Source code for statsmodels.sandbox.stats.runs

'''runstest

formulas for mean and var of runs taken from SAS manual NPAR tests, also idea
for runstest_1samp and runstest_2samp

Description in NIST handbook and dataplot doesn't explain their expected
values, or variance

Note:
There are (at least) two definitions of runs used in literature. The classical
definition which is also used here, is that runs are sequences of identical
observations separated by observations with different realizations.
The second definition allows for overlapping runs, or runs where counting a
run is also started after a run of a fixed length of the same kind.


TODO
* add one-sided tests where possible or where it makes sense

'''

from __future__ import print_function
import numpy as np
from scipy import stats
from scipy.special import comb
import warnings

[docs]class Runs(object): '''class for runs in a binary sequence Parameters ---------- x : array_like, 1d data array, Notes ----- This was written as a more general class for runs. This has some redundant calculations when only the runs_test is used. TODO: make it lazy The runs test could be generalized to more than 1d if there is a use case for it. This should be extended once I figure out what the distribution of runs of any length k is. The exact distribution for the runs test is also available but not yet verified. ''' def __init__(self, x): self.x = np.asarray(x) self.runstart = runstart = np.nonzero(np.diff(np.r_[[-np.inf], x, [np.inf]]))[0] self.runs = runs = np.diff(runstart) self.runs_sign = runs_sign = x[runstart[:-1]] self.runs_pos = runs[runs_sign==1] self.runs_neg = runs[runs_sign==0] self.runs_freqs = np.bincount(runs) self.n_runs = len(self.runs) self.n_pos = (x==1).sum()
[docs] def runs_test(self, correction=True): '''basic version of runs test Parameters ---------- correction: bool Following the SAS manual, for samplesize below 50, the test statistic is corrected by 0.5. This can be turned off with correction=False, and was included to match R, tseries, which does not use any correction. pvalue based on normal distribution, with integer correction ''' self.npo = npo = (self.runs_pos).sum() self.nne = nne = (self.runs_neg).sum() #n_r = self.n_runs n = npo + nne npn = npo * nne rmean = 2. * npn / n + 1 rvar = 2. * npn * (2.*npn - n) / n**2. / (n-1.) rstd = np.sqrt(rvar) rdemean = self.n_runs - rmean if n >= 50 or not correction: z = rdemean else: if rdemean > 0.5: z = rdemean - 0.5 elif rdemean < 0.5: z = rdemean + 0.5 else: z = 0. z /= rstd pval = 2 * stats.norm.sf(np.abs(z)) return z, pval
[docs]def runstest_1samp(x, cutoff='mean', correction=True): '''use runs test on binary discretized data above/below cutoff Parameters ---------- x : array_like data, numeric cutoff : {'mean', 'median'} or number This specifies the cutoff to split the data into large and small values. correction: bool Following the SAS manual, for samplesize below 50, the test statistic is corrected by 0.5. This can be turned off with correction=False, and was included to match R, tseries, which does not use any correction. Returns ------- z_stat : float test statistic, asymptotically normally distributed p-value : float p-value, reject the null hypothesis if it is below an type 1 error level, alpha . ''' if cutoff == 'mean': cutoff = np.mean(x) elif cutoff == 'median': cutoff = np.median(x) xindicator = (x >= cutoff).astype(int) return Runs(xindicator).runs_test(correction=correction)
[docs]def runstest_2samp(x, y=None, groups=None, correction=True): '''Wald-Wolfowitz runstest for two samples This tests whether two samples come from the same distribution. Parameters ---------- x : array_like data, numeric, contains either one group, if y is also given, or both groups, if additionally a group indicator is provided y : array_like (optional) data, numeric groups : array_like group labels or indicator the data for both groups is given in a single 1-dimensional array, x. If group labels are not [0,1], then correction: bool Following the SAS manual, for samplesize below 50, the test statistic is corrected by 0.5. This can be turned off with correction=False, and was included to match R, tseries, which does not use any correction. Returns ------- z_stat : float test statistic, asymptotically normally distributed p-value : float p-value, reject the null hypothesis if it is below an type 1 error level, alpha . Notes ----- Wald-Wolfowitz runs test. If there are ties, then then the test statistic and p-value that is reported, is based on the higher p-value between sorting all tied observations of the same group This test is intended for continuous distributions SAS has treatment for ties, but not clear, and sounds more complicated (minimum and maximum possible runs prevent use of argsort) (maybe it's not so difficult, idea: add small positive noise to first one, run test, then to the other, run test, take max(?) p-value - DONE This gives not the minimum and maximum of the number of runs, but should be close. Not true, this is close to minimum but far away from maximum. maximum number of runs would use alternating groups in the ties.) Maybe adding random noise would be the better approach. SAS has exact distribution for sample size <=30, doesn't look standard but should be easy to add. currently two-sided test only This has not been verified against a reference implementation. In a short Monte Carlo simulation where both samples are normally distribute, the test seems to be correctly sized for larger number of observations (30 or larger), but conservative (i.e. reject less often than nominal) with a sample size of 10 in each group. See Also -------- runs_test_1samp Runs RunsProb ''' x = np.asarray(x) if not y is None: y = np.asarray(y) groups = np.concatenate((np.zeros(len(x)), np.ones(len(y)))) # note reassigning x x = np.concatenate((x, y)) gruni = np.arange(2) elif not groups is None: gruni = np.unique(groups) if gruni.size != 2: # pylint: disable=E1103 raise ValueError('not exactly two groups specified') #require groups to be numeric ??? else: raise ValueError('either y or groups is necessary') xargsort = np.argsort(x) #check for ties x_sorted = x[xargsort] x_diff = np.diff(x_sorted) # used for detecting and handling ties if x_diff.min() == 0: print('ties detected') #replace with warning x_mindiff = x_diff[x_diff > 0].min() eps = x_mindiff/2. xx = x.copy() #don't change original, just in case xx[groups==gruni[0]] += eps xargsort = np.argsort(xx) xindicator = groups[xargsort] z0, p0 = Runs(xindicator).runs_test(correction=correction) xx[groups==gruni[0]] -= eps #restore xx = x xx[groups==gruni[1]] += eps xargsort = np.argsort(xx) xindicator = groups[xargsort] z1, p1 = Runs(xindicator).runs_test(correction=correction) idx = np.argmax([p0,p1]) return [z0, z1][idx], [p0, p1][idx] else: xindicator = groups[xargsort] return Runs(xindicator).runs_test(correction=correction)
class TotalRunsProb(object): '''class for the probability distribution of total runs This is the exact probability distribution for the (Wald-Wolfowitz) runs test. The random variable is the total number of runs if the sample has (n0, n1) observations of groups 0 and 1. Notes ----- Written as a class so I can store temporary calculations, but I don't think it matters much. Formulas taken from SAS manual for one-sided significance level. Could be converted to a full univariate distribution, subclassing scipy.stats.distributions. *Status* Not verified yet except for mean. ''' def __init__(self, n0, n1): self.n0 = n0 self.n1 = n1 self.n = n = n0 + n1 self.comball = comb(n, n1) def runs_prob_even(self, r): n0, n1 = self.n0, self.n1 tmp0 = comb(n0-1, r//2-1) tmp1 = comb(n1-1, r//2-1) return tmp0 * tmp1 * 2. / self.comball def runs_prob_odd(self, r): n0, n1 = self.n0, self.n1 k = (r+1)//2 tmp0 = comb(n0-1, k-1) tmp1 = comb(n1-1, k-2) tmp3 = comb(n0-1, k-2) tmp4 = comb(n1-1, k-1) return (tmp0 * tmp1 + tmp3 * tmp4) / self.comball def pdf(self, r): r = np.asarray(r) r_isodd = np.mod(r, 2) > 0 r_odd = r[r_isodd] r_even = r[~r_isodd] runs_pdf = np.zeros(r.shape) runs_pdf[r_isodd] = self.runs_prob_odd(r_odd) runs_pdf[~r_isodd] = self.runs_prob_even(r_even) return runs_pdf def cdf(self, r): r_ = np.arange(2,r+1) cdfval = self.runs_prob_even(r_[::2]).sum() cdfval += self.runs_prob_odd(r_[1::2]).sum() return cdfval class RunsProb(object): '''distribution of success runs of length k or more (classical definition) The underlying process is assumed to be a sequence of Bernoulli trials of a given length n. not sure yet, how to interpret or use the distribution for runs of length k or more. Musseli also has longest success run, and waiting time distribution negative binomial of