Hi,
Thanks James. Here is how I would do it -- groupCount is not needed.
g.idx(index_name)[[key:value]].both.loop(1){it.loops < depth}.count()
Note: Be wary of this query. Make sure the branch factor of your graph is
sufficiently small or the depth to which you are exploring is sufficiently
small. With a large branch and depth, you can easily touch everything in your
graph if your graph has "natural statistics." (
http://en.wikipedia.org/wiki/Scale-free_network )
Also, if you want to get fancy, what I like to do, is unroll my loops to
increase performance. Given that the "while" construct of your loop step is
simply "< depth," you can append an appropriate number of .both steps.
traversal = g.idx(index_name)[[key:value]];
for(i in 0..depth) {
traversal = traversal.both;
}
traversal.count();
Finally, 'both' is for undirected traversals. Use 'out' for outgoing traversals
(follow the direction of the arrows) and 'in' for incoming traversals.
https://github.com/tinkerpop/gremlin/wiki/Gremlin-Steps
HTH,
Marko.
http://markorodriguez.com
>
> Xavier Shay wrote:
>>
>> "For all nodes in a particular index, how many other nodes are they
>> connected to at depth X?"
>>
>
> Marko will be able to improve upon this, but try something like this (this
> is untested)...
>
> m = [:]
> depth = 10
> index_name = "vertices"
> index_key = "name"
> index_nodes = g.idx(index_name).get(index_key,Neo4jTokens.QUERY_HEADER +
> "*").
> index_nodes._().both.groupCount(m).loop(2){it.loops < depth}
> m.size()
>
> - James
>
>
> --
> View this message in context:
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