axis=axis) res = arctan2(sin_sum, cos_sum) mask_nan = ~np.isnan(res) if mask_nan.ndim > 0: mask = res[mask_nan] < 0 else: mask = res < 0 if mask.ndim > 0: mask_nan[mask_nan] = mask res[mask_nan] += 2*pi elif mask: res += 2*pi # Set output to NaN if no samples went into the mean if nmask is not None: if nmask.all(): res = np.full(shape=res.shape, fill_value=np.nan) else: # Find out if any of the axis that are being averaged consist # entirely of NaN. If one exists, set the result (res) to NaN nshape = 0 if axis is None else axis smask = nmask.shape[nshape] == nmask.sum(axis=axis) if smask.any(): res[smask] = np.nan return res*(high - low)/2.0/pi + low def circvar(samples, high=2*pi, low=0, axis=None, nan_policy='propagate'): """Compute the circular variance for samples assumed to be in a range. Parameters ---------- samples : array_like Input array. high : float or int, optional High boundary for the sample range. Default is ``2*pi``. low : float or int, optional Low boundary for the sample range. Default is 0. axis : int, optional Axis along which variances are computed. The default is to compute the variance of the flattened array. nan_policy : {'propagate', 'raise', 'omit'}, optional Defines how to handle when input contains nan. 'propagate' returns nan, 'raise' throws an error, 'omit' performs the calculations ignoring nan values. Default is 'propagate'. Returns ------- circvar : float Circular variance. See Also -------- circmean : Circular mean. circstd : Circular standard deviation. Notes ----- This uses the following definition of circular variance: ``1-R``, where ``R`` is the mean resultant vector. The returned value is in the range [0, 1], 0 standing for no variance, and 1 for a large variance. In the limit of small angles, this value is similar to half the 'linear' variance. References ---------- .. [1] Fisher, N.I. *Statistical analysis of circular data*. Cambridge University Press, 1993. Examples -------- >>> import numpy as np >>> from scipy.stats import circvar >>> import matplotlib.pyplot as plt >>> samples_1 = np.array([0.072, -0.158, 0.077, 0.108, 0.286, ... 0.133, -0.473, -0.001, -0.348, 0.131]) >>> samples_2 = np.array([0.111, -0.879, 0.078, 0.733, 0.421, ... 0.104, -0.136, -0.867, 0.012, 0.105]) >>> circvar_1 = circvar(samples_1) >>> circvar_2 = circvar(samples_2) Plot the samples. >>> fig, (left, right) = plt.subplots(ncols=2) >>> for image in (left, right): ... image.plot(np.cos(np.linspace(0, 2*np.pi, 500)), ... np.sin(np.linspace(0, 2*np.pi, 500)), ... c='k') ... image.axis('equal') ... image.axis('off') >>> left.scatter(np.cos(samples_1), np.sin(samples_1), c='k', s=15) >>> left.set_title(f"circular variance: {np.round(circvar_1, 2)!r}") >>> right.scatter(np.cos(samples_2), np.sin(samples_2), c='k', s=15) >>> right.set_title(f"circular variance: {np.round(circvar_2, 2)!r}") >>> plt.show() """ samples, sin_samp, cos_samp, mask = _circfuncs_common(samples, high, low, nan_policy=nan_policy) if mask is None: sin_mean = sin_samp.mean(axis=axis) cos_mean = cos_samp.mean(axis=axis) else: nsum = np.asarray(np.sum(~mask, axis=axis).astype(float)) nsum[nsum == 0] = np.nan sin_mean = sin_samp.sum(axis=axis) / nsum cos_mean = cos_samp.sum(axis=axis) / nsum # hypot can go slightly above 1 due to rounding errors with np.errstate(invalid='ignore'): R = np.minimum(1, hypot(sin_mean, cos_mean)) res = 1. - R return res def circstd(samples, high=2*pi, low=0, axis=None, nan_policy='propagate', *, normalize=False): """ Compute the circular standard deviation for samples assumed to be in the range [low to high]. Parameters ---------- samples : array_like Input array. high : float or int, optional High boundary for the sample range. Default is ``2*pi``. low : float or int, optional Low boundary for the sample range. Default is 0. axis : int, optional Axis along which standard deviations are computed. The default is to compute the standard deviation of the flattened array. nan_policy : {'propagate', 'raise', 'omit'}, optional Defines how to handle when input contains nan. 