如何根据 SQL 的汇总结果计算每日平均值？

Question

我正在输出一些数据，我想提取一些数字的日平均值。

如您所见，我想做的是计算行数 received/results（考虑行 ID），然后将其除以日值以获得每日平均值。（30/1 ) , (64/2) 等我已经尝试了所有方法，但我用这个运行挡住了墙。

就目前而言，我猜想使这项工作需要某种子查询。我只是不知道如何获取用于除法的日期（行号 1、2、3、4 等）。

SELECT calendar_date, SUM(NY_dayscore * cAttendance) 
FROM  vw_Appointments 
WHERE status = 'Confirmed'
Group by calendar_date

尝试使用 distinct 进行计数，但无济于事

SUM(NY_dayscore * cAttendance) ) / count(distinct calendar_date)

我的原始代码很长，cba 到 post 全部。所以只是尝试 post 一个小的示例代码来获得关于这个问题的指导。

Answer 1

我觉得你需要sum(showed) over (..)/row_number() over (..)

WITH Table1(date, showed) AS
(
 SELECT '2019-01-02', 30 UNION ALL
 SELECT '2019-01-03', 34 UNION ALL
 SELECT '2019-01-03', 41 UNION ALL
 SELECT '2019-01-04', 48
)
SELECT date, 
       sum(showed) over (order by date) / 
       row_number() over (order by date)
       as daily_average
 FROM Table1
GROUP BY showed, date;

date        daily_average
2019-01-02  30
2019-01-03  52
2019-01-03  35
2019-01-04  38

Demo

Answer 2

你是不是ROW_NUMBER()不见了？

SELECT 
  calendar_date, 
  SUM(NY_dayscore * cAttendance) / (ROW_NUMBER() OVER (ORDER BY calendar_date ASC)) AS average
FROM  vw_Appointments 
WHERE status = 'Confirmed'
GROUP BY calendar_date
ORDER BY calendar_date

Answer 3

在 SQL Server 2012+ 中，您将使用累积平均值：

select calendar_date, sum(NY_dayscore * cAttendance),
       avg(sum(NY_dayscore * cAttendance)) over (order by calendar_date) as running_average
from vw_appointments a
where status = 'Confirmed'
group by calendar_date
order by calendar_date;

在 SQL Server 2008 中，这更困难：

with a as (
      select calendar_date, sum(NY_dayscore * cAttendance) as showed
      from vw_appointments a
      where status = 'Confirmed'
      group by calendar_date
     )
select a.*, a2.running_average
from a outer apply
     (select avg(showed) as running_average
      from a a2
      where a2.calendar_date <= a.calendar_date
     ) a2
order by calendar_date;

如何根据 SQL 的汇总结果计算每日平均值？

How to calculate daily average from aggregate results with SQL?

sql

sql-server

sql-server-2008