Function to Calculate Median in SQL Server

For newbies like myself who are learning the very basics, I personally find this example easier to follow, as it is easier to understand exactly what's happening and where median values are coming from...

select
( max(a.[Value1]) + min(a.[Value1]) ) / 2 as [Median Value1]
,( max(a.[Value2]) + min(a.[Value2]) ) / 2 as [Median Value2]

from (select
datediff(dd,startdate,enddate) as [Value1]
,xxxxxxxxxxxxxx as [Value2]
from dbo.table1
)a


In absolute awe of some of the codes above though!!!

This is as simple an answer as I could come up with. Worked well with my data. If you want to exclude certain values just add a where clause to the inner select.

SELECT TOP 1
ValueField AS MedianValue
FROM
(SELECT TOP(SELECT COUNT(1)/2 FROM tTABLE)
ValueField
FROM
tTABLE
ORDER BY
ValueField) A
ORDER BY
ValueField DESC

My original quick answer was:

select max(my_column) as [my_column], quartile
from (select my_column, ntile(4) over (order by my_column) as [quartile]
from my_table) i
--where quartile = 2
group by quartile

This will give you the median and interquartile range in one fell swoop. If you really only want one row that is the median then uncomment the where clause.

When you stick that into an explain plan, 60% of the work is sorting the data which is unavoidable when calculating position dependent statistics like this.

I've amended the answer to follow the excellent suggestion from Robert Ševčík-Robajz in the comments below:

;with PartitionedData as
(select my_column, ntile(10) over (order by my_column) as [percentile]
from my_table),
MinimaAndMaxima as
(select min(my_column) as [low], max(my_column) as [high], percentile
from PartitionedData
group by percentile)
select
case
when b.percentile = 10 then cast(b.high as decimal(18,2))
else cast((a.low + b.high) as decimal(18,2)) / 2
end as [value], --b.high, a.low,
b.percentile
from MinimaAndMaxima a
join MinimaAndMaxima b on (a.percentile -1 = b.percentile) or (a.percentile = 10 and b.percentile = 10)
--where b.percentile = 5

This should calculate the correct median and percentile values when you have an even number of data items. Again, uncomment the final where clause if you only want the median and not the entire percentile distribution.

Justin's example above is very good. But that Primary key need should be stated very clearly. I have seen that code in the wild without the key and the results are bad.

The complaint I get about the Percentile_Cont is that it wont give you an actual value from the dataset.
To get to a "median" that is an actual value from the dataset use Percentile_Disc.


SELECT SalesOrderID, OrderQty,
PERCENTILE_DISC(0.5)
WITHIN GROUP (ORDER BY OrderQty)
OVER (PARTITION BY SalesOrderID) AS MedianCont
FROM Sales.SalesOrderDetail
WHERE SalesOrderID IN (43670, 43669, 43667, 43663)
ORDER BY SalesOrderID DESC

The following solution works under these assumptions:

- No duplicate values
- No NULLs

Code:


IF OBJECT_ID('dbo.R', 'U') IS NOT NULL
DROP TABLE dbo.R

CREATE TABLE R (
A FLOAT NOT NULL);

INSERT INTO R VALUES (1);
INSERT INTO R VALUES (2);
INSERT INTO R VALUES (3);
INSERT INTO R VALUES (4);
INSERT INTO R VALUES (5);
INSERT INTO R VALUES (6);

-- Returns Median®
select SUM(A) / CAST(COUNT(A) AS FLOAT)
from R R1
where ((select count(A) from R R2 where R1.A > R2.A) =
(select count(A) from R R2 where R1.A < R2.A)) OR
((select count(A) from R R2 where R1.A > R2.A) + 1 =
(select count(A) from R R2 where R1.A < R2.A)) OR
((select count(A) from R R2 where R1.A > R2.A) =
(select count(A) from R R2 where R1.A < R2.A) + 1) ;

DECLARE @Obs int
DECLARE @RowAsc table
(
ID INT IDENTITY,
Observation FLOAT
)
INSERT INTO @RowAsc
SELECT Observations FROM MyTable
ORDER BY 1
SELECT @Obs=COUNT(*)/2 FROM @RowAsc
SELECT Observation AS Median FROM @RowAsc WHERE ID=@Obs

I try with several alternatives, but due my data records has repeated values, the ROW_NUMBER versions seems are not a choice for me. So here the query I used (a version with NTILE):


SELECT distinct
CustomerId,
(
MAX(CASE WHEN Percent50_Asc=1 THEN TotalDue END) OVER (PARTITION BY CustomerId) +
MIN(CASE WHEN Percent50_desc=1 THEN TotalDue END) OVER (PARTITION BY CustomerId)
)/2 MEDIAN
FROM
(
SELECT
CustomerId,
TotalDue,
NTILE(2) OVER (
PARTITION BY CustomerId
ORDER BY TotalDue ASC) AS Percent50_Asc,
NTILE(2) OVER (
PARTITION BY CustomerId
ORDER BY TotalDue DESC) AS Percent50_desc
FROM Sales.SalesOrderHeader SOH
) x
ORDER BY CustomerId;

Even better:

SELECT @Median = AVG(1.0 * val)
FROM
(
SELECT o.val, rn = ROW_NUMBER() OVER (ORDER BY o.val), c.c
FROM dbo.EvenRows AS o
CROSS JOIN (SELECT c = COUNT(*) FROM dbo.EvenRows) AS c
) AS x
WHERE rn IN ((c + 1)/2, (c + 2)/2);

From the master Himself, [Itzik Ben-Gan][1]!


[1]:

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Building on Jeff Atwood's answer above here it is with GROUP BY and a correlated subquery to get the median for each group.

SELECT TestID,
(
(SELECT MAX(Score) FROM
(SELECT TOP 50 PERCENT Score FROM Posts WHERE TestID = Posts_parent.TestID ORDER BY Score) AS BottomHalf)
+
(SELECT MIN(Score) FROM
(SELECT TOP 50 PERCENT Score FROM Posts WHERE TestID = Posts_parent.TestID ORDER BY Score DESC) AS TopHalf)
) / 2 AS MedianScore,
AVG(Score) AS AvgScore, MIN(Score) AS MinScore, MAX(Score) AS MaxScore
FROM Posts_parent
GROUP BY Posts_parent.TestID

MS SQL Server 2012 (and later) has the PERCENTILE_DISC function which computes a specific percentile for sorted values. PERCENTILE_DISC (0.5) will compute the median -

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