0

我正在尝试计算 igraph 中图形的全局效率,但我不确定我是否正确使用了该模块。我认为有一个解决方案可能有点道理,但它在 r 中,我无法破译他们在说什么。

我曾尝试以 networkx 方式编写代码,试图模仿他们计算全局效率的方式,但到目前为止我还没有成功。我正在使用 igraph,因为我正在处理大图。任何帮助将不胜感激:D

这是我尝试过的:

import igraph
import pandas as pd
import numpy as np
from itertools import permutations

datasafe = pd.read_csv("b1.csv", index_col=0)
D = datasafe.values
g = igraph.Graph.Adjacency((D > 0).tolist())
g.es['weight'] = D[D.nonzero()]

def efficiency_weighted(g):
    weights = g.es["weight"][:]
    eff = (1.0 / np.array(g.shortest_paths_dijkstra(weights=weights)))
    return eff

def global_efficiecny_weighted(g):
    n=180.0
    denom=n*(n-1)
    g_eff = sum(efficiency_weighted(g) for u, v in permutations(g, 2))
    return g_eff

global_efficiecny_weighted(g)

我收到的错误消息是:- TypeError: 'Graph' object is not iterable

4

1 回答 1

1

假设您想要所有节点的节点效率,那么您可以这样做:

import numpy as np
from igraph import *
np.seterr(divide='ignore')

# Example using a random graph with 20 nodes
g = Graph.Erdos_Renyi(20,0.5)

# Assign weights on the edges. Here 1s everywhere
g.es["weight"] = np.ones(g.ecount())

def nodal_eff(g):

    weights = g.es["weight"][:]
    sp = (1.0 / np.array(g.shortest_paths_dijkstra(weights=weights)))
    np.fill_diagonal(sp,0)
    N=sp.shape[0]
    ne= (1.0/(N-1)) * np.apply_along_axis(sum,0,sp)

    return ne

eff = nodal_eff(g)
print(eff)
#[0.68421053 0.81578947 0.73684211 0.76315789 0.76315789 0.71052632
# 0.81578947 0.81578947 0.81578947 0.73684211 0.71052632 0.68421053
# 0.71052632 0.81578947 0.84210526 0.76315789 0.68421053 0.68421053
# 0.78947368 0.76315789]

要获得全局,只需执行以下操作:

np.mean(eff)
于 2019-06-17T20:44:30.787 回答