使用neo4j查找与给定节点具有关系的节点集合的有效方法 [英] Efficient way to find node set having relationships to given nodes using neo4j

问题描述

对于给定的两个节点来说，找到一组共同节点（具有已定义的关系）是否有效？

例如，节点 A1 ， B1 ， C1 - C4 与关系关系 x 和 y ：

  A1 --x  - > C1 
 A1 --x  - > C2 
 A1 --x  - > C3 
 B1 --y  - > C2 
 B1 --y  - > C3 
 B1 --y  - > C4 
  

为 A1（x）和 B1（y）会是 [C2，C3] 。


解决方案

在很多情况下，可以利用域的结构来提高性能。假设您知道一般情况下，与<$ c $的数量相比，您的 A 实体具有更少的 x 关系c> y B 个实体之间的关系。然后，您可以遍历A节点的两个步骤，并查看 B 节点出现的位置，并过滤出 C 节点这样。下面是这种方法的一些代码：



  Set< Node> found = new HashSet< Node>（）; 
 for（关系firstRel：a1.getRelationships（Reltypes.x，Direction.OUTGOING））
 {
节点cNode = firstRel.getEndNode（）; 
 for（relationship secondRel：cNode.getRelationships（Reltypes.y，Direction.INCOMING））
 {
 Node bNode = secondRel.getStartNode（）; 
 if（bNode.equals（b1））
 {
 found.add（cNode）; 
休息; 
} 
} 
} 
  

另一种方法是开始两个线程可以扫描任何一方的关系。第三种方法是创建一个专门的索引来帮助回答这种查询，这显然会影响插入性能。

Is there an efficient way with given two nodes to find a set of their common nodes (with defined relationships).

For example, having nodes A1, B1, C1-C4 connected with relationships x and y:

A1 --x--> C1
A1 --x--> C2
A1 --x--> C3
B1 --y--> C2
B1 --y--> C3
B1 --y--> C4

a common node set for A1(x) and B1(y) would be [C2, C3].

解决方案

In many cases the structure of the domain can be leveraged to improve performance. Let's say that you know that in general your A entities have less x relationships compared to the number of y relationships on the B entities. Then you could traverse two steps from the A node and see where the B node shows up, and filter out the C nodes this way. Here's some code for this approach:

Set<Node> found = new HashSet<Node>();
for ( Relationship firstRel : a1.getRelationships( Reltypes.x, Direction.OUTGOING ) )
{
    Node cNode = firstRel.getEndNode();
    for ( Relationship secondRel : cNode.getRelationships( Reltypes.y, Direction.INCOMING ) )
    {
        Node bNode = secondRel.getStartNode();
        if ( bNode.equals( b1 ) )
        {
            found.add( cNode );
            break;
        }
    }
}

Another way would be to start two threads that scan the relationships from either side.

A third approach would be to create a specialized index that would help answering this kind of queries, which would obviously hurt insert performance.

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