这将是我的方法。

MATCH (l:Listener)-[lo:LOGS]->(s:Scrobble)-[f:FEATURES]->(t:Track)<-[p:PERFORMS]-(a:Artist)
WITH l, a, count(*) AS numberOfScrobbles
WITH l, numberOfScrobbles, collect(a.title) AS artists
ORDER BY numberOfScrobbles DESC
WITH l, collect(numberOfScrobbles) AS scrobbleCountList, collect(artists) AS artistsList
RETURN l.name as listnerName, 
scrobbleCountList[0] as highestScrobbleCount, 
artistsList[0] as artists

第二行计算每个listner-artist对的scrobble数。

l        a                  numberOfScrobbles
Nathan   Lang Lang          4
Nathan   Martha Argerich    3
Nathan   Sviatoslav Richter 4

第三行和第四行收集了所有艺术家，这些艺术家为每个听众提供了该数量的Scrobble，并按Scrobble计数降序排序。

l       numberOfScrobbles artists
Nathan  4                 ["Lang Lang", "Sviatoslav Richter"]
Nathan  3                 ["Martha Argerich"]

第五行将numberOfScrobbles和艺术家收集到列表中，这样每个listner就有一行。

l       scrobbleCountList artistsList
Nathan  [4, 3]            [["Lang Lang", "Sviatoslav Richter"], ["Martha Argerich"]]

最后三行返回listner名称以及scrobbleCountList和artistsList中的第一个值。

l      highestScrobbleCount artists
Nathan 4                    ["Lang Lang", "Sviatoslav Richter"]