英文:
Neo4j How to get equal subset of data for all relationship types from a graph?
问题
我使用以下查询来获取所有节点ID及其对应的关系类型，但我只需要每种关系类型的前100条记录，我使用了LIMIT子句，但它没有起作用。
MATCH (s)-[r]->(t)
WHERE type(r) IN ['LIVES_IN', 'FROM', 'STUDIES']
RETURN toString(id(s)) as source, toString(id(t)) AS target, type(r) as type
如何使用Cypher查询获取每种关系类型的前100条记录，提前谢谢。
英文:
I use below query to get all node ID's and their corresponding relationship types, but I need only 100 records for each relationship types, I used LIMIT clause, but it didn't work.
MATCH (s)-[r]->(t)
WHERE type(r) IN ['LIVES_IN', 'FROM', 'STUDIES']
RETURN toString(id(s)) as source, toString(id(t)) AS target, type(r) as type
How can I get 100 records for each type of relationship using the cypher query, thanks in advance.
答案1
得分: 1
你可以使用 APOC 函数 apoc.agg.slice 来高效地仅获取每个 type 的前 100 个结果（而无需先获取每个 type 的所有结果）：
MATCH (s)-[r]->(t)
WHERE TYPE(r) IN ['NEXT', 'BEGIN', 'END']
RETURN TYPE(r) AS type, apoc.agg.slice({source: TOSTRING(ID(s)), target: TOSTRING(ID(t))}, 0, 100) AS res
上述查询为每个 type 返回一行，以及包含 100 个 source 和目标 values 的列表。
要将每个结果单独返回在不同的行中：
MATCH (s)-[r]->(t)
WHERE TYPE(r) IN ['NEXT', 'BEGIN', 'END']
WITH TYPE(r) AS type, apoc.agg.slice({source: TOSTRING(ID(s)), target: TOSTRING(ID(t))}, 0, 100) AS res
UNWIND res AS row
RETURN type, row.source AS source, row.target AS target
英文:
You can use the APOC function apoc.agg.slice to efficiently get just the first 100 results for each type (without having to get all the results for each type first):
MATCH (s)-[r]->(t)
WHERE TYPE(r) IN ['NEXT', 'BEGIN', 'END']
RETURN TYPE(r) AS type, apoc.agg.slice({source: TOSTRING(ID(s)), target: TOSTRING(ID(t))}, 0, 100) AS res
The abover query returns a row for each type, along with a list of 100 source and target values.
To get every result on a separate row:
MATCH (s)-[r]->(t)
WHERE TYPE(r) IN ['NEXT', 'BEGIN', 'END']
WITH TYPE(r) AS type, apoc.agg.slice({source: TOSTRING(ID(s)), target: TOSTRING(ID(t))}, 0, 100) AS res
UNWIND res AS row
RETURN type, row.source AS source, row.target AS target
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