Gremlin/Groovy groupCount结果

Question

我对groupCount和使用生成的节点执行更多查询感到有点困惑。我一直在Neo4j控制台中执行此操作。例如，使用TinkerGraph数据集：

gremlin> g = TinkerGraphFactory.createTinkerGraph()
==> tinkergraph[vertices:6 edges:6]

gremlin> g.V.getClass()
==> class com.tinkerpop.gremlin.groovy.GremlinGroovyPipeline

gremlin> m = [:]

gremlin> g.V.out.groupCount(m)
==> v[2]
==> v[4]
==> v[3]
==> v[3]
==> v[5]
==> v[3]

gremlin> m
==> v[2]=1
==> v[4]=1
==> v[3]=3
==> v[5]=1

gremlin> m.getClass()
==> class java.util.LinkedHashMap

gremlin> m = m.keySet()
==> v[2]
==> v[4]
==> v[3]
==> v[5]

gremlin> m.getClass()
==> class java.util.HashMap$KeySet

gremlin> m.outE
==> [StartPipe, OutEdgesPipe]
==> [StartPipe, OutEdgesPipe]
==> [StartPipe, OutEdgesPipe]
==> [StartPipe, OutEdgesPipe]

gremlin> m.outE.map
==> [StartPipe, OutEdgesPipe, PropertyMapPipe]
==> [StartPipe, OutEdgesPipe, PropertyMapPipe]
==> [StartPipe, OutEdgesPipe, PropertyMapPipe]
==> [StartPipe, OutEdgesPipe, PropertyMapPipe]

如何将m用作GremlinGroovyPipeline对象？我期待一个类似于下面这样的结果：

gremlin> m.outE
==> e[7][1-knows->2]
==> e[8][1-knows->4]
==> e[9][1-created->3]
==> e[12][6-created->3]
==> e[10][4-created->5]
==> e[11][4-created->3]

Answer 1

我在一个与这个问题完全无关的论坛帖子上随机找到了答案。

尽管如此，答案还是在这里：

gremlin> m.keySet()_().outE.map
==> {weight=1.0}
==> {weight=0.4}

只需在keySet()方法调用后添加文本_()，似乎就可以将其直接转换回GremlinGrrovyPipeline对象：

gremlin> m.keySet()_().getClass()
==> class com.tinkerpop.gremlin.groovy.GremlinGroovyPipeline