'''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
'''
import numpy as np
from scipy import stats
[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
'''
npo = len(self.runs_pos)
nne = len(self.runs_neg)
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
from scipy import stats
pval = 2 * stats.norm.sf(np.abs(z))
return z, pval
[docs]def runstest_1samp(x, cutoff='mean'):
'''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. This
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()
[docs]def runstest_2samp(x, y=None, groups=None):
'''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
groups : {'mean', 'median'} or number
This specifies the cutoff to split the data into large and small
values. This
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 prvent 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
See Also
--------
runs_test_1samp
Runs
RunsProb
'''
x = np.asarray(x)
if not y is None:
y = np.asarray(y)
x = np.concatenate((x, y))
groups = np.concatenate((np.zeros(len(x)), np.ones(len(y))))
gruni = np.arange(1)
elif not groups is None:
gruni = np.unique(groups)
if gruni.size != 2:
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)
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()
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()
idx = np.argmax([p0,p1])
return [z0, z1][idx], [p0, p1][idx]
else:
xindicator = groups[xargsort]
return Runs(xindicator).runs_test()
from scipy import comb
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 cochran_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.
what's this ? Test that the number of successes is the same for each case.
The alternative is that at least two treatements come from different
populations.
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
others ????
currently some test output, table and expected
Notes
-----
not verified,
In Wikipedia terminology, rows are blocks and N should be large for
the chisquare distribution to be a good approximation; columns are
treatments.
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
'''
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, exact='auto', correction=True):
'''McNemar test
Parameters
----------
x, y : array_like
two paired data samples
exact : bool or 'auto'
correction : bool
If true then a continuity corection is used for the approximate
chisquare distribution.
Returns
-------
stat : float or int
The test statistic is the chisquare statistic in the case of large
samples or 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
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 the continuity correction.
'''
n1 = np.sum(x < y)
n2 = np.sum(x > y)
if exact or (exact=='auto' and n1+n2<25):
stat = min(n1,n2)
pval = stats.binom.sf(min(n1,n2), n1+n2, 0.5)
else:
corr = int(correction)
stat = (np.abs(n1-n2)-corr)**2 / (1. * (n1+n2))
df = 1
pval = stats.chi2.sf(stat,1)
return stat, pval
from numpy.testing import assert_almost_equal, assert_array_almost_equal
def test_cochransq():
#example from dataplot docs, Conovover p. 253
#http://www.itl.nist.gov/div898/software/dataplot/refman1/auxillar/cochran.htm
x = np.array([[1, 1, 1],
[1, 1, 1],
[0, 1, 0],
[1, 1, 0],
[0, 0, 0],
[1, 1, 1],
[1, 1, 1],
[1, 1, 0],
[0, 0, 1],
[0, 1, 0],
[1, 1, 1],
[1, 1, 1]])
res_qstat = 2.8
res_pvalue = 0.246597
assert_almost_equal(cochran_q(x), [res_qstat, res_pvalue])
#equivalence of mcnemar and cochranq for 2 samples
a,b = x[:,:2].T
assert_almost_equal(mcnemar(a,b, exact=False, correction=False),
cochran_q(x[:,:2]))
def test_runstest():
#comparison numbers from R, tseries, runs.test
#currently only 2-sided used
x = np.array([1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1])
z_twosided = 1.386750
pvalue_twosided = 0.1655179
z_greater = 1.386750
pvalue_greater = 0.08275893
z_less = 1.386750
pvalue_less = 0.917241
print Runs(x).runs_test(correction=False)
assert_array_almost_equal(np.array(Runs(x).runs_test(correction=False)),
[z_twosided, pvalue_twosided], decimal=6)
if __name__ == '__main__':
x = np.array([1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1])
print Runs(x).runs_test()
print runstest_1samp(x, cutoff='mean')
print runstest_2samp(np.arange(16,0,-1), groups=x)
print TotalRunsProb(7,9).cdf(11)
print median_test_ksample(np.random.randn(100), np.random.randint(0,2,100))
print cochran_q(np.random.randint(0,2,(100,8)))
test_runstest()
test_cochransq()