neo4j - Cypher:限制选择节点的返回链接数

Question

我在一个看似简单的查询上失败了。

算法:

1. 按某个值(desc)对节点进行排序:

   MATCH (n) WHERE has(n.II_VAL) WITH n, n.II_VAL as ShInflInd order by ShInflInd desc
   

2. 对于每个节点，返回它的直接邻居(按第二个值排序):

   MATCH (n)-[r]->(m) with n, m, m.VALUE as SubsOpRev order by SubsOpRev desc
   

挑战是返回前 100 个 n 个节点，并且每个节点只有 10 个关系 r(如果可能的话在一行中)。

编辑:

很抱歉我的表述不明确以至于人们误解了我。

详细来说，以下查询为我提供了排名靠前的节点:

MATCH (n) WHERE HAS(n.II_VAL)
WITH n, n.`II_VAL` AS ShInflInd ORDER BY ShInflInd DESC
RETURN n.NAME LIMIT 100;

+--------------------------------------+
| n.NAME                               |
+--------------------------------------+
| "PUBLIC"                             |
| "BARCLAYS PLC"                       |

现在我可以将子查询添加到此添加链接:

MATCH (n) WHERE HAS(n.II_VAL)
WITH n, n.`II_VAL` AS ShInflInd ORDER BY ShInflInd DESC LIMIT 100
MATCH (n)-[r]->(m) WHERE HAS(m.VALUE)
WITH r, n, m, m.VALUE AS SubsOpRev
RETURN n.NAME, r.WEIGHT_MERGED, m.NAME, SubsOpRev LIMIT 10;

+----------------------------------------------------------------------------------------+
| n.NAME   | r.WEIGHT_MERGED    | m.NAME                                     | SubsOpRev |
+----------------------------------------------------------------------------------------+
| "PUBLIC" | 0.66               | "VBS MUTUAL BANK"                          | 2630      |
| "PUBLIC" | 0.2923             | "STRATCORP LIMITED"                        | 10842     |

现在我想要的是，在返回“PUBLIC”的 10 个链接(可能按 r.WEIGHT_MERGED 或 SubsOpRev 排序)后，查询返回第二个节点(“BARCLAYS PLC”)及其 10 个链接等。

我试过:

MATCH (n) WHERE HAS(n.II_VAL)
WITH n, n.`II_VAL` AS ShInflInd ORDER BY ShInflInd DESC
MATCH (n)-[r]->(m) WHERE HAS(m.VALUE)
WITH r, n, m, m.VALUE AS SubsOpRev
RETURN collect([n.NAME, r.WEIGHT_MERGED, m.NAME, SubsOpRev])[0..10];

导致:

+------------------------------------------------------------------------------------------------------------------------------------------+
| collect([n.NAME, r.WEIGHT_MERGED, m.NAME, SubsOpRev])[0..3]                                                                              |
+------------------------------------------------------------------------------------------------------------------------------------------+
| [["PUBLIC",0.66,"VBS MUTUAL BANK",2630],["PUBLIC",0.2923,"STRATCORP LIMITED",10842], ...

这意味着我仍然坚持使用“PUBLIC”。

稍微修改查询会使事情变得更糟，因为它返回完全不同的数据:

MATCH (n) WHERE HAS(n.II_VAL)
WITH n, n.`II_VAL` AS ShInflInd ORDER BY ShInflInd DESC
MATCH (n)-[r]->(m) WHERE HAS(m.VALUE)
WITH r, n, m, m.VALUE AS SubsOpRev
RETURN n.NAME, collect([r.WEIGHT_MERGED, m.NAME, SubsOpRev])[0..10] LIMIT 3;

+------------------------------------------------------------------------------+
| n.NAME                | collect([r.WEIGHT_MERGED, m.NAME, SubsOpRev])[0..10] |
+------------------------------------------------------------------------------+
| "RS-INVEST AS"        | [[0.5,"VERUCO EIENDOM AS",100]]                      |
| "DBM"                 | [[0.1435,"CHELYABINSKOBLGAZ",6752]]                  |

理想情况下，查询应该产生类似于

| [["PUBLIC",0.66,"VBS MUTUAL BANK",2630],["PUBLIC",0.2923,"STRATCORP LIMITED",10842], ... |
| [["BARCLAYS PLC",x,"XYZ",y], ... |

Answer 1

您只需要限制然后继续您的查询。为了提供可重现的示例，让我们切换到 Neo4j 附带的电影数据集。假设您想要获取数据库中最老的 3 部电影，然后是每部电影中最老的 2 位 Actor 。

MATCH (m:Movie)
WITH m ORDER BY m.released LIMIT 3
MATCH (p:Person)-[:ACTED_IN]->(m)
WITH m, p ORDER BY p.born
WITH m, COLLECT(p.name)[0..2] AS oldest
RETURN m.title, oldest;

这会产生:

   | m.title                         | oldest                              
---+---------------------------------+--------------------------------------
 1 | Something's Gotta Give          | ['Jack Nicholson', 'Diane Keaton']
 2 | Top Gun                         | ['Tom Skerritt', 'Kelly McGillis']
 3 | One Flew Over the Cuckoo's Nest | ['Jack Nicholson', 'Danny DeVito']

所以你会想要这样的东西:

MATCH (n) WHERE HAS(n.II_VAL)
WITH n ORDER BY n.II_VAL DESC LIMIT 100
MATCH (n)-[r]->(m) WHERE HAS(m.VALUE)
WITH n, r, m ORDER BY m.VALUE DESC
RETURN n.NAME, COLLECT([r.WEIGHT_MERGED, m.NAME, m.VALUE])[0..10];