Re: [Neo4j] Gremlin Query - Help
Marko,
I am thinking of using this statement.
g.V.ifThenElse{it.out('hasCentre').filter{it == aCentre}.hasNext()}{}{println
${it} has center unknown}
to get nodes that DO NOT have a relationship, is this the best way to do it? I
guess so, if this is the case, I am now adding support to the .NET graph client
to support ifThenElse expressions. I am wondering though, as I am designing the
interface for it, can and ifThenElse have nested ifThenElse expressions as
well? I guess it and _() can be used within them?
Cheers
-Original Message-
From: user-boun...@lists.neo4j.org [mailto:user-boun...@lists.neo4j.org] On
Behalf Of Marko Rodriguez
Sent: Saturday, 3 December 2011 4:10 AM
To: Neo4j user discussions
Subject: Re: [Neo4j] Gremlin Query - Help
Hi,
Scenario 1:
I would like to get all Nodes that do not have a relationship to another
node. What is the best way to do this Gremlin?
Root = NodeA = NodeB
Root = NodeC
Output should be NodeC
I don't understand the problem.
Scenario 2:
Root = User - Centre
I would like to get all centres for userA, and then get all users who
are also linked to the same centres of the userA
userA.out('hasCentre').in('hasCentre').except([userA])
Scenario 3:
User = Centre
I would like to get all users that do not have a link to a Centre, and if
this is the case, do a projection that returns a fake centre with property
Unknown
g.V.ifThenElse{it.out('hasCentre').filter{it ==
aCentre}.hasNext()}{}{println ${it} has center unknown}
If you have an index of your Users, then its more efficient to hit that index,
then iterate through all vertices (g.V). E.g.
g.idx(T.v)[[type:'User']].ifThenElse
Scenario 4:
REFERRAL = PERSON
I would like to combine a query that gets all Referrals that are
linked to a person and project a table result (contains referrals and
person property values (this is easy to do) However, I then want to do a
SPLIT query, that gets referrals without the persons, and then MERGE both
back into the table projection? So table projecton might look like this.
I guess this is done with .table().it{}cap, but what I am not sure, is
how to do a split and merge in parallel and then get that merge projected
into the same table.
I don't quite understand your problem, but here is an example of splitting and
merging.
g.v(1).out('knows').copySplit(_().out('knows').name,
_().out('created').name).fairMerge
This will get all the people that v[1] knows and then generate two parallel
pipeline. Each friend of v[1] is copied to each of the 2 parallel pipelines.
One pipeline will get their friend's names and one will get their created
project's names. It will then merge those two parallel pipelines into a single
stream. You can do either fairMerge or exhaustMerge depending on the merging
algorithm you want. fairMerge is more memory efficient. I need to write more
about split/merge in the Gremlin documentation.
https://github.com/tinkerpop/gremlin/wiki/Split-Merge-Pattern -- will
work on it :)
HTH,
Marko.
http://markorodriguez.com
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Re: [Neo4j] Gremlin Query - Help
Hi,
Scenario 1:
I would like to get all Nodes that do not have a relationship to another
node. What is the best way to do this Gremlin?
Root = NodeA = NodeB
Root = NodeC
Output should be NodeC
I don't understand the problem.
Scenario 2:
Root = User - Centre
I would like to get all centres for userA, and then get all users who are
also linked to the same centres of the userA
userA.out('hasCentre').in('hasCentre').except([userA])
Scenario 3:
User = Centre
I would like to get all users that do not have a link to a Centre, and if
this is the case, do a projection that returns a fake centre with property
Unknown
g.V.ifThenElse{it.out('hasCentre').filter{it ==
aCentre}.hasNext()}{}{println ${it} has center unknown}
If you have an index of your Users, then its more efficient to hit that index,
then iterate through all vertices (g.V). E.g.
g.idx(T.v)[[type:'User']].ifThenElse
Scenario 4:
REFERRAL = PERSON
I would like to combine a query that gets all
Referrals that are linked to a person and project a table result (contains
referrals and person property values (this is easy to do)
However, I then want to do a SPLIT query, that gets referrals without the
persons, and then MERGE both back into the table projection? So table
projecton might look like this.
I guess this is done with .table().it{}cap, but what I am not sure, is
how to do a split and merge in parallel and then get that merge projected
into the same table.
I don't quite understand your problem, but here is an example of splitting and
merging.
g.v(1).out('knows').copySplit(_().out('knows').name,
_().out('created').name).fairMerge
This will get all the people that v[1] knows and then generate two parallel
pipeline. Each friend of v[1] is copied to each of the 2 parallel pipelines.
One pipeline will get their friend's names and one will get their created
project's names. It will then merge those two parallel pipelines into a single
stream. You can do either fairMerge or exhaustMerge depending on the merging
algorithm you want. fairMerge is more memory efficient. I need to write more
about split/merge in the Gremlin documentation.
https://github.com/tinkerpop/gremlin/wiki/Split-Merge-Pattern -- will
work on it :)
HTH,
Marko.
http://markorodriguez.com
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Re: [Neo4j] Gremlin Query - Help
Marko,
Thank you very much, I really appreciate it, I will start going through your
response here. I have also added the copy/split aggregate functionality into
our Gremlin Client. I was not aware of the if then else as well, will extend
our client to support this.
