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101 lines
2.7 KiB
101 lines
2.7 KiB
# This file is part of MAMMULT: Metrics And Models for Multilayer Networks
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#
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# This program is free software: you can redistribute it and/or modify
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# it under the terms of the GNU General Public License as published by
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# the Free Software Foundation, either version 3 of the License, or (at
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# your option) any later version.
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#
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# This program is distributed in the hope that it will be useful, but
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# WITHOUT ANY WARRANTY; without even the implied warranty of
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# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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# General Public License for more details.
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#
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# You should have received a copy of the GNU General Public License
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# along with this program. If not, see <http://www.gnu.org/licenses/>.
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####
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##
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##
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## Get a file as input, whose n-th line corresponds to the value of a
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## certain property of node n, and rank nodes according to their
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## properties, taking into account ranking ties properly.
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##
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## The output is a file whose n-th line is the "ranking" of the n-th
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## node according to the given property. (notice that rankings could
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## be fractional, due to the tie removal algorithm)
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##
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## The rank of a node is set to "0" (ZERO) if the corresponding
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## property is smaller than a value given as second parameter
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##
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import sys
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import math
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if len(sys.argv) < 3:
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print "Usage: %s <filein> <thresh>" % sys.argv[0]
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sys.exit(1)
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thresh = float(sys.argv[2])
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lines = open(sys.argv[1], "r").readlines()
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ranking = []
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n=0
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for l in lines:
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if l[0] == "#" or l.strip(" \n").split(" ") == []:
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continue
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v = [float(x) if "." in x or "e" in x else int(x) for x in l.strip(" \n").split(" ")][0]
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if v >= thresh:
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ranking.append((v,n))
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else:
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ranking.append((0,n))
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n +=1
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ranking.sort(reverse=True)
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#print ranking
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new_ranking = {}
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v0, n0 = ranking[0]
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old_value = v0
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tot_rankings = 1
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stack = [n0]
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l=1.0
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for v,n in ranking[1:]:
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l += 1
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##print stack, tot_rankings
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if v != old_value: ### There is a new rank
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# We first compute the rank for all the nodes in the stack and then we set it
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if old_value == 0:
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new_rank_value = 0
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else:
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new_rank_value = 1.0 * tot_rankings / len(stack)
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##print new_rank_value
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for j in stack:
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new_ranking[j] = new_rank_value
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old_value = v
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tot_rankings = l
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stack = [n]
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else: # One more value with the same rank, keep it for the future
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stack.append(n)
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tot_rankings += l
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if v == 0 :
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new_rank_value = 0
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else:
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new_rank_value = 1.0 * tot_rankings / len(stack)
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#print new_rank_value
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for j in stack:
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new_ranking[j] = new_rank_value
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#print new_ranking
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keys = new_ranking.keys()
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keys.sort()
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for k in keys:
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print new_ranking[k]
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