Gremlin: division after groupCount

0
votes

I am using Gremlin to query Neptune.

I have 2 counts

  1. g.V().hasLabel(*).outE.inV().groupCount().by('name') result is like : 'a':2, 'b':4
  2. g.V().hasLabel(*).count() 4

How can I write a single query to get the numbers that result 1 divided by result 2? i.e. 'a': 0.5, 'b': 1

1
graph-databasesgremlinamazon-neptune

1 Answers

2
votes

I can think of a few ways, but I guess using match() is the easiest:

g.V().hasLabel(*).
  union(count(),
        out().groupCount().by('name')).fold().
  match(__.as('values').limit(local, 1).as('c'),
        __.as('values').tail(local, 1).unfold().as('kv'),
        __.as('kv').select(values).math('_/c').as('v')).
  group().
    by(select('kv').by(keys)).
    by(select('v'))

A similar query on the modern graph:

gremlin> g = TinkerFactory.createModern().traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
gremlin> g.V().union(count(),
......1>             out().groupCount().by(label)).fold().
......2>   match(__.as('values').limit(local, 1).as('c'),
......3>         __.as('values').tail(local, 1).unfold().as('kv'),
......4>         __.as('kv').select(values).math('_/c').as('v')).
......5>   group().
......6>     by(select('kv').by(keys)).
......7>     by(select('v'))
==>[software:0.6666666666666666,person:0.3333333333333333]

The next one is probably harder to understand, but would be my personal favorite (because a) I don't like match() and b) it doesn't rely on the order of the results returned by union()):

gremlin> g.V().
......1>   groupCount('a').
......2>     by(constant('c')).
......3>   out().
......4>   groupCount('b').
......5>     by(label).
......6>   cap('a','b').as('x').
......7>   select('a').select('c').as('c').
......8>   select('x').select('b').unfold().
......9>   group().
.....10>     by(keys).
.....11>     by(select(values).math('_/c'))
==>[software:0.6666666666666666,person:0.3333333333333333]