如何在一个脱节的子图中获取所有节点 - neo4j / py2neo

Question

If I have a neo4j database and I want to query to get all nodes that are in a one specific disjointed sub-graph. (in py2neo or in cypher)

If I have groups of nodes, the nodes in each group are connected by relationship within that group, but are not connects between groups. Can I query for one node and get all the nodes in that node's group?

Answer 1

[UPDATED]

Original Answer

If by "group of nodes" you mean a "disjoint subgraph", here is how you can get all the nodes in the disjoint subgraph (with relationships of any type) that contains a specific node (say, the Neo node):

MATCH (n { name: "Neo" })
OPTIONAL MATCH p=(n)-[*]-(m)
RETURN REDUCE(s = [n], x IN COLLECT(NODES(p)) |
  REDUCE(t = s, y IN x | CASE WHEN y IN t THEN t ELSE t + y END )) AS nodes;

This query uses an OPTIONAL MATCH to find the nodes "related" to the Neo node, so that if that node has no relationships, the query would still be able to return a result.

The two (nested) REDUCE clauses make sure that the returned collection only has distinct nodes.

The outer REDUCE clause initialized the result collection with the n node, since it must always be in the disjoint subgraph, even if there are no other nodes.

Better Answer

MATCH p=(n { name: "Neo" })-[r*0..]-(m)
WITH NODES(p) AS nodes
UNWIND nodes AS node
RETURN DISTINCT node

This simpler query (which returns node rows) uses [r*0..] to allow 0-length paths (ie, n does not need to have any relationships -- and m can be the same as n ). It uses UNWIND to turn the nodes node collection(s) into rows, and then uses DISTINCT to eliminate duplicates.

Alternate Solution (does not work yet, due to bug)

This alternate solution below (which returns node rows) should also have worked, except that there is currently a bug ( that I just reported ) that causes all identifiers to be forgotten after a query unwinds a NULL (which can happen, for instance, if an OPTIONAL MATCH fails to find a match). Due to this bug, if the Neo node has no relationships, the query below currently returns no results. Therefore, you have to use the query above until the bug is fixed.

MATCH (n { name: "Neo" })
OPTIONAL MATCH p=(n)-[*]-(m)
WITH n, NODES(p) AS nodes
UNWIND nodes AS node
RETURN DISTINCT (
  CASE
  WHEN node IS NULL THEN n
  ELSE node END ) AS res;

Answer 2

It might help if you added a diagram or some sample data. However, if I'm understanding your data model correctly you are defining "groups" of nodes as all nodes connected by a certain relationship? So to get all members of "group1" nodes (let's define that as all nodes connected by a group1 relationship) and not any nodes connected to group2 you could use a query like this:

MATCH (a:Person)-[:GROUP1]-(b:Person)
WHERE NOT exists((a)-[:GROUP2]-())
RETURN a

You can also use Node labels to define groups of Nodes.