替代联合以及where语句来对存储期限和按列进行归类SQL Impala

Question

给一个像这样的表：

+----+-----------+------------------+
| 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

而且我将重复使用年龄范围和逻辑稍有不同的多次，以输出一个表，该表显示每个年龄组的频率最高，频率最高的代码以及频率。

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并使用并集？

欢呼声

Answer 1

您可以将聚合与case表达式一起使用来定义组：

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

如果您希望每个年龄段使用最频繁的代码，则可以在查询顶部使用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