NetworkX - 删除节点并重新连接边缘

Question

我在图中有一个节点，它充当一种“临时连接器”节点。我想删除该节点并更新图中的边，以便其所有直接前辈都指向其直接后继者。

在 networkx 中是否有内置功能可以做到这一点，还是我需要推出自己的解决方案？

例子:

我有图 1 > 2 > 3 .我想删除节点 2最后得到图形 1 > 3 .

这是我目前的做法:

In [230]: g = nx.DiGraph()

In [231]: g.add_edges_from([(1,2),(2,3)])

In [232]: g.edges()
Out[232]: [(1, 2), (2, 3)]

In [233]: predecessors = g.predecessors(2)

In [234]: successors = g.successors(2)

In [235]: new_edges = [(p,s) for p in predecessors for s in successors]

In [236]: new_edges
Out[236]: [(1, 3)]

In [237]: g.remove_node(2)

In [238]: g.add_edges_from(new_edges)

In [239]: g.nodes()
Out[239]: [1, 3]

In [240]: g.edges()
Out[240]: [(1, 3)]

Answer 1

我尝试在一个巨大的复杂图形上使用 mjvaak 的答案，但它没有用，因为它用不再存在的节点创建边。我通过简单地从 g0 中取出边缘来修复它。
所以我改变了:

edges = g.edges(node)

为了:

edges = g0.edges(node)

因此，固定代码如下:

def simplifyGraph(G):
''' Loop over the graph until all nodes of degree 2 have been removed and their incident edges fused '''

g = G.copy()

while any(degree==2 for _, degree in g.degree):

    g0 = g.copy() #<- simply changing g itself would cause error `dictionary changed size during iteration` 
    for node, degree in g.degree():
        if degree==2:

            if g.is_directed(): #<-for directed graphs
                a0,b0 = list(g0.in_edges(node))[0]
                a1,b1 = list(g0.out_edges(node))[0]

            else:
                edges = g0.edges(node)
                edges = list(edges.__iter__())
                a0,b0 = edges[0]
                a1,b1 = edges[1]

            e0 = a0 if a0!=node else b0
            e1 = a1 if a1!=node else b1

            g0.remove_node(node)
            g0.add_edge(e0, e1)
    g = g0

return g

感谢 mjkvaak 的解决方案!它只需要对更大的图形稍作修改。