order k and geometric of order k need to compare with Godpole need a MonteCarlo function to do some quick tests before doing more ''' def pdf(self, x, k, n, p): '''distribution of success runs of length k or more Parameters ---------- x : float count of runs of length n k : int length of runs n : int total number of observations or trials p : float probability of success in each Bernoulli trial Returns ------- pdf : float probability that x runs of length of k are observed Notes ----- not yet vectorized References ---------- Muselli 1996, theorem 3 ''' q = 1-p m = np.arange(x, (n+1)//(k+1)+1)[:,None] terms = (-1)**(m-x) * comb(m, x) * p**(m*k) * q**(m-1) \ * (comb(n - m*k, m - 1) + q * comb(n - m*k, m)) return terms.sum(0) def pdf_nb(self, x, k, n, p): pass #y = np.arange(m-1, n-mk+1 ''' >>> [np.sum([RunsProb().pdf(xi, k, 16, 10/16.) for xi in range(0,16)]) for k in range(16)] [0.99999332193894064, 0.99999999999999367, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0] >>> [(np.arange(0,16) * [RunsProb().pdf(xi, k, 16, 10/16.) for xi in range(0,16)]).sum() for k in range(16)] [6.9998931510341809, 4.1406249999999929, 2.4414062500000075, 1.4343261718749996, 0.83923339843749856, 0.48875808715820324, 0.28312206268310569, 0.1629814505577086, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0] >>> np.array([(np.arange(0,16) * [RunsProb().pdf(xi, k, 16, 10/16.) for xi in range(0,16)]).sum() for k in range(16)])/11 array([ 0.63635392, 0.37642045, 0.22194602, 0.13039329, 0.07629395, 0.04443255, 0.02573837, 0.0148165 , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. ]) >>> np.diff([(np.arange(0,16) * [RunsProb().pdf(xi, k, 16, 10/16.) for xi in range(0,16)]).sum() for k in range(16)][::-1]) array([ 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0.16298145, 0.12014061, 0.20563602, 0.35047531, 0.59509277, 1.00708008, 1.69921875, 2.85926815]) '''
[docs]def median_test_ksample(x, groups): '''chisquare test for equality of median/location This tests whether all groups have the same fraction of observations above the median. Parameters ---------- x : array_like data values stacked for all groups groups : array_like group labels or indicator Returns ------- stat : float test statistic pvalue : float pvalue from the chisquare distribution others ???? currently some test output, table and expected ''' x = np.asarray(x) gruni = np.unique(groups) xli = [x[groups==group] for group in gruni] xmedian = np.median(x) counts_larger = np.array([(xg > xmedian).sum() for xg in xli]) counts = np.array([len(xg) for xg in xli]) counts_smaller = counts - counts_larger nobs = counts.sum() n_larger = (x > xmedian).sum() n_smaller = nobs - n_larger table = np.vstack((counts_smaller, counts_larger)) #the following should be replaced by chisquare_contingency table expected = np.vstack((counts * 1. / nobs * n_smaller, counts * 1. / nobs * n_larger)) if (expected < 5).any(): print('Warning: There are cells with less than 5 expected' \ 'observations. The chisquare distribution might not be a good' \ 'approximation for the true distribution.') #check ddof return stats.chisquare(table.ravel(), expected.ravel(), ddof=1), table, expected
[docs]def cochrans_q(x): '''Cochran's Q test for identical effect of k treatments Cochran's Q is a k-sample extension of the McNemar test. If there are only two treatments, then Cochran's Q test and McNemar test are equivalent. Test that the probability of success is the same for each treatment. The alternative is that at least two treatments have a different probability of success. Parameters ---------- x : array_like, 2d (N,k) data with N cases and k variables Returns ------- q_stat : float test statistic pvalue : float pvalue from the chisquare distribution Notes ----- In Wikipedia terminology, rows are blocks and columns are treatments. The number of rows N, should be large for the chisquare distribution to be a good approximation. The Null hypothesis of the test is that all treatments have the same effect. References ---------- http://en.wikipedia.org/wiki/Cochran_test SAS Manual for NPAR TESTS ''' warnings.warn("Deprecated, use stats.cochrans_q instead", DeprecationWarning) x = np.asarray(x) gruni = np.unique(x) N, k = x.shape count_row_success = (x==gruni[-1]).sum(1, float) count_col_success = (x==gruni[-1]).sum(0, float) count_row_ss = count_row_success.sum() count_col_ss = count_col_success.sum() assert count_row_ss == count_col_ss #just a calculation check #this is SAS manual q_stat = (k-1) * (k * np.sum(count_col_success**2) - count_col_ss**2) \ / (k * count_row_ss - np.sum(count_row_success**2)) #Note: the denominator looks just like k times the variance of the #columns #Wikipedia uses a different, but equivalent expression ## q_stat = (k-1) * (k * np.sum(count_row_success**2) - count_row_ss**2) \ ## / (k * count_col_ss - np.sum(count_col_success**2)) return q_stat, stats.chi2.sf(q_stat, k-1)
