Can you simply print the values given by the following? cluster = gt.local_clustering(g) loc_cluster = [c for c in cluster] print(loc_cluster) In my opinion, this is the correct way to calculate local clustering coefficients. Regards Snehal On Wed, Oct 19, 2016 at 5:15 PM, P-M <pmj27@cam.ac.uk> wrote:

I have computed the local clustering coefficient for my network and then created a histogram as follows:

g = gt.load_graph('graph_no_multi.gt')

#"The clustering coefficient is normalized only for _simple_ graphs, with at #most one edge between nodes." gt.remove_parallel_edges(g)

#create new property map clustering = g.new_vertex_property("float")

#calculate clustering coefficient gt.local_clustering(g,prop=clustering,undirected=False)

#Make propery map internal g.vp.clust = clustering #Initiliase dictionary containing list of clustering coefficients for given degree clust_k_hist={} for v in g.vertices(): k = v.out_degree() c=g.vp.clust[v] if k in clust_k_hist: clust_k_hist[k].append(c) else: clust_k_hist[k]=[c]

If I now however type `print max(clust_k_hist[0])` I get an answer of `2` which surprises me slightly (similarly for `k=1` I get a value of `2`). Firstly I wasn't expecting a clustering coefficient greater than `1` but also for a degree of `0` I would expect to find only clustering coefficients of `0`. The documentation states that the outdegree is being used in calculating the local clustering coefficient so I am using the outdegree for compiling the histogram.

Have I gone wrong somewhere?

-- View this message in context: http://main-discussion-list- for-the-graph-tool-project.982480.n3.nabble.com/Local- Clustering-Coefficient-for-k-1-tp4026696p4026772.html Sent from the Main discussion list for the graph-tool project mailing list archive at Nabble.com. _______________________________________________ graph-tool mailing list graph-tool@skewed.de https://lists.skewed.de/mailman/listinfo/graph-tool

-- Snehal Madhukar Shekatkar Pune India

On 19.10.2016 12:45, P-M wrote:

If I now however type `print max(clust_k_hist[0])` I get an answer of `2` which surprises me slightly (similarly for `k=1` I get a value of `2`). Firstly I wasn't expecting a clustering coefficient greater than `1` but also for a degree of `0` I would expect to find only clustering coefficients of `0`. The documentation states that the outdegree is being used in calculating the local clustering coefficient so I am using the outdegree for compiling the histogram.

Have I gone wrong somewhere?

Is your graph really directed? I.e. what does g.is_directed() say? -- Tiago de Paula Peixoto <tiago@skewed.de>

On 19.10.2016 15:59, P-M wrote:

Yup, g.is_directed() returns True.

I can't reproduce the problem. Please post an example of a graph that shows the strange behavior. -- Tiago de Paula Peixoto <tiago@skewed.de>

On 19.10.2016 16:30, P-M wrote:

I will try my best. Are you aware of an easy way of extracting a relevant section of the graph? (The whole thing is in the million of vertices so probably too large to post here.)

The clustering coefficient only depends on the local neighborhood of a node. So you can just extract a subgraph that contains the node in question and its neighbors. -- Tiago de Paula Peixoto <tiago@skewed.de>

Attached is a graph which exhibits the behaviour. graph2.gt <http://main-discussion-list-for-the-graph-tool-project.982480.n3.nabble.com/file/n4026784/graph2.gt> Here is a screenshot of what graph-tool returns for me: <http://main-discussion-list-for-the-graph-tool-project.982480.n3.nabble.com/file/n4026784/Capture.png> And here is a visualisation of the graph: <http://main-discussion-list-for-the-graph-tool-project.982480.n3.nabble.com/file/n4026784/clust.png> -- View this message in context: http://main-discussion-list-for-the-graph-tool-project.982480.n3.nabble.com/... Sent from the Main discussion list for the graph-tool project mailing list archive at Nabble.com.

Hmm. I could reproduce this. Looks like gt somehow ignores that out_degree is 0 and is counting it to be 3 (which is actually in-degree). Then the maximum number of links among neighbours would be at most 3C2 = 3 but actually 6 are present. So it simply takes the ration 6/3 = 2. Tiago would be able to say more about this though. Best Snehal On Wed, Oct 19, 2016 at 9:47 PM, P-M <pmj27@cam.ac.uk> wrote:

Here is a more helpful visualisation with the vertices labelled: <http://main-discussion-list-for-the-graph-tool-project. 982480.n3.nabble.com/file/n4026785/clust.png>

So the issue lies with vertex 0 which has an outdegree of 0 and an indegree of 3 but a clustering coefficient of 2.

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-- Snehal Madhukar Shekatkar Pune India

Great; thank you very much! Do you have any idea when it might make its way into the apt-get repository? Running Ubuntu 14.04 I recall building the package myself having been a major pain due to Boost. -- View this message in context: http://main-discussion-list-for-the-graph-tool-project.982480.n3.nabble.com/... Sent from the Main discussion list for the graph-tool project mailing list archive at Nabble.com.

On 21.10.2016 17:37, P-M wrote:

Could I just briefly follow up on this? I realise this may not be a priority for you (which is fine) but if you have a rough estimate of when it make its way into the stable release (days, weeks, months) that would greatly help me to decide whether to embark on spending a lot of time on trying to compile the git version it or whether it is just a matter of waiting a few more days.

It is difficult to make a prediction, so I would rather not make one. If you need to use it ASAP, you are better off compiling it yourself. -- Tiago de Paula Peixoto <tiago@skewed.de>