how to aggregate by edge property

[Martin Junghanns]

Hi,

I'm currently looking into Cypher and Gremlin and I need a little help with grouping in gremlin

in Cypher I write

START n=node(42)  

MATCH n-[:FRIEND_OF*1..2]-()<-[r:REVIEWED_BY]-p  

RETURN p.title, count(p), sum(r.rating) as h  

ORDER BY h DESC;

So, I'd like to know which products where reviewed by my friends and their friends and order them by their rating (which is an edge property).

In Gremlin my first guess was:

m=[:]

v.both('FRIEND_OF').both('FRIEND_OF').inE('REVIEWED_BY').as('r').outV.groupBy(m){it.title}{r.rating.sum()}

But as may already see, the second closure doesn't work as expected because I cannot refer to 'r' again. I also tried to get back to the edges by using {it.outE('REVIEWED_BY').retain(r).rating.sum()} but this didn't work because of the same reason.

A little hint would be very appreciated. Thank you guys in advance.

Martin

[Martin Junghanns]

It works when I store the edges

> r =[]

> m=[:]

> v.both('FRIEND_OF').both('FRIEND_OF').inE('REVIEWED_BY').aggregate(r)

> v.both('FRIEND_OF').both('FRIEND_OF').in('REVIEWED_BY').groupBy(m){it.title}{it.outE('REVIEWED_BY').retain(r).rating}{it.sum()}

is it possible to do this in a single statement?

Martin

[Stephen Mallette]

I tried to convert your problem to the toy graph:

gremlin> g = TinkerGraphFactory.createTinkerGraph()                                

==>tinkergraph[vertices:6 edges:6]

gremlin> m=[:];g.V.outE.groupBy(m){it.inV.next().name}{it.weight}{it.sum()}

==>e[7][1-knows->2]

==>e[8][1-knows->4]

==>e[9][1-created->3]

==>e[12][6-created->3]

==>e[10][4-created->5]

==>e[11][4-created->3]

gremlin> m

==>lop=1.0000000149011612

==>ripple=1.0

==>josh=1.0

==>vadas=0.5

The difference in my approach is to groupBy on the edge and traverse to inV in the first groupBy closure.  Then to order you could just orderMap:

gremlin> g.V.outE.groupBy{it.inV.next().name}{it.weight}{it.sum().doubleValue()}.cap.orderMap(T.decr)

==>lop

==>ripple

==>josh

==>vadas

HTH,

Stephen

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[Martin Junghanns]

Nice solution! Thank you.

Martin