怎么用一条查询语句输出多个聚合结果

[zaobao]

• 问题的具体描述

假设这样的场景，点player之间有两种边，follow和like
我想通过一条查询语句输出，一个player 分别follow了哪些 player 、like了哪些player
返回包括三个字段：playerID, followIDs, likeIDs

统计一种边可以使用GROUP BY

GO FROM "player100" OVER follow \
        YIELD $^.id AS playerID, $$.id as followID \
        | GROUP BY $-.playerID \
        YIELD $-.playerID  as playerID, collect($-.followID) AS followIDs;

如果统计两种边，是不是要写成这样

GO FROM "player100" OVER follow \
        YIELD follow._src AS playerID, follow._dst AS followID \
        | GROUP BY $-.playerID \
        YIELD $-.playerID  as playerID, collect($-.followID) AS followIDs \
        | GO FROM playerID OVER like \
       YIELD $-.playerID  as playerID,  $-.followIDs AS followIDs, like._dst AS likeID \
        | GROUP BY $-.playerID,  $-.followIDs \
        YIELD $-.playerID  as playerID,  $-.followIDs AS followIDs, collect($-.likeID) AS likeIDs \

感觉有点浪费性能，有没有更加优雅的写法

[kyle]

@zaobao

过一条查询语句输出，一个player 分别follow了哪些 player

MATCH (f)<-[:follow]-(v)-[:like]->(l)
WHERE id(v)=="vid"
RETURN id(v) AS playerId, collect(distinct id(f)) AS followerIds, collect(distinct id(l)) AS likerIds

go from "vid" over follow yield $^.id as playerId, $$.id as followerId 
| go from $-.playerId over like yield $-.playerId as playerId, $-.followerId as followerId, $$.id as likerId
| yield $-.playerId as playerId, collect(distinct $-.followerId) as followerIds, collect(distinct $-.likerId) as likerIds