Convert python-igraph graph to networkx

Question

Recently I've been working with python-igraph package and all my code is based on graphs I create using igraph. Right now, I need to calculate some measures for my graph which apparently are implemented in networkx and not in igraph such as (katz_centrality_numpy, edge_betweenness_centrality, ...). I am wondering if there is a way to convert one graph to another between these two packages and to avoid reading from files again since my files are huge and have to repeat the same process alot.

By the way, when I pass the igraph graph to a networkx function I receive the following error:

TypeError: 'Graph' object is not iterable

Thanks :)

You can create a new networkx graph - just take the edges of your igraph and generate a networkx graph with them — Ella Sharakanski, Apr 21 '15 at 21:59

Ella Sharakanski · Accepted Answer · 2015-04-22T13:28:09.080

5

You can initiate a networkx graph with edges:

Graph([(1,2), (3,4)])

See the documentation.

EDIT:

This is how to use it (Thank you nimafl for the code):

graph is the igraph graph and we create G which is a networkx graph.

import networkx
A = graph.get_edgelist()
G = networkx.DiGraph(A) # In case your graph is directed
G = networkx.Graph(A) # In case you graph is undirected

edited Apr 22 '15 at 13:28

answered Apr 21 '15 at 22:49

Ella Sharakanski

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Thanks Ella, this helped me figure out how to approach my problem :) – ahajib Apr 21 '15 at 23:37
1

Actually if you had degree 0 nodes you would lose those nodes passing from igraph to networkx – Tommaso Guerrini Apr 16 '20 at 12:57

Ebrahim Byagowi · Answer 2 · 2017-08-08T11:57:27.227

As I try to store names of nodes/edges on both igraph or nx, this is my one-liner version which also transfers nodes names while transferring from igraph object, g, to nx, G, the result:

G = nx.from_edgelist([(names[x[0]], names[x[1]])
                      for names in [g.vs['name']] # simply a let
                      for x in g.get_edgelist()], nx.DiGraph())

Also if you need the reverse way, have a look at this answer.

score 1 · Answer 3 · answered Apr 21 '15 at 23:36

Ok so I figured it out myself. Here is what you should do. Assuming that your python.igraph object is called graph we create a networkx graph called G as following:

import networkx as netx

A = [edge.tuple for edge in graph.es]
# In case your graph is directed
G = netx.DiGraph(A)
# In case you graph is undirected
G = netx.Graph(A)

graph.es returns the graph edge list and then add all of them to A and using matrix A we create a graph in networkx.

Good luck with your codes :)

Instead of writing `[edge.tuple for edge in graph.es]`, you could simply write `graph.get_edgelist()` - it is probably a tad bit faster. — Tamás, Apr 22 '15 at 11:41

score 1 · Answer 4 · answered Jan 25 '23 at 17:47

1

Now you can directly translate from igraph to networkx using:

G = graph.to_networkx()

where "graph" is an igraph graph, for the sake of redundancy.

https://igraph.org/python/doc/api/igraph.Graph.html#to_networkx

answered Jan 25 '23 at 17:47

user2383054

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score 0 · Answer 5 · answered Jan 05 '20 at 04:56

The top answer works nicely, but just in case one needs to preserve the order of the nodes (so that mNgraph.nodes is sorted), and one's igraph.Graph has no node names, consider using something like

mNgraph = networkx.Graph()
mNgraph.add_nodes_from(range(mIgraph.vcount()))
mNgraph.add_edges_from(mIgraph.get_edgelist())

Convert python-igraph graph to networkx

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