Hi,
I would do it like this:
m = [:]
g.v(162).in('R_PartOf').loop(1){m.put(it.object, it.loops); true} >> -1
m.sort{a, b -> a.value <=> b.value}.keySet as List
In short, fill up a Map (m) with key being the vertex and value being the
number of "hops" (or times through the loop). Then sort the map by the number
of times through the loop and return the keySet. Since you wanted a List (not a
Set), then "as List" the Set result.
HTH,
Marko.
http://markorodriguez.com
On Nov 4, 2011, at 10:25 AM, baldric wrote:
> I'm very new to much of this, and have a particularly ingrained relational
> slant to my career, unfortunately.
> I'm in the process of doing a proof of concept for a product, using neo4j,
> and gremlin over REST (we're developing in c#) What I'm trying to do at the
> moment is load a graph (representing a sort of flexible taxonomy), into a
> flattened list. So, given a starting node, find all associated nodes across
> a particular relationship type, no matter how far away. I will ultimately
> need this in a list, and I want the list sorted by how far away each node is
> (am I correct in referring to this as "hops"?)
> Anyway, my best attempt at the unsorted query so far looks something like
> this: (by the way, the "unique" part worries me a little....)
>
> x = []; g.v(162).in('R_PartOf').aggregate(x).loop(2){true} >> -1; x.unique()
>
> (Please try not to laugh too much! :) )
>
> How would I alter this to ensure it is sorted by "distance" from starting
> node?
>
> Thanks in advance!
>
> --
> View this message in context:
> http://neo4j-community-discussions.438527.n3.nabble.com/Gremlin-how-to-flatten-a-tree-and-sort-tp3480586p3480586.html
> Sent from the Neo4j Community Discussions mailing list archive at Nabble.com.
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