495 lines
19 KiB
Python
495 lines
19 KiB
Python
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import numpy as np
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from collections import namedtuple
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from scipy import special
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from scipy import stats
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from scipy.stats._stats_py import _rankdata
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from ._axis_nan_policy import _axis_nan_policy_factory
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def _broadcast_concatenate(x, y, axis):
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'''Broadcast then concatenate arrays, leaving concatenation axis last'''
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x = np.moveaxis(x, axis, -1)
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y = np.moveaxis(y, axis, -1)
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z = np.broadcast(x[..., 0], y[..., 0])
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x = np.broadcast_to(x, z.shape + (x.shape[-1],))
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y = np.broadcast_to(y, z.shape + (y.shape[-1],))
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z = np.concatenate((x, y), axis=-1)
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return x, y, z
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class _MWU:
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'''Distribution of MWU statistic under the null hypothesis'''
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def __init__(self, n1, n2):
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self._reset(n1, n2)
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def set_shapes(self, n1, n2):
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n1, n2 = min(n1, n2), max(n1, n2)
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if (n1, n2) == (self.n1, self.n2):
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return
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self.n1 = n1
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self.n2 = n2
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self.s_array = np.zeros(0, dtype=int)
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self.configurations = np.zeros(0, dtype=np.uint64)
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def reset(self):
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self._reset(self.n1, self.n2)
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def _reset(self, n1, n2):
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self.n1 = None
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self.n2 = None
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self.set_shapes(n1, n2)
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def pmf(self, k):
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# In practice, `pmf` is never called with k > m*n/2.
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# If it were, we'd exploit symmetry here:
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# k = np.array(k, copy=True)
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# k2 = m*n - k
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# i = k2 < k
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# k[i] = k2[i]
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pmfs = self.build_u_freqs_array(np.max(k))
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return pmfs[k]
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def cdf(self, k):
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'''Cumulative distribution function'''
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# In practice, `cdf` is never called with k > m*n/2.
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# If it were, we'd exploit symmetry here rather than in `sf`
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pmfs = self.build_u_freqs_array(np.max(k))
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cdfs = np.cumsum(pmfs)
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return cdfs[k]
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def sf(self, k):
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'''Survival function'''
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# Note that both CDF and SF include the PMF at k. The p-value is
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# calculated from the SF and should include the mass at k, so this
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# is desirable
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# Use the fact that the distribution is symmetric and sum from the left
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kc = np.asarray(self.n1*self.n2 - k) # complement of k
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i = k < kc
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if np.any(i):
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kc[i] = k[i]
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cdfs = np.asarray(self.cdf(kc))
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cdfs[i] = 1. - cdfs[i] + self.pmf(kc[i])
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else:
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cdfs = np.asarray(self.cdf(kc))
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return cdfs[()]
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# build_sigma_array and build_u_freqs_array adapted from code
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# by @toobaz with permission. Thanks to @andreasloe for the suggestion.
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# See https://github.com/scipy/scipy/pull/4933#issuecomment-1898082691
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def build_sigma_array(self, a):
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n1, n2 = self.n1, self.n2
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if a + 1 <= self.s_array.size:
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return self.s_array[1:a+1]
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s_array = np.zeros(a + 1, dtype=int)
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for d in np.arange(1, n1 + 1):
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# All multiples of d, except 0:
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indices = np.arange(d, a + 1, d)
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# \epsilon_d = 1:
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s_array[indices] += d
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for d in np.arange(n2 + 1, n2 + n1 + 1):
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# All multiples of d, except 0:
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indices = np.arange(d, a + 1, d)
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# \epsilon_d = -1:
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s_array[indices] -= d
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# We don't need 0:
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self.s_array = s_array
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return s_array[1:]
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def build_u_freqs_array(self, maxu):
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"""
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Build all the array of frequencies for u from 0 to maxu.
