Question

我们有数百万个顶点和数千万个边来形成有向图。有人知道如何在两个给定顶点之间获取Titan中的子图吗？

Answer 1

TinkerPop提供了两种子图方法：

有subgraph step和
有subgraph strategy

使用subgraph步骤＆＃34;弹出＆＃34;子图成单独的图形实例。您可以在以下示例中看到图表＆＃34; g＆＃34;被提交到一个新的Graph实例＆＃34; know＆＃34;边缘：

gremlin> subGraph = g.E().hasLabel('knows').subgraph('subGraph').cap('subGraph').next() //(1)
==>tinkergraph[vertices:3 edges:2]
gremlin> sg = subGraph.traversal()
==>graphtraversalsource[tinkergraph[vertices:3 edges:2], standard]
gremlin> sg.E() //(2)
==>e[7][1-knows->2]
==>e[8][1-knows->4]

如果您只想遍历图表的子集，则可以使用Subgraph策略：

gremlin> graph = TinkerFactory.createTheCrew()
==>tinkergraph[vertices:6 edges:14]
gremlin> g = graph.traversal().withStrategies(SubgraphStrategy.build().
           vertices(or(hasNot('location'),properties('location').count().is(gt(3)))).
           edges(hasLabel('develops')).
           vertexProperties(or(hasLabel(neq('location')),hasNot('endTime'))).create())
==>graphtraversalsource[tinkergraph[vertices:6 edges:14], standard]
gremlin> g.V().valueMap(true)
==>[id:1,label:person,name:[marko],location:[santa fe]]
==>[id:8,label:person,name:[matthias],location:[seattle]]
==>[id:10,label:software,name:[gremlin]]
==>[id:11,label:software,name:[tinkergraph]]
gremlin> g.E().valueMap(true)
==>[id:13,label:develops,since:2009]
==>[id:14,label:develops,since:2010]
==>[id:21,label:develops,since:2012]

根据你问题的标题，听起来你只想要两个顶点的图形，它们之间有边缘。我想这听起来像你想要使用subgraph步骤。当然，获得这个结果似乎比这更容易：

gremlin> g = TinkerFactory.createModern().traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
gremlin> g.V(1).bothE().where(otherV().hasId(2))
==>e[7][1-knows->2]

具有边缘列表在某种意义上构成子图，特别是在两个已知顶点的情况下。

鉴于两个特定的顶点，如何在Titan中获取它们之间的子图？

1 个答案: