限制朋友的朋友数会导致Cypher查询

Question

我有爱丽丝，鲍勃，查理和丹的书籍偏好的加权neo4j图表：

  CREATE (alice:Person {name:'Alice'})
  CREATE (bob:Person {name: 'Bob'})
  CREATE (charlie:Person {name: 'Charlie'})
  CREATE (dan:Person {name: 'Dan'})
  CREATE (b00:Book {title:'Book 00'})
  CREATE (b01:Book {title:'Book 01'})
  CREATE (b02:Book {title:'Book 02'})
  CREATE (b03:Book {title:'Book 03'})
  CREATE (b04:Book {title:'Book 04'})
  CREATE (b05:Book {title:'Book 05'})
  CREATE (b06:Book {title:'Book 06'})
  CREATE (b07:Book {title:'Book 07'})
  CREATE (b08:Book {title:'Book 08'})
  CREATE (b09:Book {title:'Book 09'})
  CREATE (b10:Book {title:'Book 10'})

  CREATE (alice)-[:LOVES {how_much:0.1}]->(b01)
  CREATE (alice)-[:LOVES {how_much:0.2}]->(b02)
  CREATE (alice)-[:LOVES {how_much:0.3}]->(b03)

  CREATE (bob)-[:LOVES {how_much:0.2}]->(b02)
  CREATE (bob)-[:LOVES {how_much:0.4}]->(b04)
  CREATE (bob)-[:LOVES {how_much:0.5}]->(b05)
  CREATE (bob)-[:LOVES {how_much:0.6}]->(b06)
  CREATE (bob)-[:LOVES {how_much:0.7}]->(b07)

  CREATE (charlie)-[:LOVES {how_much:0.3}]->(b03)
  CREATE (charlie)-[:LOVES {how_much:0.8}]->(b08)
  CREATE (charlie)-[:LOVES {how_much:0.9}]->(b09)
  CREATE (charlie)-[:LOVES {how_much:1.0}]->(b10)

  CREATE (dan)-[:LOVES {how_much:0.1}]->(b01)
  CREATE (dan)-[:LOVES {how_much:0.8}]->(b08)

我想以这样的方式向Alice推荐书籍，我只考虑每个人的前2 LOVES个关系。 因此，对于爱丽丝来说，我只会看看＃02;书02＆＃34;和＆＃34;书03＆＃34; （前2本书）。这些书将把我和鲍勃和查理联系起来（而不是丹，她通过她最不喜欢的书与爱丽丝联系）。 反过来，鲍勃将把爱丽丝连接到＃06; Book 06＆＃34;和＆＃34; Book 07＆＃34 ;;查理将把爱丽丝连接到 ＆＃34; Book 09＆＃34;和＆＃34;第10册＆＃34;。最后，我想看到四条建议： ＆＃34;书06＆＃34;和＆＃34; Book 07＆＃34;通过鲍勃;和＆＃34; Book 09＆＃34;和＆＃34;第10册＆＃34;通过查理。什么是最有效和最有效的Neo4j查询此任务？

这是我尝试做的事情：

  MATCH (alice:Person {name:'Alice'})-[la:LOVES]->(ba:Book)<-[lf:LOVES]-(friend:Person)
  WITH alice, la
  ORDER BY la.how_much DESC
  WITH alice, COLLECT(la)[0..2] as alice_likes
  UNWIND alice_likes as la
  return *

这是我如何将查询限制在前2爱丽丝的喜欢。但是，我不知道如何 继续这项任务。我尝试执行以下操作，但它并不像我期望的那样工作：

  MATCH (alice:Person {name:'Alice'})-[la:LOVES]->(ba:Book)<-[lf:LOVES]-(friend:Person)
  WITH alice, la
  ORDER BY la.how_much DESC
  WITH alice, COLLECT(la)[0..2] as alice_likes
  UNWIND alice_likes as la


  MATCH (alice)-[la]->(ba:Book)<-[lf:LOVES]-(friend:Person)
  WITH alice, la, lf, friend
  ORDER by friend.name,  lf.how_much DESC
  WITH friend, collect(lf)[0..2] as friend_likes, collect(la)[0..2] as alice_likes
  RETURN friend.name, friend_likes, alice_likes

Answer 1

这应该有效：

MATCH (alice:Person { name:'Alice' })-[la:LOVES]->(:Book)<-[:LOVES]-(friend:Person)
WITH alice, la, friend
ORDER BY la.how_much DESC 
LIMIT 2
MATCH (bf:Book)<-[lf:LOVES]-(friend:Person)
WITH alice, lf, bf, friend
ORDER BY lf.how_much DESC 
RETURN alice, friend, COLLECT(bf)[..2] AS books;

ALice可能只有一个节点，因此我们可以使用LIMIT 2为她获得前2本书。

然而，可能有多个朋友＆＃34;谁也喜欢这两本书，所以你不能用LIMIT 2来获得每本书的2本书（相反，你只能为他们作为一组获得前两本书）。为了获得您想要的结果，此查询汇总了每位朋友的热门书籍，然后选择前2名（同样，为每位朋友）。