First commit of MAMMULT code

1 files changed, 74 insertions, 0 deletions

diff --git a/structure/correlations/rank_nodes.py b/structure/correlations/rank_nodes.py
new file mode 100644
index 0000000..6985e88
--- /dev/null
+++ b/structure/correlations/rank_nodes.py
@@ -0,0 +1,74 @@
+####
+##
+##
+## 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)
+##
+##
+
+
+import sys
+import math
+
+
+if len(sys.argv) < 2:
+    print "Usage: %s <filein>" % sys.argv[0]
+    sys.exit(1)
+
+
+
+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]
+    ranking.append((v,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
+        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
+
+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]