Neo4j Cypher “至少 2 个导演和至少 5 个演员”

Question

Which film has at least 2 directors and at least 5 actors? So far got this but it gives me the wrong answer. It gives too many movies and counts.

MATCH (a:Person)-[:ACTED_IN]->(m:Movie)<-[:DIRECTED]-(d:Person) WITH m, count(a) AS numAct, count(d) AS numDir WHERE numAct >= 5 AND numDir >= 2 RETURN  m.title, numDir AS director_count, numAct AS actor_count

What's wrong with it?

Answer 1

The way the aggregation part works is that it groups by the movie node 'm'. In the part,

...WITH m, count(a) AS numAct, count(d) AS numDir...

it will collect the movies, actors and directors as separate rows, one row for each combination of movie, actor and director and then counts the rows. If you run,

MATCH (a:Person)-[:ACTED_IN]->(m:Movie)<-[:DIRECTED]-(d:Person) 
RETURN m, a, d

you will see exactly the rows that the count aggregation will work on

To solve your problem, you will have to update the cypher, breaking up the aggregation. One way is,

MATCH (a:Person)-[:ACTED_IN]->(m:Movie) WITH m, count(a) AS numAct
MATCH (m)<-[:DIRECTED]-(d:Person) WITH m, numAct, count(d) as numDir
WHERE numAct >= 5 AND numDir >= 2 
RETURN  m.title, numDir AS director_count, numAct AS actor_count

Updated after @Graphileon's observation

Answer 2

I guess the shortest way is

MATCH (m:Movie)
WHERE size([(m)<-[:DIRECTED]-(p:Person) | p]) >= 2
      AND size([(m)<-[:ACTED_IN]-(p:Person) | p]) >= 5
RETURN m