Apologise, will explain Scenario One.
I have a User Node that has a relationship LINKED_TO_CENTRE
I would like to get a query that can give me all Users that do not have a
relationship LINKED_TO_CENTRE. I was thinking of the Except/Retain pattern for
this, e.g. get all users as aggregate A then get all users linked as aggregate
b - then use except/retain or something along these lines.
On Another note, is there a way to get unique()._() to preserve all previous
pipes and placeholders, the reason is I want to use unique() before doing a
table projection. And unique breaks closures? ( I have a separate post on this
in the mail list)
Marko, if you interested here is how the .NET code looks like that I am
currently trying to write to generate a report. Its not yet finished as I need
to take into consideration your feedback, but I am sure I will get their
eventually.
//ToDo Chained 'As' statements are reversed, bug in gremlin plugin with
chained as - https://github.com/neo4j/community/issues/114
var resultSet = graphClient
.RootNode
.OutAgency(Hosts.TypeKey, a = a.Key ==
userIdentifier.AgencyKey)
.InUser(UserBelongsTo.TypeKey, u = u.Username ==
userIdentifier.Username)
.OutCentre(UserLinkedToCentre.TypeKey)
.As(Centre)
.InUser(UserLinkedToCentre.TypeKey)
//.GremlinDistinct() - breaks pipes for table output
.As(UserGivenName)
.As(UserFamilyName)
.InReferral(CreatedBy.TypeKey, r = r.Completed == false)
.As(ReferralId)
.OutReferralWhoSection(ReferralHasWhoSection.TypeKey)
.As(ReferralDate)
.OutReferralParticipant(HasParticipant.TypeKey)
.As(ParticipantName)
.InReferralWhoSection(HasParticipant.TypeKey)
.InReferral(ReferralHasWhoSection.TypeKey)
//.GremlinDistinct() - breaks pipes for table output
.TableReferralByGroup, Centre, User, User, Referral, Referral,
ReferralParticipant(
centre = centre.Name,
user = user.FamilyName,
user = user.GivenName,
referral = referral.UniqueId,
referral2 = referral2.DateInitiatedUtc,
participant = participant.Name
);
Thank You
Much appreciated
-Original Message-
From: user-boun...@lists.neo4j.org [mailto:user-boun...@lists.neo4j.org] On
Behalf Of Marko Rodriguez
Sent: Saturday, 3 December 2011 4:10 AM
To: Neo4j user discussions
Subject: Re: [Neo4j] Gremlin Query - Help
Hi,
Scenario 1:
I would like to get all Nodes that do not have a relationship to another
node. What is the best way to do this Gremlin?
Root = NodeA = NodeB
Root = NodeC
Output should be NodeC
I don't understand the problem.
Scenario 2:
Root = User - Centre
I would like to get all centres for userA, and then get all users who
are also linked to the same centres of the userA
userA.out('hasCentre').in('hasCentre').except([userA])
Scenario 3:
User = Centre
I would like to get all users that do not have a link to a Centre, and if
this is the case, do a projection that returns a fake centre with property
Unknown
g.V.ifThenElse{it.out('hasCentre').filter{it ==
aCentre}.hasNext()}{}{println ${it} has center unknown}
If you have an index of your Users, then its more efficient to hit that index,
then iterate through all vertices (g.V). E.g.
g.idx(T.v)[[type:'User']].ifThenElse
Scenario 4:
REFERRAL = PERSON
I would like to combine a query that gets all Referrals that are
linked to a person and project a table result (contains referrals and
person property values (this is easy to do) However, I then want to do a
SPLIT query, that gets referrals without the persons, and then MERGE both
back into the table projection? So table projecton might look like this.
I guess this is done with .table().it{}cap, but what I am not sure, is
how to do a split and merge in parallel and then get that merge projected
into the same table.
I don't quite understand your problem, but here is an example of splitting and
merging.
g.v(1).out('knows').copySplit(_().out('knows').name,
_().out('created').name).fairMerge
This will get all the people that v[1] knows and then generate two parallel
pipeline. Each friend of v[1] is copied to each of the 2 parallel pipelines.
One pipeline will get their friend's names and one will get their created
project's names. It will then merge those two
[Neo4j] Gremlin Query - Help
Hi,
Scenario 1:
I would like to get all Nodes that do not have a relationship to another node.
What is the best way to do this Gremlin?
Root = NodeA = NodeB
Root = NodeC
Output should be NodeC
Scenario 2:
Root = User - Centre
I would like to get all centres for userA, and then get all users who are also
linked to the same centres of the userA
Scenario 3:
User = Centre
I would like to get all users that do not have a link to a Centre, and if this
is the case, do a projection that returns a fake centre with property Unknown
Scenario 4:
REFERRAL = PERSON
I would like to combine a query that gets all
Referrals that are linked to a person and project a table result (contains
referrals and person property values (this is easy to do)
However, I then want to do a SPLIT query, that gets referrals without the
persons, and then MERGE both back into the table projection? So table projecton
might look like this.
I guess this is done with .table().it{}cap, but what I am not sure, is how
to do a split and merge in parallel and then get that merge projected into the
same table.
ReferralId, Name
1, Bob
2, Jill
3, null
4, null
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