Unable to calculate median - SQL Server 2017 - sql

Question

I am trying to computer the median number of transactions in each category.
A few notes (as the dataset below is a small snippet of a much larger dataset):
An employee can belong to multiple categories
Each transaction's median should be > 0
Not every person appears in every category
The data is set up like this:
| Person | Category | Transaction |
|:-------:|:--------:|:-----------:|
| PersonA | Sales | 27 |
| PersonB | Sales | 75 |
| PersonC | Sales | 87 |
| PersonD | Sales | 36 |
| PersonE | Sales | 70 |
| PersonB | Buys | 60 |
| PersonC | Buys | 92 |
| PersonD | Buys | 39 |
| PersonA | HR | 59 |
| PersonB | HR | 53 |
| PersonC | HR | 98 |
| PersonD | HR | 54 |
| PersonE | HR | 70 |
| PersonA | Other | 46 |
| PersonC | Other | 66 |
| PersonD | Other | 76 |
| PersonB | Other | 2 |
An ideal output would look like:
| Category | Median | Average |
|:--------:|:------:|:-------:|
| Sales | 70 | 59 |
| Buys | 60 | 64 |
| HR | 59 | 67 |
| Other | 56 | 48 |
I can get the average by:
SELECT
Category,
AVG(Transaction) AS Average_Transactions
FROM
table
GROUP BY
Category
And that works great!
This post tried to help me find the median. What I wrote was:
SELECT
Category,
PERCENTILE_DISC(0.5) WITHIN GROUP (ORDER BY Transaction) OVER (PARTITION BY Category) AS Median_Transactions
FROM
table
GROUP BY
Category
But I get an error:
Msg 8120: Column 'Transactions' is invalid in the select list because it is not contained in either an aggregate function or the **GROUP BY** clause
How can I fix this?

Answer 1

You can do what you want using SELECT DISTINCT:
SELECT DISTINCT Category,
PERCENTILE_DISC(0.5) WITHIN GROUP (ORDER BY Transaction) OVER (PARTITION BY Category) AS Median_Transactions
FROM table;
Unfortunately, SQL Server doesn't offer the PERCENTILE_ functions as window functions and doesn't have a MEDIAN() aggregation function. You can also do this using subqueries and counts.

Answer 2

It's not optimal but this is your solution
SELECT DISTINCT
category,
PERCENTILE_DISC(0.5)WITHIN GROUP(ORDER BY val) OVER (PARTITION BY category) AS Median_Transactions,
AVG(val) OVER (PARTITION BY d.category) [AVG]
FROM #data d;

Answer 3

I don't think this is pretty but it works. I didn't spend time on polishing it
with
avg_t as
( select category, avg(sales) as avg_sales
from sample
group by 1),
mn as
( select category, avg(sales) as median_sales
from (
select category, sales ,
row_number() over (partition by category order by sales asc) as r ,
count(person) over (partition by category) as total_count
from sample
) mn_sub
where (total_count % 2 = 0 and r in ( (total_count/2), ((total_count/2)+1)) ) or
(total_count % 2 <> 0 and r = ((total_count+1)/2))
group by 1
)
select avg_t.category, avg_t.avg_sales, mn.median_sales
from avg_t
inner join mn
on avg_t.category=mn.category