'propagate' returns nan, 'raise' throws an error, 'omit' performs the calculations ignoring nan values. Default is 'propagate'. normalize : boolean, optional If True, the returned value is equal to ``sqrt(-2*log(R))`` and does not depend on the variable units. If False (default), the returned value is scaled by ``((high-low)/(2*pi))``. Returns ------- circstd : float Circular standard deviation. See Also -------- circmean : Circular mean. circvar : Circular variance. Notes ----- This uses a definition of circular standard deviation from [1]_. Essentially, the calculation is as follows. .. code-block:: python import numpy as np C = np.cos(samples).mean() S = np.sin(samples).mean() R = np.sqrt(C**2 + S**2) l = 2*np.pi / (high-low) circstd = np.sqrt(-2*np.log(R)) / l In the limit of small angles, it returns a number close to the 'linear' standard deviation. References ---------- .. [1] Mardia, K. V. (1972). 2. In *Statistics of Directional Data* (pp. 18-24). Academic Press. :doi:`10.1016/C2013-0-07425-7`. Examples -------- >>> import numpy as np >>> from scipy.stats import circstd >>> import matplotlib.pyplot as plt >>> samples_1 = np.array([0.072, -0.158, 0.077, 0.108, 0.286, ... 0.133, -0.473, -0.001, -0.348, 0.131]) >>> samples_2 = np.array([0.111, -0.879, 0.078, 0.733, 0.421, ... 0.104, -0.136, -0.867, 0.012, 0.105]) >>> circstd_1 = circstd(samples_1) >>> circstd_2 = circstd(samples_2) Plot the samples. >>> fig, (left, right) = plt.subplots(ncols=2) >>> for image in (left, right): ... image.plot(np.cos(np.linspace(0, 2*np.pi, 500)), ... np.sin(np.linspace(0, 2*np.pi, 500)), ... c='k') ... image.axis('equal') ... image.axis('off') >>> left.scatter(np.cos(samples_1), np.sin(samples_1), c='k', s=15) >>> left.set_title(f"circular std: {np.round(circstd_1, 2)!r}") >>> right.plot(np.cos(np.linspace(0, 2*np.pi, 500)), ... np.sin(np.linspace(0, 2*np.pi, 500)), ... c='k') >>> right.scatter(np.cos(samples_2), np.sin(samples_2), c='k', s=15) >>> right.set_title(f"circular std: {np.round(circstd_2, 2)!r}") >>> plt.show() """ samples, sin_samp, cos_samp, mask = _circfuncs_common(samples, high, low, nan_policy=nan_policy) if mask is None: sin_mean = sin_samp.mean(axis=axis) # [1] (2.2.3) cos_mean = cos_samp.mean(axis=axis) # [1] (2.2.3) else: nsum = np.asarray(np.sum(~mask, axis=axis).astype(float)) nsum[nsum == 0] = np.nan sin_mean = sin_samp.sum(axis=axis) / nsum cos_mean = cos_samp.sum(axis=axis) / nsum # hypot can go slightly above 1 due to rounding errors with np.errstate(invalid='ignore'): R = np.minimum(1, hypot(sin_mean, cos_mean)) # [1] (2.2.4) res = sqrt(-2*log(R)) if not normalize: res *= (high-low)/(2.*pi) # [1] (2.3.14) w/ (2.3.7) return res class DirectionalStats: def __init__(self, mean_direction, mean_resultant_length): self.mean_direction = mean_direction self.mean_resultant_length = mean_resultant_length def __repr__(self): return (f"DirectionalStats(mean_direction={self.mean_direction}," f" mean_resultant_length={self.mean_resultant_length})") def directional_stats(samples, *, axis=0, normalize=True): """ Computes sample statistics for directional data. Computes the directional mean (also called the mean direction vector) and mean resultant length of a sample of vectors. The directional mean is a measure of "preferred direction" of vector data. It is analogous to the sample mean, but it is for use when the length of the data is irrelevant (e.g. unit vectors). The mean resultant length is a value between 0 and 1 used to quantify the dispersion of directional data: the smaller the mean resultant length, the greater the dispersion. Several definitions of directional variance involving the mean resultant length are given in [1]_ and [2]_. Parameters ---------- samples : array_like Input array. Must be at least two-dimensional, and the last axis of the input must correspond with the dimensionality of the vector space. When the input is exactly two dimensional, this means that each row of the data is a vector observation. axis : int, default: 0 Axis along which the directional mean is computed. normalize: boolean, default: True If True, normalize the input to ensure that each observation is a unit vector. It the observations are already unit vectors, consider setting this to False to avoid unnecessary computation. Returns ------- res : DirectionalStats An object containing attributes: mean_direction : ndarray Directional mean. mean_resultant_length : ndarray The mean resultant length [1]_. See Also -------- circmean: circular mean; i.e. directional mean for 2D *angles* circvar: circular variance; i.e. directional variance for 2D *angles* Notes ----- This uses a definition of directional mean from [1]_. Assuming the observations are unit vectors, the calculation is as follows. .. code-block:: python mean = samples.mean(axis=0) mean_resultant_length = np.linalg.norm(mean) mean_direction = mean / mean_resultant_length This definition is appropriate for *directional* data (i.e. vector data for which the magnitude of each observation is irrelevant) but not for *axial* data (i.e. vector data for which the magnitude and *sign* of each observation is irrelevant). Several definitions of directional variance involving the mean resultant length ``R`` have been proposed, including ``1 - R`` [1]_, ``1 - R**2`` [2]_, and ``2 * (1 - R)`` [2]_. Rather than choosing one, this function returns ``R`` as attribute `mean_resultant_length` so the user can compute their preferred measure of dispersion. References ---------- .. [1] Mardia, Jupp. (2000). *Directional Statistics* (p. 163). Wiley. .. [2] https://en.wikipedia.org/wiki/Directional_statistics Examples -------- >>> import numpy as np >>> from scipy.stats import directional_stats >>> data = np.array([[3, 4], # first observation, 2D vector space ... [6, -8]]) # second observation >>> dirstats = directional_stats(data) >>> dirstats.mean_direction array([1., 0.]) In contrast, the regular sample mean of the vectors would be influenced by the magnitude of each observation. Furthermore, the result would not be a unit vector. >>> data.mean(axis=0) array([4.5, -2.]) An exemplary use case for `directional_stats` is to find a *meaningful* center for a set of observations on a sphere, e.g. geographical locations. >>> data = np.array([[0.8660254, 0.5, 0.], ... [0.8660254, -0.5, 0.]]) >>> dirstats = directional_stats(data) >>> dirstats.mean_direction array([1., 0., 0.]) The regular sample mean on the other hand yields a result which does not lie on the surface of the sphere. >>> data.mean(axis=0) array([0.8660254, 0., 0.]) The function also returns the mean resultant length, which can be used to calculate a directional variance. For example, using the definition ``Var(z) = 1 - R`` from [2]_ where ``R`` is the mean resultant length, we can calculate the directional variance of the vectors in the above example as: >>> 1 - dirstats.mean_resultant_length 0.13397459716167093 """ samples = np.asarray(samples) if samples.ndim < 2: raise ValueError("samples must at least be two-dimensional. " f"Instead samples has shape: {samples.shape!r}") samples = np.moveaxis(samples, axis, 0) if normalize: vectornorms = np.linalg.norm(samples, axis=-1, keepdims=True) samples = samples/vectornorms mean = np.mean(samples, axis=0) mean_resultant_length = np.linalg.norm(mean, axis=-1, keepdims=True) mean_direction = mean / mean_resultant_length return DirectionalStats(mean_direction, mean_resultant_length.squeeze(-1)[()]) def false_discovery_control(ps, *, axis=0, method='bh'): """Adjust p-values to control the false discovery rate. The false discovery rate (FDR) is the expected