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[英]Cypher query to find nodes that are not related to other node by property
[英]Cypher merge nodes with same property and collected the other property
我有這種結構的節點
(g:Giocatore { nome, match, nazionale})
(nome:'Del Piero', match:'45343', nazionale:'ITA')
(nome:'Messi', match:'65324', nazionale:'ARG')
(nome:'Del Piero', match:'18235', nazionale:'ITA')
屬性“ match”是唯一的(匹配的ID),而有多個“ nome”具有相同的名稱。 我想將所有節點合並為相同的“ nome”,並創建一個不同的“ match”集合,如下所示
(nome:'Del Piero', match:[45343,18235], nazionale:'ITA')
(nome:'Messi', match:'65324', nazionale:'ARG')
我也嘗試過apoc庫,但是沒有任何效果。 任何想法?
您可以嘗試以下查詢:
MATCH (n:Giocatore)
WITH n.nome AS nome, collect(n) AS node2Merge
WITH node2Merge, extract(x IN node2Merge | x.match) AS matches
CALL apoc.refactor.mergeNodes(node2Merge) YIELD node
SET node.match = matches
在這里,我使用APOC合並節點,但是隨后我在節點列表上進行了映射轉換以具有match
數組,並在合並的節點上進行了設置。
我不知道您是否有許多Giocatore
節點,因此此查詢可能會執行OutOfMemory異常,因此您將必須對查詢進行批處理。 例如,您可以將MATCH (n:Giocatore) WHERE n.nome STARTS WITH 'A'
替換為第一行,其中MATCH (n:Giocatore) WHERE n.nome STARTS WITH 'A'
開頭,然后對每個字母重復此操作,或者也可以使用apoc.periodic.iterate
過程:
CALL apoc.periodic.iterate(
'MATCH (n:Giocatore) WITH n.nome AS nome, collect(n) AS node2Merge RETURN node2Merge, extract(x IN node2Merge | x.match) AS matches',
'CALL apoc.refactor.mergeNodes(node2Merge) YIELD node
SET node.match = matches',
{batchSize:1000,parallel:true,retries:3,iterateList:true}
) YIELD batches, total
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