计算所有节点neo4j之间的相似度-为节点对获取不同的值

Question

我的数据库中有两种节点：

1. 用户
2. 媒体

还有一种关系-“喜欢”

两个节点之间的关系描述如下：

（：USER）-[：LIKES]->（：MEDIA）

我正在尝试根据每个节点对之间共享的媒体数来计算所有“ USER”节点之间的相似度（Jaccard相似度）

然后将这种相似性存储为“ ISSIMILAR”关系。 “ ISSIMILAR”关系具有一个称为“相似性”的属性，该属性存储节点之间的相似性

这是我的查询：

Match(u:User)

WITH COLLECT(u) as users

UNWIND users as user

MATCH(user:User{id:user.id})-[:LIKES]->(common_media:Media)<-[:LIKES]-(other:User)

WITH user,other,count(common_media) AS intersection, COLLECT(common_media.name) as i

MATCH(user)-[:LIKES]->(user_media:Media)

WITH user,other,intersection,i, COLLECT(user_media.name) AS s1

MATCH(other)-[:LIKES]->(other_media:Media)

WITH user,other,intersection,i,s1, COLLECT(other_media.name) AS s2

WITH user,other,intersection,s1,s2

WITH user,other,intersection,s1+filter(x IN s2 WHERE NOT x IN s1) AS union, s1,s2

WITH ((1.0*intersection)/SIZE(union)) as jaccard,user,other

MERGE(user)-[:ISSIMILAR{similarity:jaccard}]-(other)

运行此查询，我有两个问题：

1. 我希望一对节点之间只有一个“ ISSIMILAR”关系。 但是它创造了两个。
2. 此“ ISSIMILAR”关系的“相似”属性具有不同的值。这些值应相同

这是问题的可视化：

MATCH(user:User)-[r]-(o:User) return o,user,r limit 4

提前致谢

Answer 1

出现两个相似关系的问题是因为您不排除先前构造的相似关系。 您可以通过执行以下操作来避免这种情况：

...
UNWIND users as user
  UNWIND users as other 
    WITH user, other WHERE ID(user) > ID(other)
    MATCH(user)-[:LIKES]->(common_media:Media)<-[:LIKES]-(other) 
...

最后的查询可以变得更加清晰：

MATCH (u:User) WITH COLLECT(u) AS users
UNWIND users AS user
UNWIND users AS other

MATCH (user)-[:LIKES]->(common_media:Media)<-[:LIKES]-(other) WHERE ID(other) > ID(user)
WITH user, other, COLLECT(common_media) AS intersection

MATCH (user)-[:LIKES]->(user_media:Media)
WITH user, other, intersection, 
     COLLECT(user_media) AS s1

MATCH (other)-[:LIKES]->(other_media:Media)
WITH user,other,intersection, s1, 
     COLLECT(other_media) AS s2

RETURN user, other,
       (1.0 * SIZE(intersection)) / (SIZE(s1) + SIZE(s2) - SIZE(intersection)) AS jaccard

MERGE (user)-[:ISSIMILAR {similarity: jaccard}]->(other)