Question

我的简单数据库包含＆＃39;术语的节点＆＃39;和＆＃39;代码＆＃39;相互联系。有两种类型的关系。

＆＃39;术语之间的关系＆＃39;和＆＃39;代码＆＃39;调用：CODE并且是无向的（或在两个方向上均等地读取）。术语之间的关系＆＃39;叫：NT（这意味着狭义的术语）并且是有针对性的。

我想走过所有的条款＆＃39;从上到下收集所有唯一代码并计算它们。

这是我的疑问：

MATCH (a)-[:NT*]->(b), (a)-[:CODE]-(c), (b)-[:CODE]-(d)
WHERE a.btqty = 0
RETURN a.termid AS termid, a.maxlen AS maxlen, COUNT(DISTINCT c.code) + COUNT(DISTINCT d.code) AS total, COLLECT(DISTINCT c.code) + COLLECT(DISTINCT d.code) AS codes
ORDER BY termid;

这就是我得到的：

termid  maxlen  total   codes
22  2   3   ["S70","S43","S70"]
25  4   9   ["S20","S21","S54","S61","S63","S63","S21","S61","S54"]
26  2   9   ["S99","S98","S29","S13","S13","S20","S29","S14","S15"]
68  5   13  ["S38","S11","S12","S11","S12","S38","S37","S21","S36","S22","S98","S63","S58"]
123 2   3   ["S38","S12","S12"]
154 2   2   ["S58","S58"]
155 4   3   ["S63","S62","S63"]
159 2   2   ["S36","S36"]
...

我需要摆脱集合中的重复项并正确计算它们：

termid  maxlen  total   codes
22  2   2   ["S43","S70"]
25  4   5   ["S20","S21","S54","S61","S63"]
26  2   7   ["S99","S98","S29","S13","S20","S14","S15"]
68  5   10  ["S38","S11","S12","S37","S21","S36","S22","S98","S63","S58"]
123 2   2   ["S38","S12"]
154 2   1   ["S58"]
155 4   2   ["S63","S62"]
159 2   1   ["S36"]
...

我认为这是关于REDUCE功能的应用，但我不知道如何使用它。

感谢您的帮助！

Answer 1

你是对的，可以使用REDUCE来解决。在reduce中你需要检查当前元素是否已经存在于累加器中并有条件地修改它：

MATCH (a)-[:NT*]->(b), (a)-[:CODE]-(c), (b)-[:CODE]-(d)
WHERE a.btqty = 0
WITH a.termid AS termid, a.maxlen AS maxlen, 
   REDUCE(uniqueCodes=[], 
         x in COLLECT(DISTINCT c.code) + COLLECT(DISTINCT d.code) |
          CASE WHEN x IN uniqueCodes THEN uniqueCodes ELSE uniqueCodes+x END
   ) AS codes
ORDER BY termid
RETURN termid, maxlen, count(codes) as total, codes

Neo4j中的Union和Count Collections

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