通过使用最大连接组件中的节点构建子图,您将删除第二个节点:
G = nx.gnm_random_graph(n=10, m=15, seed=1)
pos = nx.spring_layout(G)
nx.set_node_attributes(G, pos, 'pos')
G.remove_edge(2, 9)
largest_cc = max(nx.connected_components(G), key=len)
G.nodes()
# NodeView((0, 1, 2, 3, 4, 5, 6, 7, 8, 9))
G = G.subgraph(largest_cc).copy()
G.nodes()
# NodeView((0, 1, 3, 4, 5, 6, 7, 8, 9))
现在通过再次添加节点2G.add_node(2, pos=pos[2]),这实质上是更新节点数据字典(保存它的内部数据结构),字面意思是dict.update:
d = dict(G.nodes(data=True))
d.update({2:{'pos':[0.3, 0.5]}})
print(d)
{0: {'pos': array([ 0.33041585, -0.07185971])},
1: {'pos': array([-0.19659528, 0.33972794])},
3: {'pos': array([ 0.22691433, -0.1802301 ])},
4: {'pos': array([0.22462413, 0.2452357 ])},
5: {'pos': array([ 0.65037774, -0.18057473])},
6: {'pos': array([-0.14587125, -0.13225175])},
7: {'pos': array([0.05279257, 0.10579408])},
8: {'pos': array([ 0.42384353, -0.46262269])},
9: {'pos': array([-0.56650162, 0.13495046])},
2: {'pos': [0.3, 0.5]}}
因此,该节点被附加为一个新字典 key/value 对,
G.add_node(2, pos=pos[2])
G.add_edge(2, 8)
G.nodes()
# NodeView((0, 1, 3, 4, 5, 6, 7, 8, 9, 2))
有Graph.add_nodes_from,但它只在节点存在的情况下更新属性(不会删除和重新添加节点),这是有道理的:
G.add_nodes_from(sorted(G.nodes(data=True)))
G.nodes()
NodeView((0, 1, 3, 4, 5, 6, 7, 8, 9, 2))
因此,要走的路是重新创建一个图,并分配排序的节点,如 warped 的回答:
H = nx.Graph()
H.add_nodes_from(sorted(G.nodes(data=True)))
H.add_edges_from(G.edges(data=True))
H.nodes()
# NodeView((0, 1, 2, 3, 4, 5, 6, 7, 8, 9))