python - 如何使用 igraph 获取最短路径上的顶点？

Question

我igraph用来生成顶点对之间的最短路径距离矩阵，但我不知道如何返回顶点。到目前为止，我有：

path_length_matrix = ig_graph.shortest_paths_dijkstra(None,None,"distance", "ALL")

我正在寻找一个函数，它返回一个路径矩阵，如距离矩阵，但我在igraph 文档中看不到任何东西，它显示了如何获取路径。

Answer 1

get_shortest_paths我相信你需要的功能。见https://igraph.org/python/doc/igraph.GraphBase-class.html#get_shortest_paths

您需要为每个源顶点单独调用它，它只会为每对节点提供一条（任意）最短路径。如果您需要所有最短路径，请参阅get_all_shortest_paths：https ://igraph.org/python/doc/igraph.GraphBase-class.html#get_all_shortest_paths

Answer 2

我这样做

from igraph import *
g = Graph([(0,1), (0,2), (2,3), (3,4), (4,2), (2,5), (5,0), (6,3), (5,6)])
g.vs["name"] = ["Alice", "Bob", "Claire", "Dennis", "Esther", "Frank", "George"]
#You could create Vertexes like g.add_vertex(name="Bill") 
path=g.get_shortest_paths("Alice",to="Frank",mode=OUT,output='vpath')
for n in path[0]:
    print("{}".format(g.vs[n]['name']))

希望这可以帮助

Answer 3

这是为加权有向图 (DAG) 找到最短路径的方法。所以这就是我想出来的：

import igraph
from igraph import *
g = Graph(directed=True)
g.add_vertices(3)
g.vs["name"]=["GO:1234567","GO:6789056","GO:5674321"]
g.es["weight"]=1
g['GO:1234567','GO:6789056']=1
g['GO:6789056','GO:5674321']=5
weight=g.es["weight"]
print weight
print g.degree(mode="in") 
print g.shortest_paths_dijkstra(source="GO:1234567", target="GO:5674321", 
    weights=weight, mode=OUT)