替代 union 和 where 语句到 bucket age 和 group by a column SQL Impala

Question

给出一个 table 赞：

+----+-----------+------------------+
| id | code      | age              |
+----+-----------+------------------+
| 1  | 315.32000 | 2.18430371791803 |
| 1  | 315.32000 | 3.18430371791803 |
| 1  | 800.00000 | 2.18430371791803 |
| 2  | 315.32000 | 5.64822705794013 |
| 3  | 800.00000 | 5.68655778752176 |
| 3  | 120.12000 | 5.70572315231258 |
| 4  | 315.32000 | 5.72488851710339 |
| 4  | 315.32000 | 5.74405388189421 |
| 5  | 120.12000 | 5.7604813374292  |
| 6  | 315.32000 | 5.77993740687426 |
| .. | ...       | ...              |
+----+-----------+------------------+

我正在使用：

SELECT code, COUNT(*) AS code_frequency,'0-10' AS age_range
FROM table 
WHERE age >= 0 AND age < 10
GROUP BY code 
ORDER BY code_frequency DESC LIMIT 1
UNION
SELECT code, COUNT(*) AS code_frequency,'10-20' AS age_range
FROM table
WHERE age >= 10 AND age < 20
GROUP BY code 
ORDER BY code_frequency DESC LIMIT 1
UNION
...
ORDER BY age_range

我用略有不同的年龄范围和逻辑多次重复此操作，以输出一个 table，显示每个年龄组的最频繁代码及其频率。

输出：

+-----------+-----------+-----------+
| code      | frequency | age_range |
+-----------+-----------+-----------+
| 315.32000 | 99832     | 0-10      |
| 800.00000 | 45223     | 10-20     |
| ...       | ...       | ...       |
+-----------+-----------+-----------+

是否有更有效的方法来实现相同的输出，而无需重复相同的代码块和使用 union？

干杯

Answer 1

您可以使用带有 case 表达式的聚合来定义组：

select
    code,
    count(*) frequency,
    case 
        when age >= 0  and age < 10 then '0-10'
        when age >= 10 and age < 20 then '10-20'
    end age_range
from mytable
group by code, age_range
order by age_range

如果您想要每个年龄段最频繁的代码，那么您可以在查询之上使用 row_number()：

select code, frequence, age_range
from (
    select 
        t.*, 
        row_number() over(partition by age_range order by frequency desc) rn
    from (
        select
            code,
            count(*) frequency,
            case 
                when age >= 0  and age < 10 then '0-10'
                when age >= 10 and age < 20 then '10-20'
            end age_range
        from mytable
        group by code, age_range
    ) t
) t
where rn = 1

或者更好：

select code, frequence, age_range
from (
    select
        code,
        count(*) frequency,
        case 
            when age >= 0  and age < 10 then '0-10'
            when age >= 10 and age < 20 then '10-20'
        end age_range,
        row_number() over(partition by min(age) order by count(*) desc) rn
    from mytable
    group by code, age_range
) t
where rn = 1

替代 union 和 where 语句到 bucket age 和 group by a column SQL Impala

Alternative to union and where statement to bucket age and group by a column SQL Impala

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pivot

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bucket

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