Sql 如何在SELECT语句中重用子查询的结果

我一直在处理一些大学课程的数据，我希望优化我的查询

我使用的数据集是英国国家警察局关于停止和搜查的数据，我试图得到种族和他们停止和搜查的比例之间的相关性

我有一个查询，它将为每个警察部队和种族组合查找搜索总数、同一部队在该种族与其他种族相比的搜索百分比、全国平均百分比以及该部队平均值与全国平均值之间的差异（我知道这是一个令人困惑的问题）

这是我当前的“有效”查询：

SELECT c1.FORCE,
       c1.ETHNICITY,
       (SELECT COUNT(*) FROM CRIMES WHERE FORCE = c1.FORCE AND ETHNICITY = c1.ETHNICITY) AS num_searches,
       (ROUND(((SELECT COUNT(*) FROM CRIMES WHERE FORCE = c1.FORCE AND ETHNICITY = c1.ETHNICITY) /
           (SELECT COUNT(*) FROM CRIMES WHERE FORCE = c1.FORCE)::DECIMAL), 4) * 100) AS percentage_of_force,
       (SELECT ROUND((COUNT(*) / 303565::DECIMAL) * 100, 4) FROM CRIMES WHERE ETHNICITY = c1.ETHNICITY GROUP BY ETHNICITY) AS national_average,
       (SELECT (ROUND(((SELECT COUNT(*) FROM CRIMES WHERE FORCE = c1.FORCE AND ETHNICITY = c1.ETHNICITY) /
           (SELECT COUNT(*) FROM CRIMES WHERE FORCE = c1.FORCE)::DECIMAL), 4) * 100) - (SELECT ROUND((COUNT(*) / 303565::DECIMAL) * 100, 4) FROM CRIMES WHERE ETHNICITY = c1.ETHNICITY GROUP BY ETHNICITY)) AS difference_from_average
FROM (SELECT * FROM CRIMES) AS c1
GROUP BY c1.FORCE, c1.ETHNICITY
ORDER BY c1.FORCE, c1.ETHNICITY;

因此，我的问题是在“选择”部分多次重用同一查询

正如您从上面的查询中所看到的，

与平均值的差异\u

仅仅是力量百分比

减去全国平均值
的结果，但是我似乎无法找到一种方法来计算这些值一次，然后在
选择
部分的其他地方重复使用它们。所以我的问题是我如何才能做到这一点

其他信息

示例输入数据

示例查询结果

我正在使用PostgreSQL v11.2。
诀窍是使用子选择：

SELECT f(a, b), a, c
FROM (SELECT g(c, d) AS a,
             h(c) AS b, 
             c, d
      FROM x) AS q;

你明白了。
有不同的方法来简化查询。您可以使用一系列CTE来预计算不同聚合级别的结果。但我认为最有效和可读的选择是使用窗口函数

所有中间计数都可以在子查询中计算，使用（…）上的
COUNT（…）
和各种
PARTITION BY
选项，如下所示：

SELECT
    force,
    ethnicity,
    COUNT(*) OVER(PARTITION BY force, ethnicity) AS cnt,
    COUNT(*) OVER(PARTITION BY force) AS cnt_force,
    COUNT(*) OVER(PARTITION BY ethnicity) AS cnt_ethnicity,
    ROW_NUMBER() OVER(PARTITION BY force, ethnicity) AS rn
FROM crimes

然后外部查询可以计算最终结果（同时对每个
元组中的第一条记录进行过滤以避免重复）

查询：

SELECT 
    force,
    ethnicity,
    cnt AS num_searches,
    ROUND(cnt / cnt_force::decimal * 100, 4) AS percentage_of_force,
    ROUND(cnt_ethnicity / 303565::decimal * 100, 4) AS national_average,
    ROUND(cnt / cnt_force::decimal * 100, 4) 
        - ROUND(cnt_ethnicity / 303565::decimal * 100, 4) AS difference_from_average
FROM (
    SELECT
        force,
        ethnicity,
        COUNT(*) OVER(PARTITION BY force, ethnicity) AS cnt,
        COUNT(*) OVER(PARTITION BY force) AS cnt_force,
        COUNT(*) OVER(PARTITION BY ethnicity) AS cnt_ethnicity,
        ROW_NUMBER() OVER(PARTITION BY force, ethnicity) AS rn
    FROM crimes
    ) x
WHERE rn = 1
ORDER BY force, ethnicity;

：

| force           | ethnicity | num_searches | percentage_of_force | national_average | difference_from_average |
| --------------- | --------- | ------------ | ------------------- | ---------------- | ----------------------- |
| metropolitan    | Black     | 6            | 46.1538             | 0.0020           | 46.1518                 |
| metropolitan    | Undefined | 1            | 7.6923              | 0.0003           | 7.6920                  |
| metropolitan    | White     | 6            | 46.1538             | 0.0043           | 46.1495                 |
| norfolk         | White     | 1            | 100.0000            | 0.0043           | 99.9957                 |
| north-yorkshire | White     | 2            | 100.0000            | 0.0043           | 99.9957                 |
| northumbria     | White     | 1            | 100.0000            | 0.0043           | 99.9957                 |
| west-yorkshire  | White     | 3            | 100.0000            | 0.0043           | 99.9957                 |

谢谢@GMB这正是我想要的答案。我真的很感谢你花时间给我这么全面的回答！
SELECT 
    force,
    ethnicity,
    cnt AS num_searches,
    ROUND(cnt / cnt_force::decimal * 100, 4) AS percentage_of_force,
    ROUND(cnt_ethnicity / 303565::decimal * 100, 4) AS national_average,
    ROUND(cnt / cnt_force::decimal * 100, 4) 
        - ROUND(cnt_ethnicity / 303565::decimal * 100, 4) AS difference_from_average
FROM (
    SELECT
        force,
        ethnicity,
        COUNT(*) OVER(PARTITION BY force, ethnicity) AS cnt,
        COUNT(*) OVER(PARTITION BY force) AS cnt_force,
        COUNT(*) OVER(PARTITION BY ethnicity) AS cnt_ethnicity,
        ROW_NUMBER() OVER(PARTITION BY force, ethnicity) AS rn
    FROM crimes
    ) x
WHERE rn = 1
ORDER BY force, ethnicity;

| force           | ethnicity | num_searches | percentage_of_force | national_average | difference_from_average |
| --------------- | --------- | ------------ | ------------------- | ---------------- | ----------------------- |
| metropolitan    | Black     | 6            | 46.1538             | 0.0020           | 46.1518                 |
| metropolitan    | Undefined | 1            | 7.6923              | 0.0003           | 7.6920                  |
| metropolitan    | White     | 6            | 46.1538             | 0.0043           | 46.1495                 |
| norfolk         | White     | 1            | 100.0000            | 0.0043           | 99.9957                 |
| north-yorkshire | White     | 2            | 100.0000            | 0.0043           | 99.9957                 |
| northumbria     | White     | 1            | 100.0000            | 0.0043           | 99.9957                 |
| west-yorkshire  | White     | 3            | 100.0000            | 0.0043           | 99.9957                 |