proportion of rejected null hypotheses that are actually true. If the null hypothesis is rejected when the *adjusted* p-value falls below a specified level, the false discovery rate is controlled at that level. Parameters ---------- ps : 1D array_like The p-values to adjust. Elements must be real numbers between 0 and 1. axis : int The axis along which to perform the adjustment. The adjustment is performed independently along each axis-slice. If `axis` is None, `ps` is raveled before performing the adjustment. method : {'bh', 'by'} The false discovery rate control procedure to apply: ``'bh'`` is for Benjamini-Hochberg [1]_ (Eq. 1), ``'by'`` is for Benjaminini-Yekutieli [2]_ (Theorem 1.3). The latter is more conservative, but it is guaranteed to control the FDR even when the p-values are not from independent tests. Returns ------- ps_adusted : array_like The adjusted p-values. If the null hypothesis is rejected where these fall below a specified level, the false discovery rate is controlled at that level. See Also -------- combine_pvalues statsmodels.stats.multitest.multipletests Notes ----- In multiple hypothesis testing, false discovery control procedures tend to offer higher power than familywise error rate control procedures (e.g. Bonferroni correction [1]_). If the p-values correspond with independent tests (or tests with "positive regression dependencies" [2]_), rejecting null hypotheses corresponding with Benjamini-Hochberg-adjusted p-values below :math:`q` controls the false discovery rate at a level less than or equal to :math:`q m_0 / m`, where :math:`m_0` is the number of true null hypotheses and :math:`m` is the total number of null hypotheses tested. The same is true even for dependent tests when the p-values are adjusted accorded to the more conservative Benjaminini-Yekutieli procedure. The adjusted p-values produced by this function are comparable to those produced by the R function ``p.adjust`` and the statsmodels function `statsmodels.stats.multitest.multipletests`. Please consider the latter for more advanced methods of multiple comparison correction. References ---------- .. [1] Benjamini, Yoav, and Yosef Hochberg. "Controlling the false discovery rate: a practical and powerful approach to multiple testing." Journal of the Royal statistical society: series B (Methodological) 57.1 (1995): 289-300. .. [2] Benjamini, Yoav, and Daniel Yekutieli. "The control of the false discovery rate in multiple testing under dependency." Annals of statistics (2001): 1165-1188. .. [3] TileStats. FDR - Benjamini-Hochberg explained - Youtube. https://www.youtube.com/watch?v=rZKa4tW2NKs. .. [4] Neuhaus, Karl-Ludwig, et al. "Improved thrombolysis in acute myocardial infarction with front-loaded administration of alteplase: results of the rt-PA-APSAC patency study (TAPS)." Journal of the American College of Cardiology 19.5 (1992): 885-891. Examples -------- We follow the example from [1]_. Thrombolysis with recombinant tissue-type plasminogen activator (rt-PA) and anisoylated plasminogen streptokinase activator (APSAC) in myocardial infarction has been proved to reduce mortality. [4]_ investigated the effects of a new front-loaded administration of rt-PA versus those obtained with a standard regimen of APSAC, in a randomized multicentre trial in 421 patients with acute myocardial infarction. There were four families of hypotheses tested in the study, the last of which was "cardiac and other events after the start of thrombolitic treatment". FDR control may be desired in this family of hypotheses because it would not be appropriate to conclude that the front-loaded treatment is better if it is merely equivalent to the previous treatment. The p-values corresponding with the 15 hypotheses in this family were >>> ps = [0.0001, 0.0004, 0.0019, 0.0095, 0.0201, 0.0278, 0.0298, 0.0344, ... 