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Assumptions:
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n1 <= n2
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maxu <= n1 * n2 / 2
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"""
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n1, n2 = self.n1, self.n2
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total = special.binom(n1 + n2, n1)
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if maxu + 1 <= self.configurations.size:
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return self.configurations[:maxu + 1] / total
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s_array = self.build_sigma_array(maxu)
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# Start working with ints, for maximum precision and efficiency:
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configurations = np.zeros(maxu + 1, dtype=np.uint64)
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configurations_is_uint = True
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uint_max = np.iinfo(np.uint64).max
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# How many ways to have U=0? 1
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configurations[0] = 1
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for u in np.arange(1, maxu + 1):
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coeffs = s_array[u - 1::-1]
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new_val = np.dot(configurations[:u], coeffs) / u
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if new_val > uint_max and configurations_is_uint:
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# OK, we got into numbers too big for uint64.
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# So now we start working with floats.
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# By doing this since the beginning, we would have lost precision.
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# (And working on python long ints would be unbearably slow)
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configurations = configurations.astype(float)
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configurations_is_uint = False
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configurations[u] = new_val
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self.configurations = configurations
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return configurations / total
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_mwu_state = _MWU(0, 0)
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def _get_mwu_z(U, n1, n2, t, axis=0, continuity=True):
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'''Standardized MWU statistic'''
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# Follows mannwhitneyu [2]
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mu = n1 * n2 / 2
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n = n1 + n2
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# Tie correction according to [2], "Normal approximation and tie correction"
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# "A more computationally-efficient form..."
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tie_term = (t**3 - t).sum(axis=-1)
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s = np.sqrt(n1*n2/12 * ((n + 1) - tie_term/(n*(n-1))))
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numerator = U - mu
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# Continuity correction.
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# Because SF is always used to calculate the p-value, we can always
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# _subtract_ 0.5 for the continuity correction. This always increases the
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# p-value to account for the rest of the probability mass _at_ q = U.
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if continuity:
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numerator -= 0.5
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# no problem evaluating the norm SF at an infinity
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with np.errstate(divide='ignore', invalid='ignore'):
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z = numerator / s
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return z
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def _mwu_input_validation(x, y, use_continuity, alternative, axis, method):
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''' Input validation and standardization for mannwhitneyu '''
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# Would use np.asarray_chkfinite, but infs are OK
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x, y = np.atleast_1d(x), np.atleast_1d(y)
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if np.isnan(x).any() or np.isnan(y).any():
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raise ValueError('`x` and `y` must not contain NaNs.')
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if np.size(x) == 0 or np.size(y) == 0:
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raise ValueError('`x` and `y` must be of nonzero size.')
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bools = {True, False}
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if use_continuity not in bools:
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raise ValueError(f'`use_continuity` must be one of {bools}.')
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alternatives = {"two-sided", "less", "greater"}
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alternative = alternative.lower()
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if alternative not in alternatives:
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raise ValueError(f'`alternative` must be one of {alternatives}.')
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axis_int = int(axis)
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if axis != axis_int:
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raise ValueError('`axis` must be an integer.')
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if not isinstance(method, stats.PermutationMethod):
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methods = {"asymptotic", "exact", "auto"}
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method = method.lower()
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if method not in methods:
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raise ValueError(f'`method` must be one of {methods}.')
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return x, y, use_continuity, alternative, axis_int, method
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def _mwu_choose_method(n1, n2, ties):
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"""Choose method 'asymptotic' or 'exact' depending on input size, ties"""
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# if both inputs are large, asymptotic is OK
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if n1 > 8 and n2 > 8:
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return "asymptotic"
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# if there are any ties, asymptotic is preferred
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if ties:
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return "asymptotic"
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return "exact"
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MannwhitneyuResult = namedtuple('MannwhitneyuResult', ('statistic', 'pvalue'))
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@_axis_nan_policy_factory(MannwhitneyuResult, n_samples=2)
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def mannwhitneyu(x, y, use_continuity=True, alternative="two-sided",
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axis=0, method="auto"):
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r'''Perform the Mann-Whitney U rank test on two independent samples.
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The Mann-Whitney U test is a nonparametric test of the null hypothesis
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that the distribution underlying sample `x` is the same as the
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distribution underlying sample `y`. It is often used as a test of
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difference in location between distributions.
