计算networkx中仅包含具有特定属性的边的节点度

Question

我图中的所有边都有一个属性，比如 'colour'。我想要图中节点的度数，但只计算 'colour'='red'

的边

G.add_edges_from([(1,2),(3,4),(4,5)], color='red')
G.add_edges_from([(1,3),(1,4),(2,3)], color='blue')

所以我想G.degree("colour of edge = red")给{1:1, 2:1, 3:1, 4:2, 5:1}

Answer 1

我找不到内置方法，但您可以使用列表理解并通过测试属性值来构建新图形，（代码片段归因于 this）：

In [162]:

G = nx.DiGraph()
G.add_edges_from([(1,2),(3,4),(4,5)], color='red')
G.add_edges_from([(1,3),(1,4),(2,3)], color='blue')
SG=nx.Graph( [ (u,v,d) for u,v,d in G.edges(data=True) if d['color'] == 'red'] )
SG.degree()
Out[162]:
{1: 1, 2: 1, 3: 1, 4: 2, 5: 1}

Answer 2

对于这种情况，这里有一种无需复制图表即可完成的方法。相反，它创建了一个新函数来计算度数。

In [1]: import networkx as nx

In [2]: from collections import defaultdict

In [3]: G = nx.Graph()

In [4]: G.add_edges_from([(1,2),(3,4),(4,5)], color='red')

In [5]: G.add_edges_from([(1,3),(1,4),(2,3)], color='blue')

In [6]: def colored_degree(G, color):
    degree = defaultdict(int)
    for u,v in ((u,v) for u,v,d in G.edges(data=True) if d['color']==color): 
        degree[u]+=1
        degree[v]+=1
    return degree
   ...: 

In [7]: colored_degree(G, 'red')
Out[7]: defaultdict(<type 'int'>, {1: 1, 2: 1, 3: 1, 4: 2, 5: 1})

In [8]: colored_degree(G, 'blue')
Out[8]: defaultdict(<type 'int'>, {1: 2, 2: 1, 3: 2, 4: 1})

计算networkx中仅包含具有特定属性的边的节点度

Calculate the degree of nodes only including edges with a specific attribute in networkx

python

networkx