1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
|
####
##
## Take as input two files, whose n^th line contains the ranking of
## element n, and compute the Kendall's \tau_b rank correlation
## coefficient
##
##
import sys
from numpy import *
def kendalltau(x,y):
initial_sort_with_lexsort = True # if True, ~30% slower (but faster under profiler!) but with better worst case (O(n log(n)) than (quick)sort (O(n^2))
n = len(x)
temp = range(n) # support structure used by mergesort
# this closure recursively sorts sections of perm[] by comparing
# elements of y[perm[]] using temp[] as support
# returns the number of swaps required by an equivalent bubble sort
def mergesort(offs, length):
exchcnt = 0
if length == 1:
return 0
if length == 2:
if y[perm[offs]] <= y[perm[offs+1]]:
return 0
t = perm[offs]
perm[offs] = perm[offs+1]
perm[offs+1] = t
return 1
length0 = length / 2
length1 = length - length0
middle = offs + length0
exchcnt += mergesort(offs, length0)
exchcnt += mergesort(middle, length1)
if y[perm[middle - 1]] < y[perm[middle]]:
return exchcnt
# merging
i = j = k = 0
while j < length0 or k < length1:
if k >= length1 or (j < length0 and y[perm[offs + j]] <= y[perm[middle + k]]):
temp[i] = perm[offs + j]
d = i - j
j += 1
else:
temp[i] = perm[middle + k]
d = (offs + i) - (middle + k)
k += 1
if d > 0:
exchcnt += d;
i += 1
perm[offs:offs+length] = temp[0:length]
return exchcnt
# initial sort on values of x and, if tied, on values of y
if initial_sort_with_lexsort:
# sort implemented as mergesort, worst case: O(n log(n))
perm = lexsort((y, x))
else:
# sort implemented as quicksort, 30% faster but with worst case: O(n^2)
perm = range(n)
perm.sort(lambda a,b: cmp(x[a],x[b]) or cmp(y[a],y[b]))
# compute joint ties
first = 0
t = 0
for i in xrange(1,n):
if x[perm[first]] != x[perm[i]] or y[perm[first]] != y[perm[i]]:
t += ((i - first) * (i - first - 1)) / 2
first = i
t += ((n - first) * (n - first - 1)) / 2
# compute ties in x
first = 0
u = 0
for i in xrange(1,n):
if x[perm[first]] != x[perm[i]]:
u += ((i - first) * (i - first - 1)) / 2
first = i
u += ((n - first) * (n - first - 1)) / 2
# count exchanges
exchanges = mergesort(0, n)
# compute ties in y after mergesort with counting
first = 0
v = 0
for i in xrange(1,n):
if y[perm[first]] != y[perm[i]]:
v += ((i - first) * (i - first - 1)) / 2
first = i
v += ((n - first) * (n - first - 1)) / 2
tot = (n * (n - 1)) / 2
if tot == u and tot == v:
return 1 # Special case for all ties in both ranks
tau = ((tot-(v+u-t)) - 2.0 * exchanges) / (sqrt(float(( tot - u )) * float( tot - v )))
# what follows reproduces ending of Gary Strangman's original stats.kendalltau() in SciPy
svar = (4.0*n+10.0) / (9.0*n*(n-1))
z = tau / sqrt(svar)
##prob = erfc(abs(z)/1.4142136)
##return tau, prob
return tau
def main():
if len(sys.argv) < 3:
print "Usage: %s <file1> <file2>" % sys.argv[0]
sys.exit(1)
x1 = []
x2= []
lines = open(sys.argv[1]).readlines()
for l in lines:
elem = [float(x) if "e" in x or "." in x else int(x) for x in l.strip(" \n").split()][0]
x1.append(elem)
lines = open(sys.argv[2]).readlines()
for l in lines:
elem = [float(x) if "e" in x or "." in x else int(x) for x in l.strip(" \n").split()][0]
x2.append(elem)
tau = kendalltau(x1,x2)
print tau
if __name__ == "__main__":
main()
|
