Question

我想总结一下neo4j集合的每个值，任何人都可以告诉我这个cypher这个我有三个集合已经MEASURES1，MEASURES2，MEASURES3

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Answer 1

我得到了答案：

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减少（s = 0，x IN MEASURES1 | s + x）AS MEASURES1，

reduce（s = 0，x IN MEASURES2 | s + x）AS MEASURES2，

reduce（s = 0，x IN MEASURES3 | s + x）AS MEASURES3

reduce（）功能将按照我们的要求运行：）

Answer 2

您可以使用reduce()功能来实现目标。看看：

WITH [14,1] AS MEASURES1, [] AS MEASURES2, [1,1,1,1,1,1,2,1,1,1,1,1] AS MEASURES3

WITH reduce(acc1 = 0, v1 in MEASURES1 | acc1 + v1) AS MEASURES1,
     reduce(acc2 = 0, v2 in MEASURES2 | acc2 + v2) AS MEASURES2,
     reduce(acc3 = 0, v3 in MEASURES3 | acc3 + v3) AS MEASURES3
RETURN MEASURES1, MEASURES2, MEASURES3

以上查询的结果：

+-----------------------------------+
| MEASURES1 | MEASURES2 | MEASURES3 |
+-----------------------------------+
| 15        | 0         | 13        |
+-----------------------------------+

Answer 3

如果您有权访问APOC Procedures，则可以使用对集合进行操作的辅助函数，包括apoc.coll.sum()，它可以完全满足您的需求：

...
RETURN apoc.coll.sum(MEASURES1) as m1Sum, apoc.coll.sum(MEASURES2) as m2Sum, apoc.coll.sum(MEASURES3) as m3Sum

如何总结neo4j集合的价值

3 个答案: