how to get median for every record? - sql

Question

There's no median function in sql server, so I'm using this wonderful suggestion:
https://stackoverflow.com/a/2026609/117700
this computes the median over an entire dataset, but I need the median per record.
My dataset is:
+-----------+-------------+
| client_id | TimesTested |
+-----------+-------------+
| 214220 | 1 |
| 215425 | 1 |
| 212839 | 4 |
| 215249 | 1 |
| 210498 | 3 |
| 110655 | 1 |
| 110655 | 1 |
| 110655 | 12 |
| 215425 | 4 |
| 100196 | 1 |
| 110032 | 1 |
| 110032 | 1 |
| 101944 | 3 |
| 101232 | 2 |
| 101232 | 1 |
+-----------+-------------+
here's the query I am using:
select client_id,
(
SELECT
(
(SELECT MAX(TimesTested ) FROM
(SELECT TOP 50 PERCENT t.TimesTested
FROM counted3 t
where t.timestested>1
and CLIENT_ID=t.CLIENT_ID
ORDER BY t.TimesTested ) AS BottomHalf)
+
(SELECT MIN(TimesTested ) FROM
(SELECT TOP 50 PERCENT t.TimesTested
FROM counted3 t
where t.timestested>1
and CLIENT_ID=t.CLIENT_ID
ORDER BY t.TimesTested DESC) AS TopHalf)
) / 2 AS Median
) TotalAvgTestFreq
from counted3
group by client_id
but it is giving my funny data:
+-----------+------------------+
| client_id | median???????????|
+-----------+------------------+
| 100007 | 84 |
| 100008 | 84 |
| 100011 | 84 |
| 100014 | 84 |
| 100026 | 84 |
| 100027 | 84 |
| 100028 | 84 |
| 100029 | 84 |
| 100042 | 84 |
| 100043 | 84 |
| 100071 | 84 |
| 100072 | 84 |
| 100074 | 84 |
+-----------+------------------+
i can i get the median for every client_id ?
I am currently trying to use this awesome query from Aaron's site:
select c3.client_id,(
SELECT AVG(1.0 * TimesTested ) median
FROM
(
SELECT o.TimesTested ,
rn = ROW_NUMBER() OVER (ORDER BY o.TimesTested ), c.c
FROM counted3 AS o
CROSS JOIN (SELECT c = COUNT(*) FROM counted3) AS c
where count>1
) AS x
WHERE rn IN ((c + 1)/2, (c + 2)/2)
) a
from counted3 c3
group by c3.client_id
unfortunately, as Richardthekiwi points out:
it's for a single median whereas this question is about a median
per-partition
i would like to know how i can join it on counted3 to get the median per partition?>

Answer 1

Note: If testFreq is an int or bigint type, you need to CAST it before taking an average, otherwise you'll get integer division, e.g. (2+5)/2 => 3 if 2 and 5 are the median records - e.g. AVG(Cast(testfreq as float)).
select client_id, avg(testfreq) median_testfreq
from
(
select client_id,
testfreq,
rn=row_number() over (partition by CLIENT_ID
order by testfreq),
c=count(testfreq) over (partition by CLIENT_ID)
from tbk
where timestested>1
) g
where rn in (round(c/2,0),c/2+1)
group by client_id;
The median is found either as the central record in an ODD number of rows, or the average of the two central records in an EVEN number of rows. This is handled by the condition rn in (round(c/2,0),c/2+1) which picks either the one or two records required.

Answer 2

try this:
select client_id,
(
SELECT
(
(SELECT MAX(testfreq) FROM
(SELECT TOP 50 PERCENT t.testfreq
FROM counted3 t
where t.timestested>1
and c3.CLIENT_ID=t.CLIENT_ID
ORDER BY t.testfreq) AS BottomHalf)
+
(SELECT MIN(testfreq) FROM
(SELECT TOP 50 PERCENT t.testfreq
FROM counted3 t
where t.timestested>1
and c3.CLIENT_ID=t.CLIENT_ID
ORDER BY t.testfreq DESC) AS TopHalf)
) / 2 AS Median
) TotalAvgTestFreq
from counted3 c3
group by client_id
I added the c3 alias to the outer CLIENT_ID references and the outer table.