Clickhouse：基于另一个数组的数组总和答案

【问题标题】：Clickhouse: sum of arrays based on another arrayClickhouse：基于另一个数组的数组总和
【发布时间】：2020-06-02 12:48:38
【问题描述】：

假设你有下表：

ID  COUNTRIES                   UNITS
1   ['UK', 'Italy', 'France']   [2000, 1000, 1000]
1   ['France', 'Italy', 'UK']   [3000, 4000, 2000]
1   ['Italy']                   [2000]

ClickHouse 中有没有一种方法可以将COUNTRIES 列中每个国家/地区的RATIOS 列相加？类似sum(UNITS) group by COUNTRIES

预期结果：

ID  COUNTRIES                   TOTAL_UNITS
1   ['UK', 'Italy', 'France']   [4000, 4000, 7000]

代表代码：

SELECT data.1 ID, data.2 COUNTRIES, data.3 UNITS
FROM (
    SELECT arrayJoin([(1, ['UK', 'Italy', 'France'], [2000, 1000, 1000]), 
                    (1, ['France', 'Italy', 'UK'], [3000, 4000, 2000]), 
                    (1, ['Italy'], [2000])]) data);

【问题讨论】：

标签： arrays clickhouse

【解决方案1】：

select sumMap(COUNTRIES, UNITS) from (

SELECT data.1 ID, data.2 COUNTRIES, data.3 UNITS
FROM (
    SELECT arrayJoin([(1, ['UK', 'Italy', 'France'], [2000, 1000, 1000]), 
                    (1, ['France', 'Italy', 'UK'], [3000, 4000, 2000]), 
                    (1, ['Italy'], [2000])]) data)

)

┌─sumMap(COUNTRIES, UNITS)───────────────────┐
│ (['France','Italy','UK'],[4000,7000,4000]) │
└────────────────────────────────────────────┘




select (sumMap(COUNTRIES, UNITS) as x).1 a , x.2  b from (

SELECT data.1 ID, data.2 COUNTRIES, data.3 UNITS
FROM (
    SELECT arrayJoin([(1, ['UK', 'Italy', 'France'], [2000, 1000, 1000]), 
                    (1, ['France', 'Italy', 'UK'], [3000, 4000, 2000]), 
                    (1, ['Italy'], [2000])]) data)

)
┌─a───────────────────────┬─b────────────────┐
│ ['France','Italy','UK'] │ [4000,7000,4000] │
└─────────────────────────┴──────────────────┘




select groupArray(COUNTRY) COUNTRIES, groupArray(s) UNITS from (
select sum(UNIT) s, COUNTRY from (

SELECT data.1 ID, data.2 COUNTRIES, data.3 UNITS
FROM (
    SELECT arrayJoin([(1, ['UK', 'Italy', 'France'], [2000, 1000, 1000]), 
                    (1, ['France', 'Italy', 'UK'], [3000, 4000, 2000]), 
                    (1, ['Italy'], [2000])]) data)
) array join UNITS as UNIT, COUNTRIES as COUNTRY
group by COUNTRY)

┌─COUNTRIES───────────────┬─UNITS────────────┐
│ ['France','Italy','UK'] │ [4000,7000,4000] │
└─────────────────────────┴──────────────────┘

【讨论】：