Quickest way to transform a graph into a complete one using networkx

Question

I want to compare the performance of two graphs, the second one being a K_n complete graph. Let's call G the first graph (the non-complete one) and H the second one (the K_n graph). I want the nodes from H to have the exact same attributes (and relative values) as the ones from G, so (apparently) I can't just run networkx.complete_graph(n). My first idea was to use this code:

H = G.copy()
nh = H.nodes()
eh = H.edges() # I create those variables in hope to save some speed
for u in nh:
    for v in filter(lambda x: x > u, nh): # it's an undirected graph, saves some speed
        if (u,v) not in eh:
            H.add_edge(u,v)

However, since my graphs have 10,000+ nodes, this becomes quite a slow process. Is there a way to speed it up?

I think the answer you've got is likely the best option, but just a couple comments on this code - testing if something is in a list is slow. Instead you would want to make eh a set. Also, I don't think there's much value in testing whether something is already an edge - I'm not sure that adding an edge which already exists is significantly slower than first testing if the edge exists. — Joel
– Joel, Commented Nov 11, 2016 at 21:39

Tommy · Accepted Answer · 2016-11-11 16:56:34Z

1

Ok, I may have found an answer:

H = networkx.complete_graph(n)
labels = nx.get_node_attributes(G,attribute)
nx.set_node_attributes(H,attribute,labels)

Does anyone have a faster alternative?

answered Nov 11, 2016 at 16:56

Tommy

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