0.0459, 0.3240, 0.4262, 0.5719, 0.6528, 0.7590, 1.000] If the chosen significance level is 0.05, we may be tempted to reject the null hypotheses for the tests corresponding with the first nine p-values, as the first nine p-values fall below the chosen significance level. However, this would ignore the problem of "multiplicity": if we fail to correct for the fact that multiple comparisons are being performed, we are more likely to incorrectly reject true null hypotheses. One approach to the multiplicity problem is to control the family-wise error rate (FWER), that is, the rate at which the null hypothesis is rejected when it is actually true. A common procedure of this kind is the Bonferroni correction [1]_. We begin by multiplying the p-values by the number of hypotheses tested. >>> import numpy as np >>> np.array(ps) * len(ps) array([1.5000e-03, 6.0000e-03, 2.8500e-02, 1.4250e-01, 3.0150e-01, 4.1700e-01, 4.4700e-01, 5.1600e-01, 6.8850e-01, 4.8600e+00, 6.3930e+00, 8.5785e+00, 9.7920e+00, 1.1385e+01, 1.5000e+01]) To control the FWER at 5%, we reject only the hypotheses corresponding with adjusted p-values less than 0.05. In this case, only the hypotheses corresponding with the first three p-values can be rejected. According to [1]_, these three hypotheses concerned "allergic reaction" and "two different aspects of bleeding." An alternative approach is to control the false discovery rate: the expected fraction of rejected null hypotheses that are actually true. The advantage of this approach is that it typically affords greater power: an increased rate of rejecting the null hypothesis when it is indeed false. To control the false discovery rate at 5%, we apply the Benjamini-Hochberg p-value adjustment. >>> from scipy import stats >>> stats.false_discovery_control(ps) array([0.0015 , 0.003 , 0.0095 , 0.035625 , 0.0603 , 0.06385714, 0.06385714, 0.0645 , 0.0765 , 0.486 , 0.58118182, 0.714875 , 0.75323077, 0.81321429, 1. ]) Now, the first *four* adjusted p-values fall below 0.05, so we would reject the null hypotheses corresponding with these *four* p-values. Rejection of the fourth null hypothesis was particularly important to the original study as it led to the conclusion that the new treatment had a "substantially lower in-hospital mortality rate." """ # Input Validation and Special Cases ps = np.asarray(ps) ps_in_range = (np.issubdtype(ps.dtype, np.number) and np.all(ps == np.clip(ps, 0, 1))) if not ps_in_range: raise ValueError("`ps` must include only numbers between 0 and 1.") methods = {'bh', 'by'} if method.lower() not in methods: raise ValueError(f"Unrecognized `method` '{method}'." f"Method must be one of {methods}.") method = method.lower() if axis is None: axis = 0 ps = ps.ravel() axis = np.asarray(axis)[()] if not np.issubdtype(axis.dtype, np.integer) or axis.size != 1: raise ValueError("`axis` must be an integer or `None`") if ps.size <= 1 or ps.shape[axis] <= 1: return ps[()] ps = np.moveaxis(ps, axis, -1) m = ps.shape[-1] # Main Algorithm # Equivalent to the ideas of [1] and [2], except that this adjusts the # p-values as described in [3]. The results are similar to those produced # by R's p.adjust. # "Let [ps] be the ordered observed p-values..." order = np.argsort(ps, axis=-1) ps = np.take_along_axis(ps, order, axis=-1) # this copies ps # Equation 1 of [1] rearranged to reject when p is less than specified q i = np.arange(1, m+1) ps *= m / i # Theorem 1.3 of [2] if method == 'by': ps *= np.sum(1 / i) # accounts for rejecting all null hypotheses i for i < k, where k is # defined in Eq. 1 of either [1] or [2]. See [3]. Starting with the index j # of the second to last element, we replace element j with element j+1 if # the latter is smaller. np.minimum.accumulate(ps[..., ::-1], out=ps[..., ::-1], axis=-1) # Restore original order of axes and data np.put_along_axis(ps, order, values=ps.copy(), axis=-1) ps = np.moveaxis(ps, -1, axis) return np.clip(ps, 0, 1)