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In networkX, I have a tree as DiGraph().

#!/usr/bin/python
# -*- coding: utf-8 -*-
import networkx as nx
t = nx.DiGraph()
t.add_edge(1,'r')
t.add_edge(2,'r')
t.add_edge(3,'r')
t.add_edge(4,2)
t.add_edge(5,2)
t.add_edge(6,5)
print t.edges()

If a take the node 2 of tree.
how I can get the subtree of 2 ?

Edit

I expected this subtree

[(4,2),(5,2),(6,5)]
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JuanPablo
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  • Can you be more specific about what you mean by "subtree of 2" in your oriented/directed tree? – Aric Oct 25 '11 at 17:00

1 Answers1

12

If you mean the subtree rooted at node 2, that's

from networkx.algorithms.traversal.depth_first_search import dfs_tree

subtree_at_2 = dfs_tree(t, 2)

Edit: it seems you've reversed the order of nodes in your edges. In a directed tree, all paths proceed from the root to a leaf, not the other way around. dfs_tree(t.reverse(), 2) gives you the tree that you want, but do change your code.

Fred Foo
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  • with this I get [(2, 'r')] and not is a subtree – JuanPablo Oct 25 '11 at 18:02
  • @JuanPablo: that certainly *is* the subtree rooted at 2 in your graph. You've reversed the edges. – Fred Foo Oct 25 '11 at 18:15
  • and in a undirected tree, how I can use dfs_tree ? – JuanPablo Oct 25 '11 at 18:22
  • @JuanPablo: in an undirected tree, the notion of a unique subtree rooted at a node simply disappears; no can do. (There are still subtrees, but not unique ones per node.) You can still apply `dfs_tree`, but it will give you a directed tree containing all nodes, rooted at the node you give as an argument. Try `nx.draw(dfs_tree(t.to_undirected(), 2))`, for example. – Fred Foo Oct 25 '11 at 18:24