2 回答

TA贡献1883条经验 获得超3个赞
干得好。这正是该edge_match选项的用途。我将创建 3 个图,前两个是同构的(即使权重有不同的名称 --- 我已经设置了比较函数来说明这一点)。第三个不是同构的。
import networkx as nx
G1 = nx.Graph()
G1.add_weighted_edges_from([(0,1,0), (0,2,1), (0,3,2)], weight = 'aardvark')
G2 = nx.Graph()
G2.add_weighted_edges_from([(0,1,0), (0,2,2), (0,3,1)], weight = 'baboon')
G3 = nx.Graph()
G3.add_weighted_edges_from([(0,1,0), (0,2,2), (0,3,2)], weight = 'baboon')
def comparison(D1, D2):
#for an edge u,v in first graph and x,y in second graph
#this tests if the attribute 'aardvark' of edge u,v is the
#same as the attribute 'baboon' of edge x,y.
return D1['aardvark'] == D2['baboon']
nx.is_isomorphic(G1, G2, edge_match = comparison)
> True
nx.is_isomorphic(G1, G3, edge_match = comparison)
> False

TA贡献1847条经验 获得超7个赞
此处使用完全相同的图表专门回答问题中的问题。请注意,我正在使用 networkx.MultiGraph 并在放置这些边时考虑一些“排序”。
import networkx as nx
G1,G2,G3,G4=nx.MultiGraph(),nx.MultiGraph(),nx.MultiGraph(),nx.MultiGraph()
G1.add_weighted_edges_from([(0, 1, 0), (0, 2, 1), (0, 3, 2)], weight='ordering')
G2.add_weighted_edges_from([(0, 1, 0), (0, 3, 1), (0, 2, 2)], weight='ordering')
G3.add_weighted_edges_from([(0, 1, 0), (0, 1, 1), (2, 3, 2)], weight='ordering')
G4.add_weighted_edges_from([(0, 1, 0), (2, 3, 1), (0, 1, 2)], weight='ordering')
def comparison(D1,D2):
return D1[0]['ordering'] == D2[0]['ordering']
nx.is_isomorphic(G1,G2, edge_match=comparison)
>True
nx.is_isomorphic(G3,G4, edge_match=comparison)
>False
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