有没有一种方法可以在python中创建包含节点和权重的networkx图的自定义规范化numpy数组

Question

我有一个networkx无向加权图，它有n个节点。我想创建一个对称的自定义numpy矩阵，使其形状为n×n。我正在尝试设计一个解决方案，该解决方案不依赖于创建节点时给定的“名称”。

每条边都有一个给定的权重。

只需应用三条规则：

1. 矩阵的对角线必须等于1

我确实意识到networkx具有内置的normalized\u laplacian\u matrix功能，但据我所知，它并没有构建我想要的功能。

Example:
    FG=nx.Graph()
    FG.add_weighted_edges_from([('A','B',0.125),('A','C',0.75),('B','D',1.2),('C','D',0.375)])

因此，这应产生以下矩阵

   ([[1.        , 0.92872692, 1.00790526, 0.        ],
    [0.92872692, 1.        , 0.        , 0.69223218],
    [1.00790526, 0.        , 1.        , 0.75124823],
    [0.        , 0.69223218, 0.75124823, 1.        ]])    

Answer 1

import networkx as nx
FG=nx.Graph()
FG.add_weighted_edges_from([('A','B',0.125),('A','C',0.75),('B','D',1.2),('C','D',0.375)])

length = len(FG.nodes())
matrix = nx.convert_matrix.to_numpy_matrix(FG)
new_matrix = np.zeros((length,length))
new_matrix = 1/np.sqrt(matrix.sum(axis=1)*matrix.sum(axis=0))
new_matrix[matrix==0] = 0
np.fill_diagonal(new_matrix,1)


new_matrix
matrix([[1.        , 0.92872692, 1.00790526, 0.        ],
        [0.92872692, 1.        , 0.        , 0.69223218],
        [1.00790526, 0.        , 1.        , 0.75124823],
        [0.        , 0.69223218, 0.75124823, 1.        ]])

import networkx as nx
FG=nx.Graph()
FG.add_weighted_edges_from([('A','B',0.125),('A','C',0.75),('B','D',1.2),('C','D',0.375)])

length = len(FG.nodes())
matrix = nx.convert_matrix.to_numpy_matrix(FG)
new_matrix = np.zeros((length,length))

import numpy as np
for i in range(length):
    for j in range(length):
        if i == j:
            new_matrix[i,j] = 1
        elif matrix[i,j] == 0:
            new_matrix[i,j] = 0
        else:
            w_1 = matrix[:,i].sum()
            w_2 = matrix[j,:].sum()
            new_matrix[i,j] = 1/np.sqrt(w_1*w_2)
new_matrix