Hi,

> > eids = [index for index, is_multiple in enumerate(graph.is_multiple()) if
> > is_multiple]
> 
> There is a small catch here; you are generating a large number of
> eids, whereas in my attempt, I first filtered out the 'unique' edges,
> then among them I find those with 'multiple' counterparts. I assumed
> this is efficient for v.large graphs. Please correct me if I am wrong.
I haven't tested it -- maybe it makes a difference for large graphs, I don't
know. You can measure it from IPython using %timeit.

> Now, I am linking another thread here for a quick question. Can I
> replace the below code with your suggestion found in:
> http://lists.nongnu.org/archive/html/igraph-help/2015-03/msg00024.html?
I'm pretty sure you can.

-- 
T.

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