def spearman_rho(m, n):
"""
Return Spearman's rho; based off stats.py
>>> x = [2, 8, 5, 4, 2, 6, 1, 4, 5, 7, 4]
>>> y = [3, 9, 4, 3, 1, 7, 2, 5, 6, 8, 3]
>>> print round(spearman_rho(x, y), 3)
0.936
"""
assert len(m) == len(n)
dsq = sum([(mi - ni) ** 2 for (mi, ni) in zip(rank(m), rank(n))])
return 1. - 6. * dsq / float(len(m) * (len(n) ** 2 - 1.))
def mann_whitney_U(n1, n2):
"""
Wilcoxan (Mann-Whitney/rank sum) U test, as generalized in:
H.B. Mann & D.R. Whitney. 1947. On a test of whether one of two random
variables is stochastically larger than the other. Annals of
Mathematical Statistics 18(1): 50-60.
NB: no adjustments are made for ties; these are both fine if your
samples are large and aren't ordinal (or small counts)
>>> from csv import DictReader
>>> from collections import defaultdict
>>> species2petal_width = defaultdict(list)
>>> for row in DictReader(open('iris.csv', 'r')):
... species = row['Species']
... width = float(row['Sepal.Width'])
... species2petal_width[species].append(width)
>>> U_results = mann_whitney_U(species2petal_width['versicolor'],
... species2petal_width['virginica'])
>>> U_results['U']
841.0
>>> round(U_results['p.U'], 4)
0.0045
"""
l1 = len(n1)
l2 = len(n2)
rank_sum = sum(rank(n1 + n2)[:len(n1)])
prod = l1 * l2
u1 = rank_sum - l1 * (l1 + 1) / 2.
u2 = prod - u1
(U, bigU) = sorted((u1, u2))
p = 2. * p_U(U, l1, l2)
return {'U': U, 'p.U': p}
def spearman_rho_tr(m, n):
"""
rho for tied ranks, checked by comparison with Pycluster
>>> x = [2, 8, 5, 4, 2, 6, 1, 4, 5, 7, 4]
>>> y = [3, 9, 4, 3, 1, 7, 2, 5, 6, 8, 3]
>>> print round(spearman_rho_tr(x, y), 3)
0.935
"""
assert len(m) == len(n), 'args must be the same length'
m = rank(m)
n = rank(n)
num = 0.
den_m = 0.
den_n = 0.
m_mean = mean(m)
n_mean = mean(n)
for (i, j) in zip(m, n):
i = i - m_mean
j = j - n_mean
num += i * j
den_m += i ** 2
den_n += j ** 2
return num / sqrt(den_m * den_n)