如何使用Gremlin

Question

我很难描述。我试图写的Gremlin查询有一个节点作为输入。然后我用类Group输入搜索所有节点。然后在每个组中只有一个Text类节点和一些Elem节点。 我想让所有Elem节点作为兄弟节点具有相同的Text节点，即使它们来自不同的Group节点。你会看到不同的分组颜色。

这就是我的意思：

到目前为止我所拥有的：

g = new OrientGraph("remote:localhost/graphdb")
v = g.v('#12:109')
v.bothE.has('@class','hasElem').outV.has('@class','Group').bothE.or(_().has('@class','hasText'), _().has('@class','hasElem').except([v])).inV().except([v])

这会将所有绿色和蓝色节点返回给我，但我不知道如何进行分组。

感谢任何帮助：）

谢谢！

Answer 1

It took 5hours but I found the query that does it:

The groupBy mapping basically takes as a key the Text node {it} and as a value the leaf from text->parent->sibling {it.inE.outV.outE.inV.hasNot('@class','Text').except([v])} that is not of class Text except for the Input. The last line m.sort{a,b -> b.value.size() <=> a.value.size()} is sorting Texts in descending popularity, or in other words the ones that have the longer list of related siblings are coming first.

g = new OrientGraph("remote:localhost/graphdb")
v = g.v('#12:109')
m = [:]  
v.bothE.has('@class','hasElem').outV.has('@class','hasElem').dedup().bothE.has('@class','hasText').inV().groupBy(m){it}{it.inE.outV.outE.inV.hasNot('@class','Text').except([v])}
m.sort{a,b -> b.value.size() <=> a.value.size()}