ranking vertices by value of vertex property
Hi, I ran PageRank on my graph, and now I need to get the top-k highest scored. Is there any faster / more straightforward way of achieving this than by creating and then sorting a new dictionary, as in the following method? def top_k_page_rank(g, pagerank_prop, k): pr_sorted_scores = sort_pagerank_scores(g, pagerank_prop) top_k = list(pr_sorted_scores.items())[0:k] return top_k def sort_pagerank_scores(g, pagerank_prop): pr_scores = dict() for v in g.vertices(): pr_scores[v] = g.vertex_properties[pagerank_prop][v] pr_sorted_scores = {k: v for k, v in sorted(pr_scores.items(), key=lambda item: item[1], reverse=True)} return pr_sorted_scores Thanks, Ioana -- Sent from: https://nabble.skewed.de/
Am 07.07.20 um 15:39 schrieb Ioana K-Hulpus:
Hi, I ran PageRank on my graph, and now I need to get the top-k highest scored. Is there any faster / more straightforward way of achieving this than by creating and then sorting a new dictionary, as in the following method?
Yes, you can access property map values as numpy arrays via the ".a" attribute, e.g. if pr is a vertex property map with the pagerank scores, then pr.a will give you a numpy array. This means you can use the argsort() method to obtain the indexes (i.e. vertices) in increasing order: pr.a.argsort() Best, Tiago -- Tiago de Paula Peixoto <tiago@skewed.de>
Great, thanks a lot! -Ioana On Tue, Jul 7, 2020 at 4:29 PM Tiago de Paula Peixoto <tiago@skewed.de> wrote:
Am 07.07.20 um 15:39 schrieb Ioana K-Hulpus:
Hi, I ran PageRank on my graph, and now I need to get the top-k highest scored. Is there any faster / more straightforward way of achieving this than by creating and then sorting a new dictionary, as in the following method?
Yes, you can access property map values as numpy arrays via the ".a" attribute, e.g. if pr is a vertex property map with the pagerank scores, then
pr.a
will give you a numpy array. This means you can use the argsort() method to obtain the indexes (i.e. vertices) in increasing order:
pr.a.argsort()
Best, Tiago
-- Tiago de Paula Peixoto <tiago@skewed.de>
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participants (3)
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Ioana Hulpus -
Ioana K-Hulpus
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Tiago de Paula Peixoto