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####
##
##
## Get a file as input, whose n-th line corresponds to the value of a
## certain property of node n, and rank nodes according to their
## properties, taking into account ranking ties properly.
##
## The output is a file whose n-th line is the "ranking" of the n-th
## node according to the given property. (notice that rankings could
## be fractional, due to the tie removal algorithm)
##
## The rank of a node is set to "0" (ZERO) if the corresponding
## property is smaller than a value given as second parameter
##
import sys
import math
if len(sys.argv) < 3:
print "Usage: %s <filein> <thresh>" % sys.argv[0]
sys.exit(1)
thresh = float(sys.argv[2])
lines = open(sys.argv[1], "r").readlines()
ranking = []
n=0
for l in lines:
if l[0] == "#" or l.strip(" \n").split(" ") == []:
continue
v = [float(x) if "." in x or "e" in x else int(x) for x in l.strip(" \n").split(" ")][0]
if v >= thresh:
ranking.append((v,n))
else:
ranking.append((0,n))
n +=1
ranking.sort(reverse=True)
#print ranking
new_ranking = {}
v0, n0 = ranking[0]
old_value = v0
tot_rankings = 1
stack = [n0]
l=1.0
for v,n in ranking[1:]:
l += 1
##print stack, tot_rankings
if v != old_value: ### There is a new rank
# We first compute the rank for all the nodes in the stack and then we set it
if old_value == 0:
new_rank_value = 0
else:
new_rank_value = 1.0 * tot_rankings / len(stack)
##print new_rank_value
for j in stack:
new_ranking[j] = new_rank_value
old_value = v
tot_rankings = l
stack = [n]
else: # One more value with the same rank, keep it for the future
stack.append(n)
tot_rankings += l
if v == 0 :
new_rank_value = 0
else:
new_rank_value = 1.0 * tot_rankings / len(stack)
#print new_rank_value
for j in stack:
new_ranking[j] = new_rank_value
#print new_ranking
keys = new_ranking.keys()
keys.sort()
for k in keys:
print new_ranking[k]
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