Question

我有一个很大的（半百万条边）加权图（非定向），我想找到两个节点 u 和 v 之间的距离。我可以使用 my_graph.shortest_paths(u, v, weights='length') 来获取距离。但是，这真的很慢。

我也可以先找到路径，然后计算它的长度。这很快，但我不明白为什么这比直接计算长度要快。

在 networkx 中我使用了 nx.shortest_path_length(my_graph u, v, weight='length')

我使用此代码来计算速度。对于任何想要运行代码的人，我将边缘列表放在 Google Drive here

import pandas as pd
import networkx as nx
import igraph
import time

# load edgelist
edgelist = pd.read_pickle('edgelist.pkl')

# create igraph
tuples = [tuple(x) for x in edgelist[['u', 'v', 'length']].values]
graph_igraph = igraph.Graph.TupleList(tuples, directed=False, edge_attrs=['length'])

# create nx graph
graph_nx = nx.from_pandas_edgelist(edgelist, source='u', target='v', edge_attr=True)


def distance_shortest_path(u, v):
    return graph_igraph.shortest_paths(u, v, weights='length')[0]

get_length = lambda edge: graph_igraph.es[edge]['length']
def distance_path_then_sum(u, v):
    path = graph_igraph.get_shortest_paths(u, v, weights='length', output='epath')[0]
    return sum(map(get_length, path))

def distance_nx(u, v):
    return nx.shortest_path_length(graph_nx, u, v, weight='length')


some_nodes = [
    'Delitzsch unt Bf',
    'Neustadt(Holst)Gbf',
    'Delitzsch ob Bf',
    'Karlshagen',
    'Berlin-Karlshorst (S)',
    'Köln/Bonn Flughafen',
    'Mannheim Hbf',
    'Neu-Edingen/Friedrichsfeld',
    'Ladenburg',
    'Heddesheim/Hirschberg',
    'Weinheim-Lützelsachsen',
    'Wünsdorf-Waldstadt',
    'Zossen',
    'Dabendorf',
    'Rangsdorf',
    'Dahlewitz',
    'Blankenfelde(Teltow-Fläming)',
    'Berlin-Schönefeld Flughafen',
    'Berlin Ostkreuz',
]

print('distance_shortest_path ', end='')
start = time.time()
for node in some_nodes:
    distance_shortest_path('Köln Hbf', node)
print('took', time.time() - start)

print('distance_nx ', end='')
start = time.time()
for node in some_nodes:
    distance_nx('Köln Hbf', node)
print('took', time.time() - start)

print('distance_path_then_sum ', end='')
start = time.time()
for node in some_nodes:
    distance_path_then_sum('Köln Hbf', node)
print('took', time.time() - start)

结果

distance_shortest_path took 46.34037733078003
distance_nx took 12.006148099899292
distance_path_then_sum took 0.9555535316467285

Answer 1

您可以在 igraph 中为此使用 shortest_paths 函数。使用非常简单，假设 G 是您的图，具有 G.es['weight'] 边权重，然后

D = G.shortest_paths(weights='weight'))

会给你一个igraph matrix D。您可以将其转换为 numpy 数组为

D = np.array(list(D))

要仅获取特定对（组）节点之间的距离，您可以指定 source 的 target 和 shortest_paths 参数。

igraph中两个顶点之间的距离

1 个答案: