Neo4j 聚合后计算

我是Neo4j的新手，我觉得自己被简单的操作绊住了，这些操作是我在常规SQL中使用子查询来解决的

如何减去两行结果？我已将结果分组，并希望返回它们之间的差异：

MATCH (seguidores:RelevantTwitterUser {location:"Madrid"})-[:FOLLOWS]->(seguidos:RelevantTwitterUser {location:"Barcelona"}) 
WITH COLLECT({origen:seguidores.location, user:seguidores.userId}) AS ROWS
MATCH (seguidores:RelevantTwitterUser {location:"Barcelona"})-[:FOLLOWS]->(seguidos:RelevantTwitterUser {location:"Madrid"}) 
WITH ROWS + COLLECT({origen:seguidores.location, user:seguidores.userId}) AS allRows
UNWIND allRows AS ROW
RETURN ROW.origen, COUNT(ROW.user)

对于输出：

使用 如果您想了解此查询的工作原理，并在需要修改此查询时避免获取错误的计数，则应阅读（如

COUNT

）上的文档。了解

分组键

（例如，最后一个

WITH

子句中的

LOC

和

count

如何影响聚合函数的行为，这一点尤为重要

回答使用

---

非常感谢两个答案。第一次是我试图表演我搞砸了工会和募捐的东西。谢谢你的回答，这很清楚。第二个比较优雅，我会读关于大小的，它有一个非常方便的语法。

WITH "Madrid" AS loc1, "Barcelona" AS loc2
MATCH (:RelevantTwitterUser{location:loc1})-[:FOLLOWS]->(:RelevantTwitterUser{location:loc2})
WITH loc1, loc2, COUNT(*) AS count1
MATCH (:RelevantTwitterUser{location:loc2})-[:FOLLOWS]->(:RelevantTwitterUser{location:loc1})
WITH loc1, loc2, count1, COUNT(*) AS count2
RETURN loc1, count1, loc2, count2, count1 - count2 AS diff

WITH "Madrid" AS loc1, "Barcelona" AS loc2
WITH loc1, loc2,
  SIZE((:RelevantTwitterUser{location:loc1})-[:FOLLOWS]->(:RelevantTwitterUser{location:loc2})) AS count1,
  SIZE((:RelevantTwitterUser{location:loc2})-[:FOLLOWS]->(:RelevantTwitterUser{location:loc1})) AS count2
RETURN loc1, count1, loc2, count2, count1 - count2 AS diff