Marko Rodriguez-2 wrote: > >> >>> "For all nodes in a particular index, how many other nodes are they >>> connected to at depth X?" > > Here is how I would do it -- groupCount is not needed. > > g.idx(index_name)[[key:value]].both.loop(1){it.loops < depth}.count() > >
Thanks Marko. A couple questions... Won't this count dupes more than once? Xavier's requirements of "how many other nodes are they connected" sounds like you should only count uniques, and that's why I am checking the size of groupCount map instead of using count(). Instead of a map you could use a Set with aggregate(), but I wasn't sure if they'd have the aggregate-loop fix yet. Also Xavier said, "For all nodes in a particular index". I took that to mean all nodes in an index, not all nodes for a particular value in an index, hence the wildcard query: index_nodes = g.idx(index_name).get(index_key,Neo4jTokens.QUERY_HEADER + "*") However, I am not sure/can't remember if you can do a wildcard query without at least one leading character. - James -- View this message in context: http://neo4j-community-discussions.438527.n3.nabble.com/Neo4j-Aggregate-queries-tp3317720p3319768.html Sent from the Neo4j Community Discussions mailing list archive at Nabble.com. _______________________________________________ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user