使用 PSQL 根据来自其他列的值生成百分比答案

【问题标题】：Generate percentage from based on values from other column using PSQL使用 PSQL 根据来自其他列的值生成百分比
【发布时间】：2019-08-27 01:59:32
【问题描述】：

我想根据表中其他列的值创建一个新列，其中包含 50%、60% 等值。从下面显示的输出中，我希望有基于“cnt”列中的值的“Desired Results”列。目前我的输入数据如下图所示

我只能从下面的查询中获取记录的数量。但是我无法生成百分比。你能帮帮我吗？

with test as
(
select subject_id,hadm_id,case 
             when valuenum between 80 and 110 then 1
             else 0
             end as "within_range"
         from labevents where itemid in ('50809','50931','51529') and 
hadm_id is not null 


) select subject_id,hadm_id,within_range,count(*) as cnt
from test group by subject_id,hadm_id,within_range

我希望输出如下所示

【问题讨论】：

这些百分比背后的逻辑是什么
如果你看上面的数据，对于 subject_id = 3，有两行，总 cnt 是 22，所以 14/22 是 64，8/22 是 36。其余的类似。对于只有一条记录的科目，则为 100 pc

标签： sql postgresql group-by percentage

【解决方案1】：

为此，您可以有两个子查询，其中一个按 hadm_id 分组，另一个不按 hadm_id 分组，并将它们都加入。

select a.* ,(a.cnt/b.cnt)*100 
from(select subject_id,hadm_id,within_range,count(*) as cnt
FROM (select subject_id,hadm_id,case 
             when valuenum between 80 and 110 then 1
             else 0
             end as "within_range"
         from labevents where itemid in ('50809','50931','51529') and 
hadm_id is not null) 
group by subject_id,hadm_id,within_range)a
INNER JOIN
(select subject_id,within_range,count(*) as cnt
FROM (select subject_id,hadm_id,case 
             when valuenum between 80 and 110 then 1
             else 0
             end as "within_range"
         from labevents where itemid in ('50809','50931','51529') and 
hadm_id is not null) 
group by subject_id,within_range)b
on a.subject_id,b.subject_id and a.within_range=b.within_range

【讨论】：

【解决方案2】：

使用窗口函数：http://www.postgresqltutorial.com/postgresql-window-function/

with cte as
     (
       select subject_id,
              hadm_id,
              case
                when valuenum between 80 and 110 then 1
                else 0
                end as "within_range"
         from labevents
        where itemid in ('50809', '50931', '51529')
          and hadm_id is not null
     ),
   subq as (
     select subject_id,
            hadm_id,
            within_range,
            count(*) as cnt
       from cte
      group by subject_id, hadm_id, within_range
   ) 
select subq.*, (cnt / sum(cnt) OVER (PARTITION BY subject_id, hadm_id)) * 100 "Desired Results" 
from subq;

【讨论】：

根据您的逻辑，您应该调整“PARTITION BY”列

【解决方案3】：

您可以将窗口函数与group by 一起使用。此外，CTE 并不是必需的，特别是因为 Postgres 允许为 group by 使用列别名：

select subject_id, hadm_id,
       (case when valuenum between 80 and 110 then 1
             else 0
        end) as within_range,
       count(*) as cnt,
       count(*) * 100.0 / sum(count(*)) over () as percentage
from labevents
where itemid in ('50809', '50931', '51529') and 
      hadm_id is not null 
group by subject_id, hadm_id, within_range

【讨论】：