Suppose one below tree-like structure in networkx
graph:
n-----n1----n11
| |----n12
| |----n13
| |----n131
|----n2 |
| |-----n21 X
| |-----n22 |
| |----n221
|----n3
n4------n41
n5
Thanks.
Graph construction:
>>> import networkx as nx
>>> G = nx.DiGraph()
>>> G.add_edges_from([('n', 'n1'), ('n', 'n2'), ('n', 'n3')])
>>> G.add_edges_from([('n4', 'n41'), ('n1', 'n11'), ('n1', 'n12'), ('n1', 'n13')])
>>> G.add_edges_from([('n2', 'n21'), ('n2', 'n22')])
>>> G.add_edges_from([('n13', 'n131'), ('n22', 'n221')])
>>> G.add_edges_from([('n131', 'n221'), ('n221', 'n131')]
>>> G.add_node('n5')
Using the out_degree function to find all the nodes with children:
>>> [k for k,v in G.out_degree().iteritems() if v > 0] ['n13', 'n', 'n131', 'n1', 'n22', 'n2', 'n221', 'n4']
Note that n131 and n221 also show up here since they both have an edge to each other. You could filter these out if you want.
All nodes without children:
>>> [k for k,v in G.out_degree().iteritems() if v == 0] ['n12', 'n11', 'n3', 'n41', 'n21', 'n5']
All orphan nodes, ie nodes with degree 0:
>>> [k for k,v in G.degree().iteritems() if v == 0] ['n5']
To get all orphan "edges", you can get the list of components of the graph, filter out the ones that don't contain n
and then keep only the ones that have edges:
>>> [G.edges(component) for component in nx.connected_components(G.to_undirected()) if len(G.edges(component)) > 0 and 'n' not in component] [[('n4', 'n41')]]
Nodes with more than 2 children:
>>> [k for k,v in G.out_degree().iteritems() if v > 2] ['n', 'n1']
If you traverse the tree, you will not get an infinite loop. NetworkX has traversal algorithms that are robust to this.
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