How to get count of intermediate vertices? - gremlin

Question

Let's say there is a tree-like structure.
Top level: Warehouse
Next level: Storage space
Last level: Stored item
I want to get count of Storage spaces and Storage items per each warehouse.
I've already tried to get the number of Stored items: it's done pretty easily using groupCount.
g.V().
hasLabel('Warehouse').
as('w').
out('HAS_SPACE').
hasLabel('Space').
as('s').
out('HAS_ITEM').
hasLabel('Item').
groupCount().by(select('w')).
unfold().
order().by(values, desc).
limit(100).
project('WarehouseName', 'ItemsCount').
by(select(keys).values('Name')).
by(select(values))
However I want to get count of 's' as well and I can't think of any fast way to achieve it. I've thought about counting traversals something like:
g.V().
hasLabel('Warehouse').
project('WarehouseName', 'SpaceCount', 'ItemCount').
by('Name').
by(out('HAS_SPACE').count()).
by(out('HAS_SPACE').out('HAS_ITEMS').count())
but it works extremely slow on large number of vertices (there are about 26M).
Is there any other way to get that count?

Answer 1

You can use groupCount as sideEffect by giving it a name:
g.V().hasLabel('Warehouse').as('w')
.out('HAS_SPACE').hasLabel('Space').as('s')
.groupCount('spaceCount').by(select('w'))
.out('HAS_ITEM').hasLabel('Item')
.groupCount('itemsCount').by(select('w'))
.count().select('spaceCount', 'itemsCount')
Note that this will return 2 maps, one for spaces and one for items.
If you need to get it as a single map you can replace last line with:
.count().union(select('spaceCount'),select('itemsCount')).unfold()
.group().by(keys).by(select(values).fold()).unfold()
The result will be a map of arrays, of which first value is spaces count and second is items count.