删除networkx有向图中入度和出度等于1的所有节点

假设我在 NetworkX 中制作了一个有向图：

import networkx as nx

G = nx.DiGraph()

n = ["A","B","C","D","E","F","H","I","J","K","L","X","Y","Z"]

e = [("A","Z"),("Z","B"),("B","Y"),("Y","C"),("C","G"),("G","H"),("G","I"),("I","J"),("K","J"),("J","L"),("F","E"),("E","D"),("D","X"),("X","C")]

G.add_nodes_from(n)

G.add_edges_from(e)

我如何删除所有入度和出度等于 1 的节点，以便我的图看起来像这样？：

import networkx as nx

G = nx.DiGraph()

n = ["A","C","F","G","H","J","K","L"]

e = [("A","C"),("C","G"),("G","H"),("G","J"),("K","J"),("J","L")

G.add_nodes_from(n)

G.add_edges_from(e)

这个想法是删除“流通”节点并保留连接性。

下面的代码可以实现您想要的功能，尽管我看不到最终结果中从 ("A", "C") 和 ("G", "J") 获得边缘的位置。

import networkx as nx

def remove_edges(g, in_degree=1, out_degree=1):
    g2=g.copy()
    d_in=g2.in_degree(g2)
    d_out=g2.out_degree(g2)
    print(d_in)
    print(d_out)
    for n in g2.nodes():
        if d_in[n]==in_degree and d_out[n] == out_degree: 
            g2.remove_node(n)
    return g2

G_trimmed = remove_edges(G)
G_trimmed.edges()
#outputs [('C', 'G'), ('G', 'H'), ('K', 'J'), ('J', 'L')]
G_trimmed.nodes()
#outputs ['A', 'C', 'G', 'F', 'H', 'K', 'J', 'L']