优化密码查询

Question

如何优化此cypher查询？它比使用Gremlin的类似查询慢3-4倍。

START movie=node:vertices(movieId="100") 
MATCH genera1<--movie<--()-[ratedRel:rated]->anotherMovie-->genera1 
WHERE ratedRel.stars > 3 
RETURN anotherMovie.title as title, anotherMovie.movieId as id, 
genera1.genera as genera, 
COUNT(anotherMovie) as count ORDER BY count(anotherMovie) DESC LIMIT 20;

我只是尝试检索那些被评为3星以上并且与开始节点具有相同类别的电影：http://markorodriguez.files.wordpress.com/2011/09/movielens-schema.png?w=350

我在控制台中运行查询，并且使用的是Neo4j 1.9

Gremlin查询：

m = [:];
x = [] as Set;
v = g.v(node_id);

v.out('hasGenera').aggregate(x).back(2).inE('rated').
filter{it.getProperty('stars') > 3}.outV.outE('rated').
filter{it.getProperty('stars') > 3}.
inV.filter{it != v}.
filter{it.out('hasGenera').toSet().equals(x)}.
groupCount(m){\"${it.id}:${it.title.replaceAll(',',' ')}\"}.iterate();

m.sort{a,b -> b.value <=> a.value}[0..24];

Answer 1

START movie=node:vertices(movieId="100") 
MATCH movie-->genera1<-anotherMovie<-[ratedRel:rated]-user
WHERE ratedRel.stars > 3 
RETURN anotherMovie.title as title, anotherMovie.movieId as id, genera1.genera as genera, 
COUNT(ratedRel) as cnt ORDER BY cnt DESC LIMIT 20;

更新:基于你的gremlin查询的，你能试试这个吗？

START movie=node:vertices(movieId="100") 
MATCH movie-[:hasGenera]->genera1<-[:hasGenera]-anotherMovie<-[ratedRel:rated]-user
WITH anotherMovie,count(ratedRel) as allVotes, sum(ratedRel) as allStars,genera1
WHERE allStars/allVotes>3 
RETURN anotherMovie.title as title, anotherMovie.movieId as id, genera1.genera as genera, 
allStars ORDER BY allStars DESC LIMIT 20;

重点是在路径中准确定义尽可能多的元素(在本例中，我们遗漏了rels名称)，并省略了user节点，我不知道如何在cypher中做到这一点，但它显然没有在gremlin代码中列出。