Calculate SQL Median - sql

Question

I was trying to implement a median from this solution (among others, but this seemed the simplest Median code): Function to Calculate Median in Sql Server
However, I'm having difficulty in its application. This is my current SQL query. My goal is to find the Median TotalTimeOnCall for CallerComplaintTypeID on a given Week, Month, and Department. I think my biggest issue is that I'm just fundamentally not understanding how to apply this Median function to achieve my results.
For example, if I needed an Average, instead, I could just change that ORDER BY to a GROUP BY and then slap an AVG(TotalTimeOnCall) instead. How do I accomplish this idea with this Median solution, instead?
This is the "raw data" query:
WITH rawData as (
SELECT
DepartmentName
,MONTH(PlacedOnLocal) AS MonthNumber
,CASE
WHEN Datepart(day, PlacedOnLocal) < 8 THEN '1'
WHEN Datepart(day, PlacedOnLocal) < 15 THEN '2'
WHEN Datepart(day, PlacedOnLocal) < 22 THEN '3'
WHEN Datepart(day, PlacedOnLocal) < 29 THEN '4'
ELSE '5'
END AS WeekNumber
,CallerComplaintTypeID
,TotalTimeOnCall
FROM [THE_RELEVANT_TABLE]
WHERE PlacedOnLocal BETWEEN '2014-09-01' AND '2014-12-31'
AND CallerComplaintTypeID IN (5,89,9,31,203)
AND TotalTimeOnCall IS NOT NULL
)
SELECT
DepartmentName,
MonthNumber,
WeekNumber,
CallerComplaintTypeID,
TotalTimeOnCall
FROM
rawData
ORDER BY DepartmentName, MonthNumber, WeekNumber, CallerComplaintTypeID
with this sample output:
DepartmentName MonthNumber WeekNumber CallerComplaintTypeID TotalTimeOnCall
Dept_01 9 1 5 654
Dept_01 9 1 5 156
Dept_01 9 1 5 21
Dept_01 9 1 5 67
Dept_01 9 1 5 13
Dept_01 9 1 5 97
Dept_01 9 1 5 87
Dept_01 9 1 5 16
this is the Median solution from above:
SELECT
(
(
SELECT MAX(TotalTimeOnCall)
FROM
(
SELECT TOP 50 PERCENT TotalTimeOnCall
FROM rawData
WHERE TotalTimeOnCall IS NOT NULL
ORDER BY TotalTimeOnCall
) AS BottomHalf
)
+
(
SELECT MIN(TotalTimeOnCall)
FROM
(
SELECT TOP 50 PERCENT TotalTimeOnCall
FROM rawData
WHERE TotalTimeOnCall IS NOT NULL
ORDER BY TotalTimeOnCall DESC
) AS TopHalf
)
) / 2 AS Median

Answer 1

Here is a simple median solution that allows you to get a median per group.
-- Example of how to get median from a set of data
;with cte_my_query as (
-- this cte simulates the query that would return your data
select '2016-01-01' as dt, 1 as val
union
select '2016-01-01' as dt, 10 as val
union
select '2016-01-01' as dt, 7 as val
union
select '2016-01-01' as dt, 16 as val
union
select '2016-01-01' as dt, 11 as val
union
select '2016-01-01' as dt, 2 as val
union
select '2016-01-01' as dt, 5 as val
union
select '2016-01-02' as dt, 6 as val
union
select '2016-01-02' as dt, 13 as val
union
select '2016-01-02' as dt, 7 as val
union
select '2016-01-02' as dt, 9 as val
union
select '2016-01-02' as dt, 18 as val
)
,cte_dates as (
-- get the distinct key we want to get median for
select distinct dt from cte_my_query
)
select dt, median.val
from cte_dates
cross apply (
-- of the top 50% (below), take the top 1, desc, which is the median value
select top 1 val from (
-- for each date, get the top 50% of the values
select top 50 percent val
from cte_my_query
where cte_dates.dt = cte_my_query.dt
order by dt
) as inner_median
order by inner_median.val desc
) median