pygraphviz或networkx:获取给定节点的所有子图的名称

Question

我想知道给定节点的子图结构。

我也同意networkx解决方案。

这是我的密码。

import pygraphviz as pgv
class Test:
subgraph1 = 'foo'
subgraph2 = 'bar'
def __init__(self):
self.G = pgv.AGraph(directed=True)
self.G.add_subgraph(label=self.subgraph1, name='cluster_' + self.subgraph1)
s1 = self.G.get_subgraph('cluster_' + self.subgraph1)
s1.add_subgraph(label=self.subgraph2, name='cluster_' + self.subgraph2)
s1.add_node('s1 node')
s2 = s1.get_subgraph('cluster_' + self.subgraph2)
s2.add_node('s2 node')
def main(self):
print(self.G.subgraph('1st node'))
print(self.G.subgraph_parent('1st node'))
print(self.G.subgraph_root('1st node'))
n = self.G.get_node('1st node')
print(n.attr['subgraph'])
print(n.attr['label'])

if __name__ == '__main__': Test().main()

我想要一个函数，它会给我返回一个子图列表，给定整个图中的任何节点。像这样的。。。

在：subgraph_structure_as_list('s2 node')

输出：['foo', 'bar']

在：subgraph_structure_as_list('s1 node')

输出：['foo']

Answer 1

您可以使用这样的递归方法：

import pygraphviz as pgv
class Test:
subgraph1 = 'foo'
subgraph2 = 'bar'
def __init__(self):
self.structure = []
self.G = pgv.AGraph(directed=True)
self.G.add_subgraph(label=self.subgraph1, name='cluster_' + self.subgraph1)
s1 = self.G.get_subgraph('cluster_' + self.subgraph1)
s1.add_subgraph(label=self.subgraph2, name='cluster_' + self.subgraph2)
s1.add_node('s1 node')
s2 = s1.get_subgraph('cluster_' + self.subgraph2)
s2.add_node('s2 node')
#  ... other methods here ...
def get_structure(self, node_name, node=None):
if node is None:
node = self.G
for subgraph in node.subgraphs():
if node_name in subgraph.nodes():
self.structure.append(subgraph.node_attr['label'])
self.get_structure(node_name, subgraph)
def subgraph_structure_as_list(self, node_name):
self.structure = []
self.get_structure(node_name)
return self.structure

你可以像这样随心所欲地使用subgraph_structure_as_list：

a = Test()
print(a.subgraph_structure_as_list('s2 node'))
print(a.subgraph_structure_as_list('s1 node'))

输出：

['foo', 'bar']
['foo']