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如何从传递边构建子图？

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我有一个带有reified关系的图表，它包含有用的信息，但出于可视化目的，我需要创建一个没有这些中间节点的子图 .

示例：

[A:Person] <--AFFILIATE-- [B:Affiliation] --COMPANY--> [C:Org]

我想制作一个这样的子图：

[A:Person] --AFFILIATED_TO--> [C:Org]

Gremlin有什么简单的方法吗？

gremlin tinkerpop3

1 回答

  • 5

    我认为你最好的选择可能是使用subgraph()步骤，因为你通常可能提取边缘引发的子图，然后在该子图上执行一些Gremlin来引入可视化边缘并删除你不想要的东西 .

    我可以使用TinkerPop打包的现代玩具图表进行演示：

    gremlin> graph = TinkerFactory.createModern()
==>tinkergraph[vertices:6 edges:6]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
gremlin> sg = g.V().outE('created').subgraph('sg').cap('sg').next() // subgraph creation
==>tinkergraph[vertices:5 edges:4]
gremlin> g = sg.traversal()
==>graphtraversalsource[tinkergraph[vertices:5 edges:4], standard]
gremlin> g.V().as('a').                                     // add special subgraph edge
......1>   out('created').as('software').
......2>   in('created').where(neq('a')).
......3>   addE('co-developer').from('a').
......4>     property('project',select('software').by('name')) 
==>e[0][1-co-developer->4]
==>e[1][1-co-developer->6]
==>e[2][4-co-developer->1]
==>e[3][4-co-developer->6]
==>e[4][6-co-developer->1]
==>e[5][6-co-developer->4]
gremlin> g.V().hasLabel('software').drop() //remove junk from subgraph
gremlin> g.E()
==>e[0][1-co-developer->4]
==>e[1][1-co-developer->6]
==>e[2][4-co-developer->1]
==>e[3][4-co-developer->6]
==>e[4][6-co-developer->1]
==>e[5][6-co-developer->4]
gremlin> g.V().has('name','marko').outE('co-developer').valueMap(true)
==>[label:co-developer,project:lop,id:0]
==>[label:co-developer,project:lop,id:1]
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