TA貢獻1883條經驗 獲得超3個贊
干得好。這正是該edge_match選項的用途。我將創建 3 個圖，前兩個是同構的（即使權重有不同的名稱 --- 我已經設置了比較函數來說明這一點）。第三個不是同構的。
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貢獻1847條經驗 獲得超7個贊
此處使用完全相同的圖表專門回答問題中的問題。請注意，我正在使用 networkx.MultiGraph 并在放置這些邊時考慮一些“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
舉報