Optional Match Queries

[Udit Sharma]

How to write a Match query where some of the graph patterns are Optional?

Example:

g.V().match(__.as('temp').has('id',"dept0"),

__.as('temp').in('wf').as('X'),

__.as('X').has('type','prof'),

__.as('X').out('to').as('Z'),

__.as('Z').in('tc').as('Y'),

__.as('Y').out('a').as('X')).select('X','Y','Z')

We want last three match patterns to be optional.

[Daniel Kuppitz]

If all 3 are optional, then as('Y').out('a').as('X') is not really relevant and can be removed. That leaves us with:

g.V().match(__.as('temp').has('id',"dept0"),

            __.as('temp').in('wf').as('X'),

            __.as('X').has('type','prof'),

            __.as('X').out('to').as('Z'),

            __.as('Z').in('tc').as('Y')).select('X','Y','Z')

Now making as('X').out('to').as('Z') optional is simple:

g.V().match(__.as('temp').has('id',"dept0"),

            __.as('temp').in('wf').as('X'),

            __.as('X').has('type','prof'),

            __.as('X').coalesce(out('to'), constant("N/A")).as('Z'),

            __.as('Z').in('tc').as('Y')).select('X','Y','Z')

However, that's an invalid traversal as we can't do as('Z').in('tc') if Z is a String value (N/A), hence the last pattern has to be optional and also take the possible String value into account:

g.V().match(__.as('temp').has('id',"dept0"),

            __.as('temp').in('wf').as('X'),

            __.as('X').has('type','prof'),

            __.as('X').coalesce(out('to'), constant("N/A")).as('Z'),

            __.as('Z').choose(__.is("N/A"),

                                constant("N/A"),

                                coalesce(__.in("tc"), constant("N/A"))).as('Y')).select('X','Y','Z')

But in general, I would say that you shouldn't use match() for patterns that are optional. As you can see, it gets cumbersome pretty quickly.

Cheers,

Daniel

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[Udit Sharma]

Thanks Daniel. I had two more questions.

1. How to modify this query if identifying non-relevant match pattern is difficult(in case of large number of patterns)?

2. How to write this query without using match()?

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[Daniel Kuppitz]

Can you please provide a small sample graph for your query? This will make it a lot easier to compare the 2 approaches.

Cheers,

Daniel

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[Udit Sharma]

Hi Daniel,

 

Sample Graph

g=TinkerGraph.open().traversal()

g.addV().property('type','student').property('id','student1').as('s1').

addV().property('type','student').property('id','student2').as('s2').

addV().property('type','student').property('id','student3').as('s3').

addV().property('type','student').property('id','student4').as('s4').

addV().property('type','course').property('id','course1').as('c1').

addV().property('type','course').property('id','course2').as('c2').

addV().property('type','course').property('id','course3').as('c3').

addV().property('type','teacher').property('id','teacher1').as('t1').

addE('takesCourse').from('s1').to('c1').

addE('takesCourse').from('s2').to('c2').

addE('takesCourse').from('s3').to('c3').

addE('teacherOf').from('t1').to('c1').

addE('teacherOf').from('t1').to('c2')

Query1

g.V().match(__.as('X').has('type','student'),

__.as('X').out('takesCourse').as('Y'),

__.as('Y').in('teacherOf').as('Z')).select('X','Y','Z')

Output

==>[X:v[0],Y:v[12],Z:v[21]]

==>[X:v[3],Y:v[15],Z:v[21]]

Expected Output: 

==>[X:v[0],Y:v[12],Z:v[21]]

==>[X:v[3],Y:v[15],Z:v[21]]

==>[X:v[6],Y:N/A,Z:N/A]

==>[X:v[9],Y:N/A,Z:N/A]

It should print 

1. All X that matches pattern 1.

2. Y and Z if both of them matches pattern 2 and 3, otherwise print N/A for both of them.

Query 2:(Your approach)

g.V().match(__.as('X').has('type','student'),

__.as('X').coalesce(out('takesCourse'), constant("N/A")).as('Y'),

__.as('Y').choose(__.is("N/A"),constant("N/A"),coalesce(__.in('teacherOf'), constant("N/A"))).as('Z')).select('X','Y','Z')

Output

==>[X:v[0],Y:v[12],Z:v[21]]

==>[X:v[3],Y:v[15],Z:v[21]]

==>[X:v[6],Y:v[18],Z:N/A]

==>[X:v[9],Y:N/A,Z:N/A]

How to modify the query to obtain the output as expected?

 

[Daniel Kuppitz]

It should print 
1. All X that matches pattern 1.
2. Y and Z if both of them matches pattern 2 and 3, otherwise print N/A for both of them.

Then coalesce() is your friend again:

gremlin> g.V().has('type','student').

           coalesce(

             __.as('X').

               out('takesCourse').as('Y').

               in('teacherOf').as('Z').

               select('X','Y','Z'),

             project('X','Y','Z').

               by().

               by(constant('N/A')).

               by(constant('N/A')))

==>[X:v[0],Y:v[12],Z:v[21]]

==>[X:v[3],Y:v[15],Z:v[21]]

==>[X:v[6],Y:N/A,Z:N/A]

==>[X:v[9],Y:N/A,Z:N/A]

Cheers,

Daniel

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[Udit Sharma]

Thanks Daniel. That was great. But, I am not able to extend your approach to other queries.

Sample Graph

g=TinkerGraph.open().traversal()

g.addV().property('type','student').property('id','student1').as('s1').

addV().property('type','course').property('id','course1').as('c1').

addV().property('type','teacher').property('id','teacher1').as('t1').

addV().property('type','teacher').property('id','teacher2').as('t2').

addV().property('type','department').property('id','department1').as('d1').

addE('worksFor').from('t1').to('d1').

addE('worksFor').from('t2').to('d1').

addE('headOf').from('t1').to('d1').

addE('teacherOf').from('t1').to('c1').

addE('takesCourse').from('s1').to('c1').

addE('advisor').from('s1').to('t1')

Original Query

g.V().match(__.as('temp').has('id',"department1"),

__.as('temp').in('worksFor','headOf').as('X'),

__.as('X').out('teacherOf').as('Z'),

__.as('Z').in('takesCourse').as('Y'),

__.as('Y').out('advisor').as('X'))

Expected Output: Last three patterns are optional

==>[temp:v[12],X:v[6],Y:v[0],Z:v[3]]

==>[temp:v[12],X:v[6],Y:v[0],Z:v[3]]

==>[temp:v[12],X:v[9] Y:N/A,Z:N/A]

[Daniel Kuppitz]

Whenever you have something that's optional, but you still want to get a value, use coalesce():

gremlin> g.V().has('id','department1').as('temp').

           in('worksFor','headOf').as('X').

           coalesce(

             out('teacherOf').as('Z').

               in('takesCourse').as('Y').

               out('advisor').where(eq('X')).

               select('temp','X','Y','Z'),

             constant('N/A').as('Y','Z').

               select('temp','X','Y','Z'))

==>[temp:v[12],X:v[9],Y:N/A,Z:N/A]

==>[temp:v[12],X:v[6],Y:v[0],Z:v[3]]

==>[temp:v[12],X:v[6],Y:v[0],Z:v[3]]

Cheers,

Daniel

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[Udit Sharma]

Thanks Daniel,

I got your point. But, I have a query regarding coalesce step:

https://groups.google.com/d/msgid/gremlin-users/6c2b45e7-6e36-4bea-9184-39d12c3ec959%40googlegroups.com