计算组内的百分比

Jon*_*Jon 19 sql

给出一个表格,用于以下命令:

select sex, count(*) from my_table group by sex;
select sex, employed, count(*) from my_table group by sex, employed;
Run Code Online (Sandbox Code Playgroud)

得到:

  sex  | count 
-------+------
male   | 1960 
female | 1801
Run Code Online (Sandbox Code Playgroud)

和:

 sex     | employed | count 
---------+----------+-------
 male    | f        |  1523 
 male    | t        |   437 
 female  | f        |  1491 
 female  | t        |   310 
Run Code Online (Sandbox Code Playgroud)

我在编写查询时遇到困难,该查询将计算每个性别组中的就业百分比.所以输出应该如下所示:

 sex     | employed | count  | percent
---------+----------+--------+-----------
 male    | f        |  1523  | 77.7% (1523/1960)
 male    | t        |   437  | 22.3% (437/1960)
 female  | f        |  1491  | 82.8% (1491/1801)
 female  | t        |   310  | 17.2% (310/1801)
Run Code Online (Sandbox Code Playgroud)

小智 27

可能为时已晚,但对于即将到来的搜索者,可能的解决方案可能是:

select sex, employed, COUNT(*) / CAST( SUM(count(*)) over (partition by sex) as float)
  from my_table
 group by sex, employed
Run Code Online (Sandbox Code Playgroud)

通过IO统计,这似乎是最有效的解决方案 - 可能取决于要查询的行数 - 在上面的数字上测试...

同样的态度可用于获得男/女比例:

select sex, COUNT(*) / CAST( SUM(count(*)) over () as float)
  from my_table
 group by sex
Run Code Online (Sandbox Code Playgroud)

此致,Jan


out*_*tis 10

您可以使用子选择和连接来执行此操作:

SELECT t1.sex, employed, count(*) AS `count`, count(*) / t2.total AS percent
  FROM my_table AS t1
  JOIN (
    SELECT sex, count(*) AS total 
      FROM my_table
      GROUP BY sex
  ) AS t2
  ON t1.sex = t2.sex
  GROUP BY t1.sex, employed;
Run Code Online (Sandbox Code Playgroud)

我无法想到其他方法.