计算两部电影（Neo4j、cypher）之间的相似性指数

在此基础上

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这一切都搞砸了，但我想不出解决这个问题的正确方法。请提供帮助。

看看我使用Neo4j控制台生成的示例：

查询应为：

MATCH (m:Movie { title: "The Matrix" })-[h1:HAS_TAG]->(t:Tag),
    (t)<-[h2:HAS_TAG]-(sm:Movie),
    (m)-[h:HAS_TAG]->(t0:Tag),
    (sm)-[H:HAS_TAG]->(t1:Tag)
WHERE m <> sm
WITH m, sm, 
    collect(DISTINCT h) AS h, 
    collect(DISTINCT H) AS H, 
    sum(h1.weight*h2.weight) AS num
WITH m, sm, num,
    sqrt(reduce(s = 0.0, x IN h | s +(x.weight^2))) AS den1, 
    sqrt(reduce(s = 0.0, x IN H | s +(x.weight^2))) AS den2
RETURN m.title, sm.title, (num/(den1*den2)) AS similarity

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我使用reduce函数来聚合不同集合中的关系值，并执行相似性指数计算。

这很有效，并给出了正确的答案

  MATCH (m:Movie { title: "The Matrix" })-[h1:HAS_TAG]->(t:Tag)<-[h2:HAS_TAG]-(sm)
  WHERE m <> sm
  WITH SUM(h1.weight * h2.weight) AS num,
     SQRT(REDUCE(xDot = 0.0, a IN COLLECT(h1)| xDot + a.weight^2)) AS xLength,
     SQRT(REDUCE(yDot = 0.0, b IN COLLECT(h2)| yDot + b.weight^2)) AS yLength, m, sm
  RETURN num, xLength, yLength

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MATCH（m:Movie{title:“The Matrix”}）-[h1:HAS_TAG]->（t:TAG）分子中的值重复，导致错误的结果。
MATCH (m:Movie { title: "The Matrix" })-[h1:HAS_TAG]->(t:Tag),
    (t)<-[h2:HAS_TAG]-(sm:Movie),
    (m)-[h:HAS_TAG]->(t0:Tag),
    (sm)-[H:HAS_TAG]->(t1:Tag)
WHERE m <> sm
WITH m, sm, 
    collect(DISTINCT h) AS h, 
    collect(DISTINCT H) AS H, 
    sum(h1.weight*h2.weight) AS num
WITH m, sm, num,
    sqrt(reduce(s = 0.0, x IN h | s +(x.weight^2))) AS den1, 
    sqrt(reduce(s = 0.0, x IN H | s +(x.weight^2))) AS den2
RETURN m.title, sm.title, (num/(den1*den2)) AS similarity

+---------------------------------------------------------------+
| m.title      | sm.title                  | similarity         |
+---------------------------------------------------------------+
| "The Matrix" | "The Matrix: Revolutions" | 3.859767091086958  |
| "The Matrix" | "The Matrix: Reloaded"    | 1.4380667053087486 |
+---------------------------------------------------------------+

  MATCH (m:Movie { title: "The Matrix" })-[h1:HAS_TAG]->(t:Tag)<-[h2:HAS_TAG]-(sm)
  WHERE m <> sm
  WITH SUM(h1.weight * h2.weight) AS num,
     SQRT(REDUCE(xDot = 0.0, a IN COLLECT(h1)| xDot + a.weight^2)) AS xLength,
     SQRT(REDUCE(yDot = 0.0, b IN COLLECT(h2)| yDot + b.weight^2)) AS yLength, m, sm
  RETURN num, xLength, yLength