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Sorting Neo4j-Gremlin-如何对.groupCount.cap()返回的结果进行排序_Sorting_Neo4j_Gremlin - Fatal编程技术网

Sorting Neo4j-Gremlin-如何对.groupCount.cap()返回的结果进行排序

Sorting Neo4j-Gremlin-如何对.groupCount.cap()返回的结果进行排序,sorting,neo4j,gremlin,Sorting,Neo4j,Gremlin,我有一个问题。 此查询: g.V.filter{it.type == 'user'}.outE.filter{it.label == 'follow'}.outV.groupCount().cap() 返回哈希映射 {v[1]=1, v[138]=3, v[140]=3, v[6]=1, v[203]=1, v[619]=7, v[131]=1, v[10]=4, v[192]=1, ...} 怎么点? 输出需要 v[619]=7,v[10]=4,v[138]=3,v[140]=3 g.V.

我有一个问题。 此查询:

g.V.filter{it.type == 'user'}.outE.filter{it.label == 'follow'}.outV.groupCount().cap()
返回哈希映射

{v[1]=1, v[138]=3, v[140]=3, v[6]=1, v[203]=1, v[619]=7, v[131]=1, v[10]=4, v[192]=1, ...}
怎么点? 输出需要 v[619]=7,v[10]=4,v[138]=3,v[140]=3

g.V.filter{it.type == 'user'}.outE.filter{it.label == 'follow'}.outV.groupCount().cap().sort{-it.value} not working
thanx

您必须下一步()将地图从cap()中取出。下面是一个来自gremlin终端的示例:

gremlin> g = TinkerGraphFactory.createTinkerGraph()
==>tinkergraph[vertices:6 edges:6]
gremlin> g.V.in.groupCount.cap().sort{-it.value}
==>{v[1]=3, v[6]=1, v[4]=2}
gremlin> g.V.in.groupCount.cap().next().sort{-it.value}
==>v[1]=3
==>v[4]=2
==>v[6]=1
您必须下一步()将贴图从cap()中取出。下面是一个来自gremlin终端的示例:

gremlin> g = TinkerGraphFactory.createTinkerGraph()
==>tinkergraph[vertices:6 edges:6]
gremlin> g.V.in.groupCount.cap().sort{-it.value}
==>{v[1]=3, v[6]=1, v[4]=2}
gremlin> g.V.in.groupCount.cap().next().sort{-it.value}
==>v[1]=3
==>v[4]=2
==>v[6]=1

排序是一个普通的Groovy构造,所以您应该能够使用普通Groovy来解决它,例如?排序是一个普通的Groovy构造,所以您应该能够使用普通Groovy来解决它,例如?