Hi Vinayak,

Try:

g.V().project('v', 'vcount', 'ecount').by(identity()).by(count()).by(bothE().count())

Best wishes,    Marc

Hi Vinayak

You could do this:

g.V().hasLabel('A').as('a').outE().as('e').inV().as('b').select('a','e','b').by(count())

That should produce something like:

==>{a=1, e=1, b=1}

Best regards,
 Graham

Graham Wallis
IBM Open Software
Internet: graham_wallis@...    
IBM, Hursley Park, Hursley, Hampshire SO21 2JN






From:        "Vinayak Bali" <vinayakbali16@...>
To:        janusgraph-users@...
Date:        23/02/2021 09:11
Subject:        [EXTERNAL] Re: [janusgraph-users] Gremlin Query to return count for nodes and edges
Sent by:        janusgraph-users@...

---

Hi Marc, I am using the following query to return the results. g.V().hasLabel('A').as('a').outE().as('e').inV().as('b').select('a','e','b').by(valueMap().by(unfold())) Want the count of
Hi Marc,

I am using the following query to return the results.
g.V().hasLabel('A').as('a').outE().as('e').inV().as('b').select('a','e','b').by(valueMap().by(unfold()))
Want the count of unique nodes in a and b together and e i.e number of edges.
Please modify this query to get the required output.

Thanks & Regards,
Vinayak

On Tue, Feb 23, 2021 at 1:08 PM <hadoopmarc@...> wrote:

Hi Vinayak,

Try:

g.V().project('v', 'vcount', 'ecount').by(identity()).by(count()).by(bothE().count())

Best wishes,    Marc

Unless stated otherwise above:
IBM United Kingdom Limited - Registered in England and Wales with number 741598.
Registered office: PO Box 41, North Harbour, Portsmouth, Hampshire PO6 3AU

Good query from @hadoopmarc and I like @cmilowka's suggestion, although I needed to modify it very slightly as follows:

g.V().hasLabel('A').union( __.count(), __.outE().count(), __.outE().inV().count() )

That has to be the shortest and neatest solution. Certainly far better than my rather basic effort below, which surely gets the prize for the longest solution :-)

g.V().hasLabel('A').aggregate('a').outE().aggregate('e').inV().aggregate('b').select('a').dedup().as('as').select('e').dedup().as('es').select('b').dedup().as('bs').select('as','es','bs').by(unfold().count())


Best regards,
 Graham

Graham Wallis
IBM Open Software
Internet: graham_wallis@...    
IBM, Hursley Park, Hursley, Hampshire SO21 2JN






From:        "cmilowka" <cmilowka@...>
To:        janusgraph-users@...
Date:        23/02/2021 22:49
Subject:        [EXTERNAL] Re: [janusgraph-users] Gremlin Query to return count for nodes and edges
Sent by:        janusgraph-users@...

---

Unless stated otherwise above:
IBM United Kingdom Limited - Registered in England and Wales with number 741598.
Registered office: PO Box 41, North Harbour, Portsmouth, Hampshire PO6 3AU

Hi Vinayak,

Speeding up your query depends on your setup. 15.000 vertices/second is already fast. Is this the janusgraph inmemory backend? Or ScyllaDB?

In a perfect world, not there yet, your query would profit from parallelization (OLAP). JanusGraph supports both the withComputer() and withComputer(SparkGraphComputer) start steps, but the former is undocumented and the performance gains of the latter are often disappointing.

Best wishes,    Marc