How to calculate daily average from aggregate results with SQL? - sql

Question

I'm working on outputting some data and I want to pull the daily average of some numbers.
As you can see, what I want to do is count the amount of rows received/results(think the row ID) and then divide it against the day value to make the daily average.(30/1) , (64/2) etc I've tried everything, but I keep running into a wall with this.
As it stands, I'm guessing to make this work a sub query of some sort is needed. I just don't know how to get the day(Row id 1,2,3,4 etc) to use for the division.
SELECT calendar_date, SUM(NY_dayscore * cAttendance)
FROM vw_Appointments
WHERE status = 'Confirmed'
Group by calendar_date
Attempted count with distinct, to no avail
SUM(NY_dayscore * cAttendance) ) / count(distinct calendar_date)
My original code is long and cba to post it all. So just attempting to post a small sample code to get guidance on the issue.

Answer 1

In SQL Server 2012+, you would use the cumulative average:
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;
In SQL Server 2008, this is more difficult:
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;

Answer 2

Is it ROW_NUMBER() that you are missing?
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

I think you need 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