将聚合除以列的总和答案

【问题标题】：Divide aggregate over sum over columns将聚合除以列的总和
【发布时间】：2020-06-10 16:12:52
【问题描述】：

以下假设是 postgres。

假设我有下表：

CREATE TABLE object (
  object_id       SERIAL  PRIMARY KEY
  object_category integer NOT NULL CHECK(value BETWEEN 1 AND 5)
);

我可以使用以下方法计算每个类别中的对象数量：

SELECT object_category,
       count(object_category) number_of_objects,
  FROM object
 GROUP BY object_category;

，但是我怎样才能找到每个类别中的对象数量与对象总数的比率呢？例如，如果上述查询返回以下结果集：

object_category | number_of_objects
-----------------------------------
              1 |                 5
              2 |                 3
              3 |                 7
              4 |                 2
              5 |                10

，我该如何返回：

object_category | object_ratio
------------------------------
              1 |        0.185
              2 |        0.111
              3 |        0.259
              4 |        0.074
              5 |        0.370

我尝试将上述总和作为子查询计算，但我无法访问 number_of_objects，因为 postgres 抱怨缺少 group-by 子句，我不太明白。有什么帮助吗？

【问题讨论】：

标签： sql postgresql

【解决方案1】：

只使用窗口函数：

SELECT object_category, COUNT(*), COUNT(*) * 1.0 / SUM(COUNT(*)) OVER ()
FROM object
GROUP BY object_category;

【讨论】：

【解决方案2】：

您可以使用WINDOW FUNCTIONS。

WITH OBJECTT
     AS (SELECT
               1 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               1 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               1 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               1 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               1 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               1 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               5 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               5 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               4 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               4 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               4 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               2 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               3 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               2 OBJECT_CATEGORY 
         UNION ALL
         SELECT
               6 OBJECT_CATEGORY )
SELECT DISTINCT
       OBJECT_CATEGORY,
       COUNT (*) OVER (PARTITION BY OBJECT_CATEGORY) / COUNT (*) OVER () AS RATIO
  FROM OBJECTT

【讨论】：