Query node path - Subset of node with property value filtering

Neo4j Graph Platform Cypher
1

Hello. I am very new in neo4j and cypher.

IMG_0102
IMG_01021585×627 72.5 KB

I have nodes between which there may be links.
Each node has a field, which I have now called a subject.
My task would be to be able to select the complete paths where the nodes following each other on the path have a certain subject.
Previously I used mongo dbt, there I used the $setIsSubset function to search for this.

Can someone give me a tutorial on how I can implement this in cypher .

MATCH p=(start:Annotation)-[:connection*1..100]->(end:Annotation)
WITH nodes(p) as path
  WHERE ANY(node in path WHERE node.SubjectId ="5f02e8254850751ffc044c8a" OR  node.SubjectId ="6493f28567d762570722e613")
RETURN path
LIMIT 150

I tried something like this, but this is obviously not a good solution.

Thanks for the help!

2

Hi @gergelyurbancsik :call_me_hand:

There's a whole documentation about the new Quantified Patter Matching syntax here.

In this case, you can start trying something like:

MATCH p=(start:Annotation)((a)-[:connection]->(b) WHERE a.SubjectId in ["5f02e8254850751ffc044c8a", "6493f28567d762570722e613"] and b.SubjectId in ["5f02e8254850751ffc044c8a", "6493f28567d762570722e613"]){1,100}(end:Annotation)
RETURN p //edited
LIMIT 150

You may like to add some conditions on your start and end node in order to avoid calculating the same thing multiple times tho.

1 Like
3

Hi @bennu_neo
First of all, thanks for the help, and the reply.
Maybe I explained it wrong, but this is not a good solution.
this will return only the two nodes for which the condition is met.
However, I want to get the full path back.

For example:
a=annotation
s=subject

a1(s0) - > a2(s1) -> a3(s2) -> a4(s3)
a5(s4) - > a2(s1) -> a3(s5) -> a4(s2)

find the graph where the subjects of the successive nodes are s1 and s2.
so this is true for the first, not for the second.
and I would like to get this result back:
a1(s0) - > a2(s1) -> a3(s2) -> a4(s3)

can you help me with this ?

4

Hello hello!

I just edited the answer to return p. The query should return path were all the elements along fulfill the where condition. It should return a path.

5

Hello @bennu_neo
maybe something is still misunderstood,
the longest path in my database:

image
image1556×487 38.9 KB

Query for this:

MATCH p=(start:Annotation)-[:connection*1..100]->(end:Annotation)

RETURN p

ORDER BY length(p) DESC

LIMIT 1

within this graph there are two nodes where the two ids follow each other: 64a592c5868c0d0007218fef, 64a592c5868c0d0007218fee.

so if I substitute this into your query, I want to see the same result as the longest path query above, i.e. the full path.
instead I get the result shown in the picture:

image
image1533×620 41.4 KB

6

Hello @gergelyurbancsik !

So you may like something like:

MATCH p =()(()-[:C]->()){,100}(b)-[:C]->(c)(()-[:C]->()){,100}()
WHERE b.name = "B"
AND c.name = "C"
RETURN p
7

Hello @bennu_neo
I think this is still not the expected result
For this query:

MATCH p =()(()-[:connection]->()){,100}(b)-[:connection]->(c) (()-[:connection]->()){,100}()
WHERE b.SubjectId ='5f02e8254850751ffc044c8a'
    AND c.SubjectId ='6493f28567d762570722e613'
RETURN p LIMIT 100

The result is:

image
image1126×582 28.2 KB

And for this query:

MATCH p=(start:Annotation)((a)-[:connection]->(b)

WHERE a.SubjectId in ["5f02e8254850751ffc044c8a", "6493f28567d762570722e613"]

and b.SubjectId in ["5f02e8254850751ffc044c8a", "6493f28567d762570722e613"]){1,100}(end:Annotation)

RETURN p

LIMIT 150

The result is:

image
image1098×553 35.2 KB

Update:
ahh, own mistake

With this query:

MATCH p =()(()-[:connection]->()){,100}(b)-[:connection]->(c) (()-[:connection]->()){,100}()

WHERE b.SubjectId in ["5f02e8254850751ffc044c8a", "6493f28567d762570722e613"]

and c.SubjectId in ["5f02e8254850751ffc044c8a", "6493f28567d762570722e613"]

RETURN p LIMIT 100

image
image1161×553 35.1 KB

but this is still not the full length of the graph.
I'll have a closer look in the database.
Thanks for your help

8

Thanks @bennu_neo

And what happens if I want to include exactly these nodes anywhere on the graph ?
What operator can be used instead of IN ?
So now if I specify that I'm looking for 3 subjects, it will return subjects where, for example, 2 of the 3 follow each other, but the 3rd is not even listed.

9

Hello hello,

The IN operator has nothing to do with position of the nodes but it only enforce the desired condition.

MATCH p =()(()-[:connection]->()){,100}(b)-[:connection*]->(c)(()-[:connection]->()){,100}()
WHERE b.SubjectId in ["5f02e8254850751ffc044c8a", "6493f28567d762570722e613"]
and c.SubjectId in ["5f02e8254850751ffc044c8a", "6493f28567d762570722e613"]
RETURN p

This one should work for distances between your 2 targets bigger than 1. However, for more intricate scenarios, as always, I may say: "It depends". You can always fallback into path predicates as ANY but be aware those are not enforces runtime but are post traversal filters, so memory can grow a lot before noticing that many of the calculated paths are not valid.

1 Like