Python 如何连接两个具有中间边的子图？

设

G1、G2、G3

为三个图。每个节点都有

N[i]

节点

让

3x3

矩阵

A

表示不相交图

G1、G2、G3

节点之间的接触概率

因此，

A_ij

表示位于

Gi

的节点与位于

Gj的节点具有边的概率

Aii
将是在
Gi
节点之间具有边缘的概率

我需要一些帮助来完成这项工作，并以这样的方式命名节点：在联合图中，我将能够看到哪些节点来自哪个
Gi

import networkx as nx
import numpy as np

A = np.array([[0.2, 0.4, 0.2], [0.4, 0.1, 0.5], [0.2, 0.5, 0.3]])
N = [20,30,40]

def get_number_edges(A,N,n):
    return int(A[n-1,n-1]*N[n-1])

G1=nx.dense_gnm_random_graph(N[0],get_number_edges(A,N,1))
G2=nx.dense_gnm_random_graph(N[1],get_number_edges(A,N,2))
G3=nx.dense_gnm_random_graph(N[2],get_number_edges(A,N,3))

C=nx.disjoint_union(nx.disjoint_union(G1,G2),G3)

我不知道如何将节点从
Gi
连接到
Gj
where
I=j
我不知道如何用一个标签来标注
C
处的节点，该标签指示了从哪个图形
Gi
产生的节点
我将首先创建完整的图
C
，然后根据需要创建子图

import itertools
import networkx as nx
import numpy as np
import matplotlib.pyplot as plt

A = np.array([[0.2, 0.4, 0.2], [0.4, 0.1, 0.5], [0.2, 0.5, 0.3]])
N = [20,30,40]

# create a square probability matrix with dimensions (total nodes, total nodes)
# and populate it with the values given in A
indices = []
ctr = 0
for n in N:
    indices.append(list(range(ctr, ctr+n)))
    ctr += n

# TODO: 
# remove this horrible triple loop; 
# have a cold and am not smart enough to do it properly now
total_nodes = sum(N)
probability = np.zeros((total_nodes, total_nodes), dtype=np.float)
for ii, sources in enumerate(indices):
    for jj, targets in enumerate(indices):
        for source, target in itertools.product(sources, targets):
            probability[source, target] = A[ii, jj]
plt.imshow(probability, cmap='gray'); plt.show()

# convert probability matrix into an adjacency matrix
adjacency = probability >= np.random.rand(*probability.shape)
plt.imshow(adjacency, cmap='gray'); plt.show()

# create networkx graph object
C = nx.from_numpy_array(adjacency)

# to get any subgraph, use the indices
G1 = C.subgraph(indices[0])
# etc

有没有办法避免这个循环，或者使用networkx API来模拟这个循环？@0x90我不知道。