Tinkerpop Gremlin深度优先搜索顺序 [英] Tinkerpop Gremlin Depth First Search Order

问题描述

我有一个非常简单的示例图，正在尝试进行深度优先查询.假设图形边缘看起来像这样

I have a very simple example graph which I am trying to get a depth first query on. Let's say the graph edges look like this

A->B
A->C
B->D
B->E
C->F
C->G

从A开始的深度优先搜索应返回

A depth first search from A should return

A-B-D-E-C-F-G

但是如果我能得到以下命令，那就更好了

But if I could get the below order it would be even better

D-E-B-A-F-G-C-A

如何创建将输出此订单的Gremlin查询?如果我做这样的事情

How can I create a Gremlin query that will output this order? If I do something like this

g.V('A').repeat(outE().inV()).emit()

我得到的A，B，C，D，E，F，G的订单优先.我不知道如何从上面得到想要的订单.

I get an order of A,B,C,D,E,F,G which is breadth first. I can't work out how to get the order I want above.

推荐答案

为其他人复制，下面是示例图:

For others to reproduce, here's the sample graph:

g = TinkerGraph.open().traversal()
g.addV().property(id, 'A').
  addV().property(id, 'B').
  addV().property(id, 'C').
  addV().property(id, 'D').
  addV().property(id, 'E').
  addV().property(id, 'F').
  addV().property(id, 'G').
  addE('link').from(V('A')).to(V('B')).
  addE('link').from(V('A')).to(V('C')).
  addE('link').from(V('B')).to(V('D')).
  addE('link').from(V('B')).to(V('E')).
  addE('link').from(V('C')).to(V('F')).
  addE('link').from(V('C')).to(V('G')).iterate()

从A开始的深度优先搜索应返回

A depth first search from A should return

A-B-D-E-C-F-G

gremlin> g.V('A').repeat(out('link')).until(__.not(outE('link'))).path().
           unfold().dedup().id().fold()
==>[A,B,D,E,C,F,G]

但是如果我能得到以下命令，那就更好了

But if I could get the below order it would be even better

D-E-B-F-G-C-A

这有点像是从后面卷起的小路.这很棘手，但可行:

This one is kinda rolling up the paths from behind. It's tricky, but doable:

gremlin> g.V('A').
           repeat(outE('link').aggregate('edges').inV()).
             until(__.not(outE('link'))).
           flatMap(
             union(identity(),
                   repeat(inE('link').where(within('edges')).as('current').
                          map(select('edges').unfold().
                                where(neq('current').and(without('done'))).
                              outV().where(without('ad')).fold()).as('bl').
                          select(last, 'current').store('done').
                            filter(outV().where(without('bl').and(without('ad')))).
                          outV().store('ad')).
                     emit())).
           id().fold()
==>[D,E,B,F,G,C,A]

但是，仅获取路径并在应用程序端进行排序可能要容易得多.

However, it's probably a lot easier to just get the paths and do the ordering on the application side.

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