I want to get a ResultSet of Objects that consist of a user and a list of all users that are not yet related to this user. The result should look like this:
[[user: [USEROBJECT], usersThatAreNotFriends: [[USEROBJECT]...]]...]
I am using the Cosmos DB Gremlin Endpoint and struggle with filtering/combining the users that are already related and all users.
My idea was:
g.V().hasLabel('user').as('user').flatMap(g.V().hasLabel('user').where(__.eq(select('user').out('isFriend')).fold()).as('usersThatAreNotFriends').select('user', 'usersThatAreNotFriends')
To set up my example use:
g.addV('user').property('id','user_1').property('partition_key','1')
g.addV('user').property('id','user_2').property('partition_key','2')
g.addV('user').property('id','user_3').property('partition_key','3')
g.addV('user').property('id','user_4').property('partition_key','4')
g.V('user_1').addE('has_relation').to(g.V('user_2'))
g.V('user_2').addE('has_relation').to(g.V('user_1'))
g.V('user_2').addE('has_relation').to(g.V('user_4'))
g.V('user_4').addE('has_relation').to(g.V('user_2'))
The expected result should be represented in simple way:
[user_1: [user_3, user_4], user_2:[user_3],
user_3:[user_1, user_2, user_3], user_4:[user_1, user_3]]
I modified your query to add the IDs as real IDs
g.addV('user').property(id,'user_1').property('partition_key','1')
g.addV('user').property(id,'user_2').property('partition_key','2')
g.addV('user').property(id,'user_3').property('partition_key','3')
g.addV('user').property(id,'user_4').property('partition_key','4')
g.V('user_1').addE('has_relation').to(g.V('user_2'))
g.V('user_2').addE('has_relation').to(g.V('user_1'))
g.V('user_2').addE('has_relation').to(g.V('user_4'))
g.V('user_4').addE('has_relation').to(g.V('user_2'))
In your example sometimes you showed the same person in the result so I am suggesting two different queries. The first includes the person as not being friends with themselves.
gremlin> g.V().as('p').
......1> project('person','not-friends').
......2> by().
......3> by(V().where(__.not(__.in('has_relation').as('p'))).fold())
==>[person:v[user_3],not-friends:[v[user_3],v[user_2],v[user_1],v[user_4]]]
==>[person:v[user_2],not-friends:[v[user_3],v[user_2]]]
==>[person:v[user_1],not-friends:[v[user_3],v[user_1],v[user_4]]]
==>[person:v[user_4],not-friends:[v[user_3],v[user_1],v[user_4]]]
If you want to avoid having the person appear in the results as not friends with themselves you can do this:
gremlin> g.V().as('p').
......1> project('person','not-friends').
......2> by().
......3> by(V().where(__.not(__.in('has_relation').as('p')).where(neq('p'))).fold())
==>[person:v[user_3],not-friends:[v[user_2],v[user_1],v[user_4]]]
==>[person:v[user_2],not-friends:[v[user_3]]]
==>[person:v[user_1],not-friends:[v[user_3],v[user_4]]]
==>[person:v[user_4],not-friends:[v[user_3],v[user_1]]]
As a sidenote to find who people are friends you can use a simple group().by() approach.
gremlin> g.V().aggregate('all').group().by().by(out().fold()).unfold()
==>v[user_3]=[]
==>v[user_2]=[v[user_1], v[user_4]]
==>v[user_1]=[v[user_2]]
==>v[user_4]=[v[user_2]]