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Parameters
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----------
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x, y : array-like
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N-d arrays of samples. The arrays must be broadcastable except along
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the dimension given by `axis`.
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use_continuity : bool, optional
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Whether a continuity correction (1/2) should be applied.
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Default is True when `method` is ``'asymptotic'``; has no effect
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otherwise.
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alternative : {'two-sided', 'less', 'greater'}, optional
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Defines the alternative hypothesis. Default is 'two-sided'.
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Let *F(u)* and *G(u)* be the cumulative distribution functions of the
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distributions underlying `x` and `y`, respectively. Then the following
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alternative hypotheses are available:
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* 'two-sided': the distributions are not equal, i.e. *F(u) ≠ G(u)* for
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at least one *u*.
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* 'less': the distribution underlying `x` is stochastically less
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than the distribution underlying `y`, i.e. *F(u) > G(u)* for all *u*.
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* 'greater': the distribution underlying `x` is stochastically greater
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than the distribution underlying `y`, i.e. *F(u) < G(u)* for all *u*.
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Note that the mathematical expressions in the alternative hypotheses
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above describe the CDFs of the underlying distributions. The directions
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of the inequalities appear inconsistent with the natural language
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description at first glance, but they are not. For example, suppose
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*X* and *Y* are random variables that follow distributions with CDFs
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*F* and *G*, respectively. If *F(u) > G(u)* for all *u*, samples drawn
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from *X* tend to be less than those drawn from *Y*.
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Under a more restrictive set of assumptions, the alternative hypotheses
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can be expressed in terms of the locations of the distributions;
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see [5] section 5.1.
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axis : int, optional
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Axis along which to perform the test. Default is 0.
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method : {'auto', 'asymptotic', 'exact'} or `PermutationMethod` instance, optional
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Selects the method used to calculate the *p*-value.
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Default is 'auto'. The following options are available.
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* ``'asymptotic'``: compares the standardized test statistic
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against the normal distribution, correcting for ties.
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* ``'exact'``: computes the exact *p*-value by comparing the observed
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:math:`U` statistic against the exact distribution of the :math:`U`
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statistic under the null hypothesis. No correction is made for ties.
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* ``'auto'``: chooses ``'exact'`` when the size of one of the samples
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is less than or equal to 8 and there are no ties;
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chooses ``'asymptotic'`` otherwise.
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* `PermutationMethod` instance. In this case, the p-value
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is computed using `permutation_test` with the provided
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configuration options and other appropriate settings.
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Returns
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-------
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res : MannwhitneyuResult
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An object containing attributes:
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statistic : float
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The Mann-Whitney U statistic corresponding with sample `x`. See
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Notes for the test statistic corresponding with sample `y`.
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pvalue : float
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The associated *p*-value for the chosen `alternative`.
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Notes
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-----
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If ``U1`` is the statistic corresponding with sample `x`, then the
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statistic corresponding with sample `y` is
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``U2 = x.shape[axis] * y.shape[axis] - U1``.
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`mannwhitneyu` is for independent samples. For related / paired samples,
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consider `scipy.stats.wilcoxon`.
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`method` ``'exact'`` is recommended when there are no ties and when either
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sample size is less than 8 [1]_. The implementation follows the algorithm
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reported in [3]_.
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Note that the exact method is *not* corrected for ties, but
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`mannwhitneyu` will not raise errors or warnings if there are ties in the
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data. If there are ties and either samples is small (fewer than ~10
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observations), consider passing an instance of `PermutationMethod`
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as the `method` to perform a permutation test.
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The Mann-Whitney U test is a non-parametric version of the t-test for
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independent samples. When the means of samples from the populations
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are normally distributed, consider `scipy.stats.ttest_ind`.
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See Also
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--------
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scipy.stats.wilcoxon, scipy.stats.ranksums, scipy.stats.ttest_ind
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References
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----------
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.. [1] H.B. Mann and D.R. Whitney, "On a test of whether one of two random
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variables is stochastically larger than the other", The Annals of
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Mathematical Statistics, Vol. 18, pp. 50-60, 1947.
