Similar Nodes

[Michka]

Hi,

i'm looking for a cypher that detects other nodes with the same relationships.

In my example i have persons and skills. This nodes are conenct with the relation has.

Person A has skills like "php" and "mysql".
Person B has skills like "php", "mysql" and "c#"
Person C has skills like "php", "c#", "ios"

Person B can be an substitute for Person A. Starting with Person A, the query should find Person B.
Unfortunately, I have no idea what s the right query for this.

Do you have any idea? Thank you very much.

[ganesanmithun32]

check jaccard similarity ,
https://neo4j.com/docs/graph-algorithms/current/labs-algorithms/jaccard/

guess this will help you

[Michka]

Thanks for the answer, but that's not what I'm looking for.
It is easy to formulate the statement in SQL, maybe the SQL says more than my text:

person A has the ID 7

SELECT personid from skillofperson where personId != 7 and
skillId in (SELECT skillid FROM `skillofperson `  where personId = 7)
group by personid
having count(skillId) = (SELECT count(skillid) FROM `skillofperson `  where personId = 7)

in response i get a list of people who have at least the skills that person7 has
i can't translate that into cypher - can you?

[andrew_bowman · Accepted solution]

We have a knowledge base article on performing MATCH intersection that may be helpful here, but with some adjustments for this particular case.

This might work for you:

MATCH (start:Person {name:'Person A'})
WITH start, size((start)-[:has]-()) as skillCount
MATCH (start)-[:has*2]-(similarPerson)
WITH start, skillCount, similarPerson, count(similarPerson) as skillsInCommon
WHERE skillCount = skillsInCommon
RETURN similarPerson.name

This should only find persons that have all of the skills that person A has (though they may have more than that).

[Michka]

Thank you!
That is what I was looking for