Graph 两个方向上都有rels的Neo4j最短路径

我用函数设置了一个图表

create (a:station {name:"a"}),
(b:station {name:"b"}),
(c:station {name:"c"}),
(d:station {name:"d"}),
(e:station {name:"e"}),
(f:station {name:"f"}),
(a)-[:CONNECTS_TO {time:8}]->(b),
(a)-[:CONNECTS_TO {time:4}]->(c),
(a)-[:CONNECTS_TO {time:10}]->(d),
(b)-[:CONNECTS_TO {time:3}]->(c),
(b)-[:CONNECTS_TO {time:9}]->(e),
(c)-[:CONNECTS_TO {time:40}]->(f),
(d)-[:CONNECTS_TO {time:5}]->(e),
(e)-[:CONNECTS_TO {time:3}]->(f)

以及使用该函数

START startStation=node:node_auto_index(name = "a"), endStation=node:node_auto_index(name = "f")
MATCH p =(startStation)-[r*]->(endStation)
WITH extract(x IN rels(p)| x.time) AS Times, length(p) AS `Number of Stops`, reduce(totalTime = 0, x IN rels(p)| totalTime + x.time) AS `Total Time`, extract(x IN nodes(p)| x.name) AS Route
RETURN Route, Times, `Total Time`, `Number of Stops`
ORDER BY `Total Time`

然后返回结果

+-------------------------------------------------------------+
| Route             | Times    | Total Time | Number of Stops |
+-------------------------------------------------------------+
| ["a","d","e","f"] | [10,5,3] | 18         | 3               |
| ["a","b","e","f"] | [8,9,3]  | 20         | 3               |
| ["a","c","f"]     | [4,40]   | 44         | 2               |
| ["a","b","c","f"] | [8,3,40] | 51         | 3               |
+-------------------------------------------------------------+

这很好，只是因为它是一个有向图，并且没有从

c->b

返回的路径（例如）

[a，c，b，e，f]

是长度为4的有效路径

所以，如果我加上反向路径

MATCH (START)-[r:CONNECTS_TO]->(END )
CREATE UNIQUE (START)<-[:CONNECTS_TO { time:r.time }]-(END )

这确实包括路径

[a，c，b，e，f]

，但也包括

[a，c，b，c，f]

使用

c

两次和

[a，c，f，e，f]

使用

f

（目的地？！）两次

---

是否有一种方法可以过滤路径，使每个路径只包含同一节点一次？

您可以在事后进行过滤，但这可能不是最快的方法

大概是这样的：

START startStation=node:node_auto_index(name = "a"), endStation=node:node_auto_index(name = "f")
MATCH p = (startStation)-[r*..4]->(endStation)

WHERE length(reduce (a=[startStation], n IN nodes(p) | CASE WHEN n IN a THEN a ELSE a + n END)) = length(nodes(p))

WITH extract(x IN rels(p)| x.time) AS Times, length(p) AS `Number of Stops`, reduce(totalTime = 0, x IN rels(p)| totalTime + x.time) AS `Total Time`, extract(x IN nodes(p)| x.name) AS Route
RETURN Route, Times, `Total Time`, `Number of Stops`
ORDER BY `Total Time`

我创建了一个带有您的问题和答案的图形列表，作为一个可执行的实时文档

---

请参见此处：

非常感谢！这正是我想要的谢谢：顺便说一句。。。从Neo4j手册来看，RESTAPI方法似乎有一个Dijkstra算法最短路径函数。这在Cypher中也可用吗？不幸的是，Cypher中还没有Dijkstra。这也是有计划的。

START startStation=node:node_auto_index(name = "a"), endStation=node:node_auto_index(name = "f")
MATCH p = (startStation)-[r*..4]->(endStation)

WHERE length(reduce (a=[startStation], n IN nodes(p) | CASE WHEN n IN a THEN a ELSE a + n END)) = length(nodes(p))

WITH extract(x IN rels(p)| x.time) AS Times, length(p) AS `Number of Stops`, reduce(totalTime = 0, x IN rels(p)| totalTime + x.time) AS `Total Time`, extract(x IN nodes(p)| x.name) AS Route
RETURN Route, Times, `Total Time`, `Number of Stops`
ORDER BY `Total Time`