Neo4j

changecpl · 08-09-2021

We can use patterns to search a subgraph like this: MATCH(a:Person{id:1})-[:Knows*1..]->(b:Person) However, I find it takes too long when the subgraph is large. Therefore I use apoc procedure: MATCH(a:Person{id:1}) CALL apoc.path.subgraphNodes(p,{ ...

changecpl · 07-28-2021

stackoverflow.com How to build dependency graphs effectively in neo4j with python？ python, neo4j, duplicates, dependency-graph asked by cpl on 10:00AM - 20 Jul 21 UTC ...

changecpl · 07-20-2021

Hi, everyone! I am new to neo4j. I use the procedure apoc.path.subgraphNodes to get all the nodes in a subgraph. MATCH (p:Pkg {name: "express", version:'4.17.1'}) CALL apoc.path.subgraphNodes(p, { relationshipFilter: "DependOn", minLevel:1,maxLevel...

changecpl · 07-24-2021

Actually I find that apoc.path.subgraphNodes is much faster than MATCH (A)-[:Relationship*1..]->(B) when the subgraph to search is big...

changecpl · 07-21-2021

Oh, Thanks a lot @tard.gabriel ！You are right. The Cypher language is easy to read ! That is so close. I want to count the nodes that the package express depends on both directly and indirectly, group by the property license 's values. I change you...

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How to expand to nodes in a subgraph with their level returned?

How to build dependency graphs effectively in neo4j with python？

How to count the nodes group by their properties' values

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Re: How to count the nodes group by their properties' values