Question

我正在使用neo4j来解决实时规范化问题。假设我有来自2个不同来源的3个位置。 1个来源45给了我2个实际上彼此重复的地方，1个来源55给了我1个正确的标识符。但是，对于任何地点标识符（重复或不重复），我想找到由Feed标识符唯一的最接近的地点集。我的数据如下：

CREATE (a: Place {feedId:45, placeId: 123, name:"Empire State", address: "350 5th Ave", city: "New York", state: "NY", zip: "10118" })
CREATE (b: Place {feedId:45, placeId: 456, name:"Empire State Building", address: "350 5th Ave", city: "New York", state: "NY"})
CREATE (c: Place {feedId:55, placeId: 789, name:"Empire State", address: "350 5th Ave", city: "New York", state: "NY", zip: "10118"})

我已通过匹配节点连接这些节点，因此我可以对数据进行一些规范化。例如：

MERGE (m1: Matching:NameAndCity { attr: "EmpireStateBuildingNewYork", cost: 5.0 })
MERGE (a)-[:MATCHES]-(m1)
MERGE (b)-[:MATCHES]-(m1)
MERGE (c)-[:MATCHES]-(m1)
MERGE (m2: Matching:CityAndZip { attr: "NewYork10118", cost: 7.0 })
MERGE (a)-[:MATCHES]-(m2)
MERGE (c)-[:MATCHES]-(m2)

当我想查找起始位置ID中最接近的匹配项时，我可以在起始节点的所有路径上运行匹配，按成本排序，即：

MATCH p=(a:Place {placeId:789, feedId:55})-[*..4]-(d:Place)
WHERE NONE (n IN nodes(p)
        WHERE size(filter(x IN nodes(p)
                          WHERE n = x))> 1)
WITH    p,
        reduce(costAccum = 0, n in filter(n in nodes(p) where has(n.cost)) | costAccum+n.cost) AS costAccum
        order by costAccum
RETURN p, costAccum

但是，由于存在多个到相同位置的路径，因此在查询时会多次复制相同的节点。是否有可能收集节点及其成本，然后只返回一个不同的子集（例如，从Feed 45和55给我最好的结果？

我如何返回一组不同的路径，按费用排名，并按Feed标识符唯一？我构建这种类型的问题是错误的吗？

请帮忙！

Answer 1

您可以收集每个地点的所有路径d，然后在每个集合中选择最佳路径（因为它们将被分类然后收集）

MATCH p=(a:Place {placeId:789, feedId:55})-[*..4]-(d:Place)
WITH d, collect(p) as paths,
        reduce(costAccum = 0, n in filter(n in nodes(p) where has(n.cost)) | costAccum+n.cost) AS costAccum
        order by costAccum
RETURN head(paths) as p, costAccum

Neo4J：如何从路径集合中查找唯一节点

1 个答案: