如何在 clickhouse 中聚合数组类型

Posted 2023-03-25

【中文标题】如何在 clickhouse 中聚合数组类型

【英文标题】：How to aggregate array type in clickhouse

【发布时间】：2020-02-23 18:15:08

【问题描述】：

这是示例表

示例表：

   id | weeklyNumber |
   ---- -------------
   1  |  [2,5,9]     |
   ------------------
   2  |  [1,10,4]    |

预期的结果应该是weeklyNumber数组的聚合结果，即 [3,15,13] (2+1, 5+10, 9+4)

我不知道该怎么做。

-----更新---- 此外， 我们有很多行下表 示例表：

   id | weeklyNumber | monthlyNumber
   ---- ------------- -------------
   1  |  [2,5,9]     | [20,50,90] 
   --------------------------------
   2  |  [1,10,4]    | [10,100,40]  

结果应该是 [2/20 + 1/10, 5/50 + 10/100, 9/90 + 4/40]。该怎么做？

【问题讨论】：

答案已更新以涵盖问题的第二部分。

【参考方案1】：

需要使用ForEach-aggregate函数组合器：

SELECT sumForEach(weeklyNumber)
FROM
(
    SELECT
        1 AS id,
        [2, 5, 9] AS weeklyNumber
    UNION ALL
    SELECT
        2 AS id,
        [1, 10, 4] AS weeklyNumber
)
/*
┌─sumForEach(weeklyNumber)─┐
│ [3,15,13]                │
└──────────────────────────┘
*/

---

在某些情况下可以使用这个查询：

SELECT arrayReduce('sumForEach', groupArray(weeklyNumber))
FROM
(
    SELECT
        1 AS id,
        [2, 5, 9] AS weeklyNumber
    UNION ALL
    SELECT
        2 AS id,
        [1, 10, 4] AS weeklyNumber
)
/*
┌─arrayReduce('sumForEach', groupArray(weeklyNumber))─┐
│ [3,15,13]                                           │
└─────────────────────────────────────────────────────┘
*/

---

更新

SELECT sumForEach(arrayMap((x, y) -> (x / y), weeklyNumber, monthlyNumber)) AS result
FROM
(
    SELECT
        1 AS id,
        [2, 5, 9] AS weeklyNumber,
        [20, 50, 90] AS monthlyNumber
    UNION ALL
    SELECT
        2 AS id,
        [1, 10, 4] AS weeklyNumber,
        [10, 100, 40] AS monthlyNumber
)
/*
┌─result────────┐
│ [0.2,0.2,0.2] │
└───────────────┘
*/