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[英]How to Find Shortest Path Back to Self in a Graph/Adjacency Matrix - NetworkX
[英]path to adjacency matrix in networkx
我正在嘗試使用networkx庫將圖形中的路徑轉換為鄰接矩陣。 我可以將整個圖轉換為鄰接矩陣:
>>>import networkx as nx
>>>DG=nx.DiGraph()
>>>DG.add_edges_from([(1,2), (2,3),(1,3)])
>>>nx.to_numpy_matrix(DG)
....matrix([[ 0., 1., 1.],
[ 0., 0., 1.],
[ 0., 0., 0.]])
但是,找到所有從節點1到節點3的簡單路徑后:
>>>list(nx.all_simple_paths(DG,1,3))
....[[1, 2, 3], [1, 3]]
我無法將它們轉換為鄰接矩陣。 我希望能夠選擇一條路徑並將其轉換為鄰接矩陣,例如,第二條路徑應返回:
....matrix([[ 0., 0., 1.],
[ 0., 0., 0.],
[ 0., 0., 0.]])
那不是鄰接矩陣。 但是您可以輕松地自己構建它,如下所示:
import networkx as nx
import numpy as np
DG=nx.DiGraph()
DG.add_edges_from([(1,2), (2,3),(1,3)])
paths = list(nx.all_simple_paths(DG,1,3))
for path in paths:
matrix = np.matrix(np.zeros((len(DG), len(DG))))
for i in range(len(path)-1):
matrix[path[i]-1], path[i+1]-1] = 1 # edit: credits to @Joel
print(matrix)
輸出:
[[ 0. 1. 0.]
[ 0. 0. 1.]
[ 0. 0. 0.]]
[[ 0. 0. 1.]
[ 0. 0. 0.]
[ 0. 0. 0.]]
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