Faster way to calculate the number of shortest paths a vertex belongs to using Networkx

Question

I am considering that the Stress of a vertex i is the number of shortest paths between all pairs of vertices that i belongs to.

I am trying to calculate it using Networkx, I've made in three ways so far. The readable, dirty, and dirtiest but none of them is fast. Actually, I would like it to be faster than the betweenness (source) present on Networkx. Is there a better way to calculate that? Thanks in advance for any suggestion, answer or comment. Following see what I did so far:

Ps.: Here is a pastie with the code ready to go if you want give it a try, thanks again.

Here is the common part on all versions:

import networkx as nx
from collections import defaultdict

Dirtiest, brace yourselves:

def stress_centrality_dirtiest(g):

  stress = defaultdict(int)

  for a in nx.nodes_iter(g):
    for b in nx.nodes_iter(g):
      if a==b:
        continue
      # pred = nx.predecessor(G,b)  # for unweighted graphs
      pred, distance = nx.dijkstra_predecessor_and_distance(g,b)  # for weighted graphs
      if not pred.has_key(a):
        return [] 
      path = [[a,0]] 
      path_length = 1
      index = 0
      while index >= 0: 
        n,i = path[index] 
        if n == b: 
          for vertex in map(lambda x:x[0], path[:index+1])[1:-1]:
            stress[vertex] += 1
        if len(pred[n]) > i: 
          index += 1 
          if index == path_length: 
            path.append([pred[n][i],0]) 
            path_length += 1 
          else: 
            path[index] = [pred[n][i],0] 
        else: 
          index -= 1 
          if index >= 0: 
            path[index][4] += 1 
  return stress

Dirty

def stress_centrality_dirty(g):

  stress = defaultdict(int)

  paths = nx.all_pairs_dijkstra_path(g)
  for item in paths.values():
    for element in item.values():
      if len(element) > 2:
        for vertex in element[1:-1]:
          stress[vertex] += 1
  return stress

Readable

def stress_centrality_readable(g):

  stress = defaultdict(int)

  paths = nx.all_pairs_dijkstra_path(g)
  for source in nx.nodes_iter(g):
    for end in nx.nodes_iter(g):
      if source == end:
        continue
      path = paths[source][end]
      if len(path) > 2:                                         # path must contains at least 3 vertices source - another node - end
        for vertex in path[1:-1]:                               # when counting the number of occurrencies, exclude source and end vertices
          stress[vertex] += 1
  return stress

Answer 1

The betweenness code you pointed to in NetworkX does almost what you want and can be adjusted easily.

In the betweenness function if you call the following (instead of _accumulate_basic) during the "accumulate" stage it should calculate the stress centrality (untested)

def _accumulate_stress(betweenness,S,P,sigma,s):
    delta = dict.fromkeys(S,0)
    while S:
        w = S.pop()
        for v in P[w]:
            delta[v] += (1.0+delta[w])
        if w != s:
            betweenness[w] += sigma[w]*delta[w]
    return betweenness

See the paper Ulrik Brandes: On Variants of Shortest-Path Betweenness Centrality and their Generic Computation. Social Networks 30(2):136-145, 2008. http://www.inf.uni-konstanz.de/algo/publications/b-vspbc-08.pdf

The stress centrality algorithm is Algorithm 12.

Answer 2

Based on the answer I have been given here, I tried to do exactly the same thing.

My attempt revolved around the use of the nx.all_shortest_paths(G,source,target) function, which produces a generator:

counts={}
for n in G.nodes(): counts[n]=0
for n in G.nodes():
    for j in G.nodes():
        if (n!=j):
            gener=nx.all_shortest_paths(G,source=n,target=j) #A generator
            print('From node '+str(n)+' to '+str(j))
            for p in gener:
                print(p) 
                for v in p: counts[v]+=1
            print('------')

I have tested this code with a NxN grid network of 100 nodes and it took me approximately 168 seconds to get the results. Now I am aware this is not the best answer as this code is not optimized, but I thought you might have wanted to know about it. Hopefully I can get some directions on how to improve my code.