【发布时间】:2019-11-01 18:36:38
【问题描述】:
看GrandStack电影工作室https://github.com/grand-stack/grand-stack-movies-workshop/blob/master/neo4j-database/answers.md的例子
此处为推荐电影提出的查询
MATCH (m:Movie) WHERE m.movieId = $movieId
MATCH (m)-[:IN_GENRE]->(g:Genre)<-[:IN_GENRE]-(movie:Movie)
WITH m, movie, COUNT(*) AS genreOverlap
MATCH (m)<-[:RATED]-(:User)-[:RATED]->(movie:Movie)
WITH movie,genreOverlap, COUNT(*) AS userRatedScore
RETURN movie ORDER BY (0.9 * genreOverlap) + (0.1 * userRatedScore) DESC LIMIT 3
这个查询会不会有偏见,因为它只会为与 ID 为 $movieId 的电影共享至少一种类型的电影计算 userRatedScore?
一个独立计算两个分数的重写查询看起来如何,这意味着它仍然会计算给定电影的 userRatedScore,即使它不与 ID 为 $movieId 的电影共享流派
【问题讨论】:
标签: neo4j cypher grandstack