如何在neo4j中查询确切路径并避免笛卡尔积? [英] How to query exact path in neo4j and avoid cartesian product?

问题描述

我有大约 3000 个节点，图中有 7000 个关系，但对于演示，我想绘制一个子图，我确切地知道我需要哪些节点.

I have around 3000 nodes with 7000 relationships in a graph but for a presentation, I'd like to plot a sub-graph of which I know exactly which nodes I need.

因此，我使用以下查询有时会为我提供正确的路径(经过长时间的等待)，有时会耗尽内存和 CPU 资源，直到我强制退出 neo4j 浏览器.

Therefore, I use the following query which sometimes gives me the correct path as a result (after a long waiting period) and sometimes depletes memory and cpu resources until I force-quit the neo4j-browser.

MATCH p1=(:DestinationNode)-[:IS_AT]->
(:CentiDegreeNode{x: 4714, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4715, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4716, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4717, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4718, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4718, y: 844})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 844}),
p2=(:DestinationNode)-[:IS_AT]->
(:CentiDegreeNode{x: 4718, y: 839})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4718, y: 840})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 840})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 841})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 842})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 844}),
p3=(:CentiDegreeNode{x: 4719, y: 844})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 845})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 846})
RETURN p1, p2, p3

我做错了什么?我将如何重新表述查询以便在几秒钟内执行它?请注意，CentiDegreeNode 的 x 和 y 已编入索引.

What am I doing wrong? How would I have to rephrase the query in order to have it executed within seconds? Note that x and y of a CentiDegreeNode are indexed.

最初我从定向关系 (-[:CONNECTED_TO]->) 开始，但这并没有更快.

Initially I started with directed relationships (-[:CONNECTED_TO]->) but this wasn't any quicker.

非常感谢！

推荐答案

当你说一个 CentiDegreeNode 的x 和 y"code> 已编入索引"，希望您的意思是这两个属性在单个索引中一起使用::CentiDegreeNode(x, y).这样性能会更好.

When you say that "x and y of a CentiDegreeNode are indexed", hopefully you meant that both properties are used together in a single index: :CentiDegreeNode(x, y). That would be more performant.

通过 WITH 子句分隔 3 条路径可能会有所帮助(这可能取决于 neo4j 的版本).此外，通过收集沿途的路径，您可以避免笛卡尔积.

Separating the 3 paths by WITH clauses might help (this could depend on the version of neo4j). Also, by collecting the paths along the way, you can avoid cartesian products.

MATCH p1=(:DestinationNode)-[:IS_AT]->
(:CentiDegreeNode{x: 4714, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4715, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4716, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4717, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4718, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4718, y: 844})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 844})
WITH COLLECT(p1) AS p1s
MATCH p2=(:DestinationNode)-[:IS_AT]->
(:CentiDegreeNode{x: 4718, y: 839})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4718, y: 840})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 840})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 841})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 842})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 843})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 844})
WITH p1s, COLLECT(p2) AS p2s
MATCH p3=(:CentiDegreeNode{x: 4719, y: 844})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 845})-[:CONNECTED_TO*1]-
(:CentiDegreeNode{x: 4719, y: 846})
RETURN p1s, p2s, COLLECT(p3) AS p3s

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