格林姆林:如何找到没有特定边的顶点? [英] Gremlin: How do you find vertices without a particular edge?

问题描述

我一直在研究Gremlin图形语言，它看起来非常强大.但是，在研究如何根据需求进行评估时，我遇到了一个我似乎无法完成的案例.

I've been looking at the Gremlin graph language, and it appears very powerful. However, whilst running through what it can do to evaluate it against requirements, I came across a case that I can't seem to complete.

假设Gremlin已启动，并使用其示例数据库:

Assuming Gremlin is started, and using its example database:

gremlin> g = TinkerGraphFactory.createTinkerGraph()
...
gremlin> g.V.out('knows')
==>v[2]
==>v[4]

因此，这显示了具有已知"边缘的顶点.

So this shows vertices that have an edge of 'knows'.

但是，我想查找不具有已知"边缘的顶点.像这样:

However, I want to find vertices that do not have edges of 'knows'. Something like:

gremlin> g.V.outNot('knows')
==>v[3]
==>v[5]
==>v[6]

如何找到这些顶点?

(编辑以使输出正确)

推荐答案

我以几种方式解释了这个问题，但这也许正是您所追求的.一种方法是:

I interpret this question several ways, but perhaps this is what you are after. One way would be to do:

gremlin> g = TinkerGraphFactory.createTinkerGraph()    
==>tinkergraph[vertices:6 edges:6]
gremlin> g.V.outE.hasNot('label','knows')
==>e[9][1-created->3]
==>e[12][6-created->3]
==>e[10][4-created->5]
==>e[11][4-created->3]
gremlin> g.V.outE.hasNot('label','knows').inV
==>v[3]
==>v[3]
==>v[5]
==>v[3]

请注意，标签和ID均被识别为属性:

Note that label and id are both recognized as properties:

gremlin> g.V.has('id',"1")
==>v[1]
gremlin> g.E.map("label","id")
==>{id=10, label=created}
==>{id=7, label=knows}
==>{id=9, label=created}
==>{id=8, label=knows}
==>{id=11, label=created}
==>{id=12, label=created}

考虑此问题的另一种方法是找到没有知道"边的顶点列表:

Another way to consider this question would be to find a list of vertices that don't have a "knows" edge:

gremlin> g.V.filter{!it.bothE('knows').hasNext()}    
==>v[3]
==>v[6]
==>v[5]

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