Question

如何在cypher查询中正确使用lowestCommonAncestor算法来实现neo4j的2.0实现？

start read=node:readID('readid:"HWI-ST884:57:1:1101:13989:75421#0"') match read-[r]->gi where r.bitscore > "35" with gi match lca=lowestCommonAncestor(gi) return lca;
SyntaxException: Invalid input '(': expected an identifier character, whitespace, node labels, a relationship pattern, ',', USING, WHERE, CREATE, DELETE, SET, REMOVE, RETURN, WITH, UNION, ';' or end of input (line 1, column 151)
"start read=node:readID('readid:"HWI-ST884:57:1:1101:13989:75421#0"') match read-[r]->gi where r.bitscore > "35" with gi match lca=lowestCommonAncestor(gi) return lca"
                                                                                                                                                        ^

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Answer 1

使用您所指的lowestCommonAncestor函数的唯一方法是通过Java API。你可以，例如写一个内部使用此函数的unmanaged extension。目前，此功能未向Cypher公开。

请注意，AncestorUtil 不是Neo4j的公共API的一部分 - 这意味着从一个版本到另一个版本可能会发生任何变化。

Answer 2

也许您可以使用常规密码来查找相同的内容？

MATCH (lca)<-[:REL*]-(n)
WHERE id(n) in {ids}
WITH lca, count(distinct n) as leaves
WHERE leaves = length({ids})
RETURN lca

neo4j的最低共同祖先

2 个答案: