neo4j - 当配对顺序在密码中不重要时查询唯一的节点对

Question

我试图根据用户在此 graph 中的共同兴趣进行比较.
我知道为什么下面的查询会产生重复的对，但想不出在密码中避免它的好方法。有没有办法在不循环密码的情况下做到这一点？

neo4j-sh (?)$ start n=node(*) match p=n-[:LIKES]->item<-[:LIKES]-other where n <> other return n.name,other.name,collect(item.name) as common, count(*) as freq order by freq desc;
==> +-----------------------------------------------+
==> | n.name | other.name | common           | freq |
==> +-----------------------------------------------+
==> | "u1"   | "u2"       | ["f1","f2","f3"] | 3    |
==> | "u2"   | "u1"       | ["f1","f2","f3"] | 3    |
==> | "u1"   | "u3"       | ["f1","f2"]      | 2    |
==> | "u3"   | "u2"       | ["f1","f2"]      | 2    |
==> | "u2"   | "u3"       | ["f1","f2"]      | 2    |
==> | "u3"   | "u1"       | ["f1","f2"]      | 2    |
==> | "u4"   | "u3"       | ["f1"]           | 1    |
==> | "u4"   | "u2"       | ["f1"]           | 1    |
==> | "u4"   | "u1"       | ["f1"]           | 1    |
==> | "u2"   | "u4"       | ["f1"]           | 1    |
==> | "u1"   | "u4"       | ["f1"]           | 1    |
==> | "u3"   | "u4"       | ["f1"]           | 1    |
==> +-----------------------------------------------+ 

Answer 1

为了避免出现 a--b 形式的重复项和 b--a ，您可以使用以下命令排除 WHERE 子句中的组合之一

WHERE ID(a) < ID(b)

进行上述查询

start n=node(*) match p=n-[:LIKES]->item<-[:LIKES]-other where ID(n) < ID(other) return n.name,other.name,collect(item.name) as common, count(*) as freq order by freq desc;