我需要对neo4j数据库中的数据进行分组，然后过滤掉每个组中除顶部n记录以外的所有内容.
I need to group the data from a neo4j database and then to filter out everything except the top n records of every group.
示例:
我有两种节点类型:订单和商品.它们之间存在添加"关系. 添加"关系具有时间戳属性.我想知道的是(对于每篇文章)在订单中的前两篇文章中有多少次.我尝试的是以下方法:
I have two node types : Order and Article. Between them there is an "ADDED" relationship. "ADDED" relationship has a timestamp property. What I want to know (for every article) is how many times it was among the first two articles added to an order. What I tried is the following approach:
获取所有订单-[ADDED]-文章
get all the Order-[ADDED]-Article
将步骤1中的结果按订单ID作为第一个排序关键字，然后按ADDED关系的时间戳作为第二个排序关键字；
sort the result from step 1 by order id as first sorting key and then by timestamp of ADDED relationship as second sorting key;
对于第2步中代表一个顺序的每个子组，仅保留前2行；
for every subgroup from step 2 representing one order, keep only the top 2 rows;
在步骤3的输出中计算不同的文章ID；
Count distinct article ids in the output of step 3;
我的问题是我陷入了第3步.是否可以为代表订单的每个子组获取前2行?
My problem is that I got stuck at step 3. Is it possible to get top 2 rows for every subgroup representing an order?
谢谢
提比留
推荐答案尝试
MATCH (o:Order)-[r:ADDED]->(a:Article) WITH o, r, a ORDER BY o.oid, r.t WITH o, COLLECT(a)[..2] AS topArticlesByOrder UNWIND topArticlesByOrder AS a RETURN a.aid AS articleId, COUNT(*) AS count结果看起来像
articleId count 8 6 2 2 4 5 7 2 3 3 6 5 0 7在此示例图使用
FOREACH(opar IN RANGE(1,15) | MERGE (o:Order {oid:opar}) FOREACH(apar IN RANGE(1,5) | MERGE (a:Article {aid:TOINT(RAND()*10)}) CREATE o-[:ADDED {t:timestamp() - TOINT(RAND()*1000)}]->a ) )更多推荐
获取neo4j中每个组的前n条记录
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