Question

更准确地说，假设您有人和动物节点，以及从:likes到Person的{{1}}关系。问题是：如何找到人们喜欢的 ALL 所有Animal？

示例：

Animal

如果杰克和玛雅是人的宇宙，那么杰克和玛雅都喜欢的动物组只包含狗。

Answer 1

使用此示例数据集：

CREATE (jake:Person {name:'Jake'}),
       (maya:Person {name:'Maya'}),
       (dog:Animal {name:'Dog'}),
       (snake:Animal {name:'Snake'}),
       (cat:Animal {name:'Cat'}),
       (jake)-[:LIKES]->(dog),
       (jake)-[:LIKES]->(cat),
       (maya)-[:LIKES]->(dog),
       (maya)-[:LIKES]->(snake)

我认为以下内容很好：

MATCH (p:Person)
MATCH (a:Animal)

WITH a, COLLECT(p) AS people
WHERE ALL(p IN people WHERE (p)-[:LIKES]->(a))
RETURN a.name

http://console.neo4j.org/r/vu3vxp

但它并不像以下那么快：

MATCH (p:Person)
WITH COUNT(p) AS people

MATCH (p:Person)-[:LIKES]->(a:Animal)
WITH a, COUNT(p) AS likes, people
WHERE likes = people
RETURN a.name

http://console.neo4j.org/r/nnrvj2

Answer 2

所有人喜欢的动物组合（称之为＃34; A＆＃34;）必须与单身人喜欢的动物组相同或相同。因此，我们可以通过测试随机人喜欢的动物来加速搜索A.

以下查询计算人数（np），获取一个人喜欢的动物组合（ca），并返回np人喜欢的那组动物（如果有的话）。

MATCH (p:Person)
WITH COUNT(p) AS np

MATCH (p:Person)-[:LIKES]->(a:Animal)
WITH p, COLLECT(a) AS ca, np
LIMIT 1

UNWIND ca AS a
MATCH (x:Person)-[:LIKES]->(a)
WITH a, COUNT(x) AS nx, np
WHERE nx = np
RETURN COLLECT(a) AS result;

Neo4j：查找所有节点指向的节点集

2 个答案: