如何在neo4j的密码查询中有两个聚合?
我对复杂的查询有问题; 以下是我的密码查询:如何在neo4j的密码查询中有两个聚合?,neo4j,cypher,Neo4j,Cypher,我对复杂的查询有问题; 以下是我的密码查询: params.put("query", "name:*"); ExecutionResult result = engine.execute( "start n=node:groups({query}) match n<-[:Members_In]-x with n,count(distinct x) as numberOfUsers where numOfUsers>avg(numOfUsers) return n.name,n
params.put("query", "name:*");
ExecutionResult result = engine.execute( "start n=node:groups({query})
match n<-[:Members_In]-x
with n,count(distinct x) as numberOfUsers
where numOfUsers>avg(numOfUsers)
return n.name,numOfUsers ", params );
params.put(“查询”,“名称:”);
ExecutionResult=engine.execute(“开始n=node:groups({query}))
匹配导航(numOfUsers)
返回n.name,numOfUsers”,参数);
n
是组名,x是每个组的用户。
如何获得组用户数的平均值,并将其与每个组用户数进行比较?
我是否可以获取组用户的平均数量,然后返回用户数高于平均值的组?
谢谢。我认为如果不重新启动,您无法在同一个查询中获得平均值和计数。下面是我的尝试:
start n=node:groups({query})
match n<-[:Members_In]-x
with n, count(distinct x) as cnt
with avg(cnt) as av
start n=node:groups({query})
match n<-[:Members_In]-x
with n, av, count(distinct x) as numOfUsers
where numOfUsers > av
return n.name, numOfUsers, av
start n=节点:组({query})
匹配