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.. [2] Mann-Whitney U Test, Wikipedia,
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http://en.wikipedia.org/wiki/Mann-Whitney_U_test
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.. [3] Andreas Löffler,
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"Über eine Partition der nat. Zahlen und ihr Anwendung beim U-Test",
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Wiss. Z. Univ. Halle, XXXII'83 pp. 87-89.
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.. [4] Rosie Shier, "Statistics: 2.3 The Mann-Whitney U Test", Mathematics
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Learning Support Centre, 2004.
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.. [5] Michael P. Fay and Michael A. Proschan. "Wilcoxon-Mann-Whitney
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or t-test? On assumptions for hypothesis tests and multiple \
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interpretations of decision rules." Statistics surveys, Vol. 4, pp.
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1-39, 2010. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2857732/
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Examples
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--------
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We follow the example from [4]_: nine randomly sampled young adults were
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diagnosed with type II diabetes at the ages below.
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>>> males = [19, 22, 16, 29, 24]
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>>> females = [20, 11, 17, 12]
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We use the Mann-Whitney U test to assess whether there is a statistically
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significant difference in the diagnosis age of males and females.
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The null hypothesis is that the distribution of male diagnosis ages is
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the same as the distribution of female diagnosis ages. We decide
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that a confidence level of 95% is required to reject the null hypothesis
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in favor of the alternative that the distributions are different.
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Since the number of samples is very small and there are no ties in the
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data, we can compare the observed test statistic against the *exact*
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distribution of the test statistic under the null hypothesis.
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>>> from scipy.stats import mannwhitneyu
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>>> U1, p = mannwhitneyu(males, females, method="exact")
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>>> print(U1)
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17.0
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`mannwhitneyu` always reports the statistic associated with the first
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sample, which, in this case, is males. This agrees with :math:`U_M = 17`
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reported in [4]_. The statistic associated with the second statistic
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can be calculated:
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>>> nx, ny = len(males), len(females)
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>>> U2 = nx*ny - U1
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>>> print(U2)
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3.0
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This agrees with :math:`U_F = 3` reported in [4]_. The two-sided
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*p*-value can be calculated from either statistic, and the value produced
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by `mannwhitneyu` agrees with :math:`p = 0.11` reported in [4]_.
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>>> print(p)
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0.1111111111111111
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The exact distribution of the test statistic is asymptotically normal, so
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the example continues by comparing the exact *p*-value against the
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*p*-value produced using the normal approximation.
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>>> _, pnorm = mannwhitneyu(males, females, method="asymptotic")
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>>> print(pnorm)
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0.11134688653314041
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Here `mannwhitneyu`'s reported *p*-value appears to conflict with the
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value :math:`p = 0.09` given in [4]_. The reason is that [4]_
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does not apply the continuity correction performed by `mannwhitneyu`;
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`mannwhitneyu` reduces the distance between the test statistic and the
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mean :math:`\mu = n_x n_y / 2` by 0.5 to correct for the fact that the
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discrete statistic is being compared against a continuous distribution.
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Here, the :math:`U` statistic used is less than the mean, so we reduce
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the distance by adding 0.5 in the numerator.
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>>> import numpy as np
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>>> from scipy.stats import norm
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>>> U = min(U1, U2)
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>>> N = nx + ny
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>>> z = (U - nx*ny/2 + 0.5) / np.sqrt(nx*ny * (N + 1)/ 12)
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>>> p = 2 * norm.cdf(z) # use CDF to get p-value from smaller statistic
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>>> print(p)
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0.11134688653314041
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If desired, we can disable the continuity correction to get a result
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that agrees with that reported in [4]_.
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>>> _, pnorm = mannwhitneyu(males, females, use_continuity=False,
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... method="asymptotic")
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>>> print(pnorm)
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0.0864107329737
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Regardless of whether we perform an exact or asymptotic test, the
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probability of the test statistic being as extreme or more extreme by
|
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chance exceeds 5%, so we do not consider the results statistically
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significant.
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Suppose that, before seeing the data, we had hypothesized that females
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would tend to be diagnosed at a younger age than males.
