Question

我的图表看起来像this

a - [：LIKES {frequency：6}] - ＆gt; ice b - [：LIKES {frequency：3}] - ＆gt;披萨等

我正在运行此查询，以找出两个人喜欢的常见食物。如果我将频率属性加起来为食物创建一个like-index，它就可以正常工作。

START b=node:node_auto_index(name='B') 
MATCH p = b-[r1:LIKES]->stuff, someone_else_too -[r2:LIKES]->stuff 
WITH b, someone_else_too, r1.frequency + r2.frequency AS likeIndex 
RETURN likeIndex ,someone_else_too 
ORDER BY likeIndex DESC

但是，我想要做的是找到两个频率值中的最小值（不是总数的最小值）。所以像这样的事情（r1.frequency，r2.frequency）。目标是消除偏斜，如果一对中的一个人对特定食物具有非常高的亲和力，则会增加类似排名。

    START b=node:node_auto_index(name='B') 
    MATCH p = b-[r1:LIKES]->stuff, someone_else_too -[r2:LIKES]->stuff 
    WITH b, someone_else_too, min(r1.frequency,r2.frequency) AS likeIndex 
    RETURN likeIndex ,someone_else_too 
    ORDER BY likeIndex DESC

但是，Cypher不会接受这种语法。任何建议或解决方法？请参阅Neo4j控制台here

Answer 1

在2.0中，您可以使用case/when：http://console.neo4j.org/r/y8sqi1

START b=node:node_auto_index(name='B') 
MATCH p = b-[r1:LIKES]->stuff, someone_else_too -[r2:LIKES]->stuff 
WITH b, someone_else_too, 
  CASE WHEN r1.frequency < r2.frequency 
    THEN r1.frequency 
    ELSE r2.frequency 
  END AS likeIndex 
RETURN likeIndex ,someone_else_too 
ORDER BY likeIndex DESC

Answer 2

有一种获得最小值的数学方法：

(r1.frequency + r2.frequency - ABS(r1.frequency - r2.frequency))/2

但它太沉重了:(

Answer 3

你可以使用 apoc.coll.min:

  START b=node:node_auto_index(name='B') 
MATCH p = b-[r1:LIKES]->stuff, someone_else_too -[r2:LIKES]->stuff 
WITH b, someone_else_too, apoc.coll.min([r1.frequency, r2.frequency]) AS likeIndex 
RETURN likeIndex ,someone_else_too 
ORDER BY likeIndex DESC

至少有两个属性

3 个答案: