获取两个节点之间的节点子图？

Question

我有这个图：

%matplotlib inline 
import networkx as nx

G = nx.Graph()

G.add_edge(1, 2)
G.add_edge(2, 3)
G.add_edge(3, 4)
G.add_edge(3, 5)
G.add_edge(4, 6)
G.add_edge(5, 6)
G.add_edge(3, 7)
G.add_edge(7, 6)
G.add_edge(6, 8)
G.add_edge(8, 9)

nx.draw(G, pos=nx.spring_layout(G), with_labels=True)

是否可以在不使用nx.subgraph(G, [3,4,5,6,7])情况下获取节点 3 和 6 之间的子图。 我的意思是，如果我知道有这个子图，但我不知道例如 5 呢？

Answer 1

我的回答和 back2basics 非常相似，但更直接的是找到两者之间的节点。 如果存在从source到target路径，则该路径将由nx.all_simple_paths(G, source=source, target=target) ，它返回路径的生成器。

import networkx as nx

G = nx.Graph()
G.add_edges_from([(1, 2), (2, 3), (3, 4), (3, 5), (4, 6), (5, 6), (3, 7), (7, 6), (6, 8), (8, 9)])

paths_between_generator = nx.all_simple_paths(G,source=3,target=6)
nodes_between_set = {node for path in paths_between_generator for node in path}
SG = G.subgraph(nodes_between_set)

nodes_between_set = ...使用“集合生成器”。 它相当于

nodes_between_set = set()
for path in paths_beween_generator:
    for node in path:
        nodes_between_set.add(node)

Answer 2

前 3 行帮助列出了制作子集所需的列表。

import networkx as nx

c_score = nx.algorithms.betweenness_centrality_subset(G,(3,), (6,))
nodes_between = [x for x in c_score if c_score[x]!=0.0]
nodes_between.extend((3,6))  #add on the ends
SG = G.subgraph(nodes_between)

nx.draw(SG, pos=nx.spring_layout(SG), with_labels=True)

一个警告：子图点被定义为在从点 3 到点 6 的路径中

Answer 3

这工作的原则是

1. 新子图中的任何节点都可以从源或目标到达，并且
2. 根据定义，路径上的任何节点至少有一个前驱和一个后继（对于有向图）或两个邻居（对于无向图）。

所以，首先我们找到我们可以到达的节点的子图，然后递归地删除没有至少一个前驱和一个后继的节点，直到只有现有的子图。

import networkx as nx
def subgraph_from_connections(G, source, target, directed = None):
    included_nodes = [x for x in G.node if nx.has_path(G, source, x) and nx.has_path(G, x, target)]
    G2 = G.subgraph(included_nodes)
    # If this is a undirected graph, we only need to know if it only has 1 neighbor
    # If this is a directed graph, then it needs at least 1 predecessor and at least 1 successor
    if directed == True or (directed is None and type(G) == nx.classes.digraph.DiGraph):
        removals = [x for x in G2.node if len(G2.predecessors(x)) == 0 or len(G2.successors(x)) == 0]
        while len(removals) > 0:
            G2.remove_nodes_from(removals)
            removals = [x for x in G.node if len(G2.predecessors(x)) == 0 or len(G2.successors(x)) == 0]
    else:
        removals = [x for x in G2.node if len(G2.neighbors(x)) < 2]
        while len(removals) > 0:
            G2.remove_nodes_from(removals)
            removals = [x for x in G2.node if len(G2.neighbors(x)) < 2]
    return G2

没有经过广泛测试，但它适用于此处概述的少数情况，其中包括 Joel 测试中的 10/11。 该算法足够快 - 我之前的 1000/10 节点随机测试为 130 毫秒（也许我根本不应该删除它）。