How to compare two arrays in Gremlin

[Alex Kumundzhiev]

Hello. I need a pattern where array contains all elements from another array. It is pretty easy in Cypher:

MATCH (:ISSUE {uid: "B"}) -[:ISSUE_FILE]-> (file)

WITH collect(file) AS issue_files_list

UNWIND issue_files_list AS file

MATCH (file) <-[:DEVICE_FILE]- (device)

WITH DISTINCT device, collect(file) AS device_file_list, issue_files_list

WITH device

WHERE all(file IN issue_files_list WHERE file IN device_file_list)

RETURN device

But with Gremlin I can't achieve needed result. I started with following (tried to compare lengths, but without success):

g.V().has('ISSUE', 'uid', 'B')

.map(

    out('ISSUE_FILE').fold()

).as('linked_files')

.unfold()

.in('DEVICE_FILE')

.dedup()

.project('uid', 'files')

.by(values('uid'))

.by(

    out('DEVICE_FILE')

    .where(

        within('linked_files')

    )

    .count()

)

.select('uid', 'files')

.where(

    select('linked_files').unfold().count().is(eq('files'))

)

[Daniel Kuppitz]

My Cypher is a little rusty, but isn't your query just doing the following?

g.V().has('ISSUE', 'uid', 'B').

  out('ISSUE_FILE').

  in('DEVICE_FILE').dedup()

Unless I'm missing something, all the folding and comparing really just ensures that each device in the result is connected to the initial issue (through a file). But obviously (?), that's always true.

Perhaps provide a sample graph if this is not the answer you're looking for.

Cheers,

Daniel

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[Alex Kumundzhiev]

Yes, the question is to ensure that every issue's file is presented in device's files, it is not always true fr every device. And return the list of such devices.

On Monday, May 4, 2020 at 6:31:20 PM UTC+4, Daniel Kuppitz wrote:

My Cypher is a little rusty, but isn't your query just doing the following?

g.V().has('ISSUE', 'uid', 'B').

  out('ISSUE_FILE').

  in('DEVICE_FILE').dedup()

Unless I'm missing something, all the folding and comparing really just ensures that each device in the result is connected to the initial issue (through a file). But obviously (?), that's always true.

Perhaps provide a sample graph if this is not the answer you're looking for.

Cheers,

Daniel

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[Alex Kumundzhiev]

In Cypher this can be done using list comprehension (https://neo4j.com/docs/cypher-manual/current/syntax/lists/#cypher-list-comprehension) that returns new list with boolean [true, true, false], then applying all() to this new list

[Alex Kumundzhiev]

Here is a sample graph in Gremlin:

g

.addV('DEVICE').as('DEVICE_1').property('uid', '1').property('partition_key', '1')

.addV('DEVICE').as('DEVICE_2').property('uid', '2').property('partition_key', '1')

.addV('FILE').as('FILE_A').property('uid', 'A').property('partition_key', '1')

.addV('FILE').as('FILE_B').property('uid', 'B').property('partition_key', '1')

.addV('FILE').as('FILE_C').property('uid', 'C').property('partition_key', '1')

.addV('FILE').as('FILE_D').property('uid', 'D').property('partition_key', '1')

.addV('FILE').as('FILE_E').property('uid', 'E').property('partition_key', '1')

.addV('FILE').as('FILE_F').property('uid', 'F').property('partition_key', '1')

.addV('ISSUE').as('ISSUE_A').property('uid', 'A').property('partition_key', '1')

.addV('ISSUE').as('ISSUE_B').property('uid', 'B').property('partition_key', '1')

.addV('ISSUE').as('ISSUE_C').property('uid', 'C').property('partition_key', '1')

.addE('DEVICE_FILE').from('DEVICE_1').to('FILE_A')

.addE('DEVICE_FILE').from('DEVICE_1').to('FILE_B')

.addE('DEVICE_FILE').from('DEVICE_1').to('FILE_C')

.addE('DEVICE_FILE').from('DEVICE_1').to('FILE_D')

.addE('DEVICE_FILE').from('DEVICE_1').to('FILE_E')

.addE('DEVICE_FILE').from('DEVICE_2').to('FILE_A')

.addE('DEVICE_FILE').from('DEVICE_2').to('FILE_B')

.addE('DEVICE_FILE').from('DEVICE_2').to('FILE_D')

.addE('DEVICE_FILE').from('DEVICE_2').to('FILE_E')

.addE('ISSUE_FILE').from('ISSUE_A').to('FILE_A')

.addE('ISSUE_FILE').from('ISSUE_A').to('FILE_B')

.addE('ISSUE_FILE').from('ISSUE_A').to('FILE_D')

.addE('ISSUE_FILE').from('ISSUE_A').to('FILE_E')

.addE('ISSUE_FILE').from('ISSUE_B').to('FILE_A')

.addE('ISSUE_FILE').from('ISSUE_B').to('FILE_B')

.addE('ISSUE_FILE').from('ISSUE_B').to('FILE_C')

.addE('ISSUE_FILE').from('ISSUE_C').to('FILE_A')

.addE('ISSUE_FILE').from('ISSUE_C').to('FILE_B')

.addE('ISSUE_FILE').from('ISSUE_C').to('FILE_F')

For Issue A it should return Device 1 and 2

For Issue B it should return only Device 1 since it's the only one with files ABC 

[Stephen Mallette]

How about this approach?

gremlin> g.V().has('ISSUE', 'uid', 'A').
......1>   out('ISSUE_FILE').aggregate('expected').
......2>   in('DEVICE_FILE').
......3>   groupCount().
......4>      by('uid').
......5>   unfold().as('e').
......6>   where('expected', eq('e')).
......7>     by(count(local)).
......8>     by(select(values)).
......9>   select(keys)
==>1
==>2
gremlin> g.V().has('ISSUE', 'uid', 'B').
......1>   out('ISSUE_FILE').aggregate('expected').
......2>   in('DEVICE_FILE').
......3>   groupCount().
......4>      by('uid').
......5>   unfold().as('e').
......6>   where('expected', eq('e')).
......7>     by(count(local)).
......8>     by(select(values)).
......9>   select(keys)
==>1

Perhaps there is a nicer way still....I just went with the direct approach of comparing counts of the files with the count of the devices we traversed to.

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[Alex Kumundzhiev]

Thanks a lot for the glimpse. But I need to return device vertices, I'm trying to do it this way:

g.V().has('ISSUE', 'uid', 'A')

.out('ISSUE_FILE').aggregate('expected')

.in('DEVICE_FILE')

.groupCount()

.by('uid')

.unfold().as('e')

.where('expected', eq('e'))

.by(count(local))

.by(select(values))

.select(keys).fold().as('device_uid')

.V().has('DEVICE', 'uid', within('device_uid'))

but it returns nothing, could you suggest something?

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[Stephen Mallette]

Just drop the by('uid') on the groupCount() - that should fix it.

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