[docs]def mcnemar(x, y=None, exact=True, correction=True): '''McNemar test Parameters ---------- x, y : array_like two paired data samples. If y is None, then x can be a 2 by 2 contingency table. x and y can have more than one dimension, then the results are calculated under the assumption that axis zero contains the observation for the samples. exact : bool If exact is true, then the binomial distribution will be used. If exact is false, then the chisquare distribution will be used, which is the approximation to the distribution of the test statistic for large sample sizes. correction : bool If true, then a continuity correction is used for the chisquare distribution (if exact is false.) Returns ------- stat : float or int, array The test statistic is the chisquare statistic if exact is false. If the exact binomial distribution is used, then this contains the min(n1, n2), where n1, n2 are cases that are zero in one sample but one in the other sample. pvalue : float or array p-value of the null hypothesis of equal effects. Notes ----- This is a special case of Cochran's Q test. The results when the chisquare distribution is used are identical, except for continuity correction. ''' warnings.warn("Deprecated, use stats.TableSymmetry instead", DeprecationWarning) x = np.asarray(x) if y is None and x.shape[0] == x.shape[1]: if x.shape[0] != 2: raise ValueError('table needs to be 2 by 2') n1, n2 = x[1, 0], x[0, 1] else: # I'm not checking here whether x and y are binary, # isn't this also paired sign test n1 = np.sum(x < y, 0) n2 = np.sum(x > y, 0) if exact: stat = np.minimum(n1, n2) # binom is symmetric with p=0.5 pval = stats.binom.cdf(stat, n1 + n2, 0.5) * 2 pval = np.minimum(pval, 1) # limit to 1 if n1==n2 else: corr = int(correction) # convert bool to 0 or 1 stat = (np.abs(n1 - n2) - corr)**2 / (1. * (n1 + n2)) df = 1 pval = stats.chi2.sf(stat, df) return stat, pval
[docs]def symmetry_bowker(table): '''Test for symmetry of a (k, k) square contingency table This is an extension of the McNemar test to test the Null hypothesis that the contingency table is symmetric around the main diagonal, that is n_{i, j} = n_{j, i} for all i, j Parameters ---------- table : array_like, 2d, (k, k) a square contingency table that contains the count for k categories in rows and columns. Returns ------- statistic : float chisquare test statistic p-value : float p-value of the test statistic based on chisquare distribution df : int degrees of freedom of the chisquare distribution Notes ----- Implementation is based on the SAS documentation, R includes it in `mcnemar.test` if the table is not 2 by 2. The pvalue is based on the chisquare distribution which requires that the sample size is not very small to be a good approximation of the true distribution. For 2x2 contingency tables exact distribution can be obtained with `mcnemar` See Also -------- mcnemar ''' warnings.warn("Deprecated, use stats.TableSymmetry instead", DeprecationWarning) table = np.asarray(table) k, k2 = table.shape if k != k2: raise ValueError('table needs to be square') #low_idx = np.tril_indices(k, -1) # this doesn't have Fortran order upp_idx = np.triu_indices(k, 1) tril = table.T[upp_idx] # lower triangle in column order triu = table[upp_idx] # upper triangle in row order stat = ((tril - triu)**2 / (tril + triu + 1e-20)).sum() df = k * (k-1) / 2. pval = stats.chi2.sf(stat, df) return stat, pval, df
if __name__ == '__main__': x1 = np.array([1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1]) print(Runs(x1).runs_test()) print(runstest_1samp(x1, cutoff='mean')) print(runstest_2samp(np.arange(16,0,-1), groups=x1)) print(TotalRunsProb(7,9).cdf(11)) print(median_test_ksample(np.random.randn(100), np.random.randint(0,2,100))) print(cochrans_q(np.random.randint(0,2,(100,8))))