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In that case, it would be natural to provide the female ages as the
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first input, and we would have performed a one-sided test using
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``alternative = 'less'``: females are diagnosed at an age that is
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stochastically less than that of males.
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>>> res = mannwhitneyu(females, males, alternative="less", method="exact")
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>>> print(res)
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|
MannwhitneyuResult(statistic=3.0, pvalue=0.05555555555555555)
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Again, the probability of getting a sufficiently low value of the
|
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|
test statistic by chance under the null hypothesis is greater than 5%,
|
||
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so we do not reject the null hypothesis in favor of our alternative.
|
||
|
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|
||
|
|
If it is reasonable to assume that the means of samples from the
|
||
|
|
populations are normally distributed, we could have used a t-test to
|
||
|
|
perform the analysis.
|
||
|
|
|
||
|
|
>>> from scipy.stats import ttest_ind
|
||
|
|
>>> res = ttest_ind(females, males, alternative="less")
|
||
|
|
>>> print(res)
|
||
|
|
TtestResult(statistic=-2.239334696520584,
|
||
|
|
pvalue=0.030068441095757924,
|
||
|
|
df=7.0)
|
||
|
|
|
||
|
|
Under this assumption, the *p*-value would be low enough to reject the
|
||
|
|
null hypothesis in favor of the alternative.
|
||
|
|
|
||
|
|
'''
|
||
|
|
|
||
|
|
x, y, use_continuity, alternative, axis_int, method = (
|
||
|
|
_mwu_input_validation(x, y, use_continuity, alternative, axis, method))
|
||
|
|
|
||
|
|
x, y, xy = _broadcast_concatenate(x, y, axis)
|
||
|
|
|
||
|
|
n1, n2 = x.shape[-1], y.shape[-1]
|
||
|
|
|
||
|
|
# Follows [2]
|
||
|
|
ranks, t = _rankdata(xy, 'average', return_ties=True) # method 2, step 1
|
||
|
|
R1 = ranks[..., :n1].sum(axis=-1) # method 2, step 2
|
||
|
|
U1 = R1 - n1*(n1+1)/2 # method 2, step 3
|
||
|
|
U2 = n1 * n2 - U1 # as U1 + U2 = n1 * n2
|
||
|
|
|
||
|
|
if alternative == "greater":
|
||
|
|
U, f = U1, 1 # U is the statistic to use for p-value, f is a factor
|
||
|
|
elif alternative == "less":
|
||
|
|
U, f = U2, 1 # Due to symmetry, use SF of U2 rather than CDF of U1
|
||
|
|
else:
|
||
|
|
U, f = np.maximum(U1, U2), 2 # multiply SF by two for two-sided test
|
||
|
|
|
||
|
|
if method == "auto":
|
||
|
|
method = _mwu_choose_method(n1, n2, np.any(t > 1))
|
||
|
|
|
||
|
|
if method == "exact":
|
||
|
|
_mwu_state.set_shapes(n1, n2)
|
||
|
|
p = _mwu_state.sf(U.astype(int))
|
||
|
|
elif method == "asymptotic":
|
||
|
|
z = _get_mwu_z(U, n1, n2, t, continuity=use_continuity)
|
||
|
|
p = stats.norm.sf(z)
|
||
|
|
else: # `PermutationMethod` instance (already validated)
|
||
|
|
def statistic(x, y, axis):
|
||
|
|
return mannwhitneyu(x, y, use_continuity=use_continuity,
|
||
|
|
alternative=alternative, axis=axis,
|
||
|
|
method="asymptotic").statistic
|
||
|
|
|
||
|
|
res = stats.permutation_test((x, y), statistic, axis=axis,
|
||
|
|
**method._asdict(), alternative=alternative)
|
||
|
|
p = res.pvalue
|
||
|
|
f = 1
|
||
|
|
|
||
|
|
p *= f
|
||
|
|
|
||
|
|
# Ensure that test statistic is not greater than 1
|
||
|
|
# This could happen for exact test when U = m*n/2
|
||
|
|
p = np.clip(p, 0, 1)
|
||
|
|
|
||
|
|
return MannwhitneyuResult(U1, p)
|