Question

networkx中是否有一个函数/方法来识别给定（可选加权）距离内的所有祖先/后代？

例如，有效产生与下面函数相同结果的东西？

import networkx
g = networkx.DiGraph()
edges_with_atts = [(1, 2, {'length':5}),
                (1, 3, {'length':11}),
                (2, 4, {'length':4}),
                (2, 5,{'length':7})]
g.add_edges_from(edges_with_atts)

def descendants_within(graph, start_node=1, constraint=10, weight='length'):
    result = set()
    for node in networkx.descendants(graph, start_node):
        if networkx.shortest_path_length(graph, start_node, node, weight) < constraint:
            result.add(node)
    return result

print(descendants_within(g))

#set([2, 4])

Answer 1

某些NetworkX最短路径算法存在“cutoff”参数。例如，在您的情况下，您可以从源节点到所有其他节点运行“单源最短路径”计算，并将搜索限制为短于指定截止长度的路径。在下面的示例中，Dijkstra算法用于计算加权网络的最短路径。

import networkx as nx
g = nx.DiGraph()
edges_with_atts = [(1, 2, {'length':5}),
                (1, 3, {'length':11}),
                (2, 4, {'length':4}),
                (2, 5,{'length':7})]
g.add_edges_from(edges_with_atts)

lengths = nx.single_source_dijkstra_path_length(g, source=1, weight='length', cutoff=10)
print(dict(lengths).keys())
# [1, 2, 4]

在networkx中有效识别给定距离内的祖先/后代

1 个答案: