Calculate Mode and Median for data in SQL [duplicate] - sql

Question

According to MSDN, Median is not available as an aggregate function in Transact-SQL. However, I would like to find out whether it is possible to create this functionality (using the Create Aggregate function, user defined function, or some other method).
What would be the best way (if possible) to do this - allow for the calculation of a median value (assuming a numeric data type) in an aggregate query?

Answer 1

If you're using SQL 2005 or better this is a nice, simple-ish median calculation for a single column in a table:
SELECT
(
(SELECT MAX(Score) FROM
(SELECT TOP 50 PERCENT Score FROM Posts ORDER BY Score) AS BottomHalf)
+
(SELECT MIN(Score) FROM
(SELECT TOP 50 PERCENT Score FROM Posts ORDER BY Score DESC) AS TopHalf)
) / 2 AS Median

Answer 2

2019 UPDATE: In the 10 years since I wrote this answer, more solutions have been uncovered that may yield better results. Also, SQL Server releases since then (especially SQL 2012) have introduced new T-SQL features that can be used to calculate medians. SQL Server releases have also improved its query optimizer which may affect perf of various median solutions. Net-net, my original 2009 post is still OK but there may be better solutions on for modern SQL Server apps. Take a look at this article from 2012 which is a great resource: https://sqlperformance.com/2012/08/t-sql-queries/median
This article found the following pattern to be much, much faster than all other alternatives, at least on the simple schema they tested. This solution was 373x faster (!!!) than the slowest (PERCENTILE_CONT) solution tested. Note that this trick requires two separate queries which may not be practical in all cases. It also requires SQL 2012 or later.
DECLARE #c BIGINT = (SELECT COUNT(*) FROM dbo.EvenRows);
SELECT AVG(1.0 * val)
FROM (
SELECT val FROM dbo.EvenRows
ORDER BY val
OFFSET (#c - 1) / 2 ROWS
FETCH NEXT 1 + (1 - #c % 2) ROWS ONLY
) AS x;
Of course, just because one test on one schema in 2012 yielded great results, your mileage may vary, especially if you're on SQL Server 2014 or later. If perf is important for your median calculation, I'd strongly suggest trying and perf-testing several of the options recommended in that article to make sure that you've found the best one for your schema.
I'd also be especially careful using the (new in SQL Server 2012) function PERCENTILE_CONT that's recommended in one of the other answers to this question, because the article linked above found this built-in function to be 373x slower than the fastest solution. It's possible that this disparity has been improved in the 7 years since, but personally I wouldn't use this function on a large table until I verified its performance vs. other solutions.
ORIGINAL 2009 POST IS BELOW:
There are lots of ways to do this, with dramatically varying performance. Here's one particularly well-optimized solution, from Medians, ROW_NUMBERs, and performance. This is a particularly optimal solution when it comes to actual I/Os generated during execution – it looks more costly than other solutions, but it is actually much faster.
That page also contains a discussion of other solutions and performance testing details. Note the use of a unique column as a disambiguator in case there are multiple rows with the same value of the median column.
As with all database performance scenarios, always try to test a solution out with real data on real hardware – you never know when a change to SQL Server's optimizer or a peculiarity in your environment will make a normally-speedy solution slower.
SELECT
CustomerId,
AVG(TotalDue)
FROM
(
SELECT
CustomerId,
TotalDue,
-- SalesOrderId in the ORDER BY is a disambiguator to break ties
ROW_NUMBER() OVER (
PARTITION BY CustomerId
ORDER BY TotalDue ASC, SalesOrderId ASC) AS RowAsc,
ROW_NUMBER() OVER (
PARTITION BY CustomerId
ORDER BY TotalDue DESC, SalesOrderId DESC) AS RowDesc
FROM Sales.SalesOrderHeader SOH
) x
WHERE
RowAsc IN (RowDesc, RowDesc - 1, RowDesc + 1)
GROUP BY CustomerId
ORDER BY CustomerId;

Answer 3

In SQL Server 2012 you should use PERCENTILE_CONT:
SELECT SalesOrderID, OrderQty,
PERCENTILE_CONT(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
See also : http://blog.sqlauthority.com/2011/11/20/sql-server-introduction-to-percentile_cont-analytic-functions-introduced-in-sql-server-2012/

Answer 4

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.

Answer 5

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!

Answer 6

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 - https://msdn.microsoft.com/en-us/library/hh231327.aspx

Answer 7

Simple, fast, accurate
SELECT x.Amount
FROM (SELECT amount,
Count(1) OVER (partition BY 'A') AS TotalRows,
Row_number() OVER (ORDER BY Amount ASC) AS AmountOrder
FROM facttransaction ft) x
WHERE x.AmountOrder = Round(x.TotalRows / 2.0, 0)

Answer 8

If you want to use the Create Aggregate function in SQL Server, this is how to do it. Doing it this way has the benefit of being able to write clean queries. Note this this process could be adapted to calculate a Percentile value fairly easily.
Create a new Visual Studio project and set the target framework to .NET 3.5 (this is for SQL 2008, it may be different in SQL 2012). Then create a class file and put in the following code, or c# equivalent:
Imports Microsoft.SqlServer.Server
Imports System.Data.SqlTypes
Imports System.IO
<Serializable>
<SqlUserDefinedAggregate(Format.UserDefined, IsInvariantToNulls:=True, IsInvariantToDuplicates:=False, _
IsInvariantToOrder:=True, MaxByteSize:=-1, IsNullIfEmpty:=True)>
Public Class Median
Implements IBinarySerialize
Private _items As List(Of Decimal)
Public Sub Init()
_items = New List(Of Decimal)()
End Sub
Public Sub Accumulate(value As SqlDecimal)
If Not value.IsNull Then
_items.Add(value.Value)
End If
End Sub
Public Sub Merge(other As Median)
If other._items IsNot Nothing Then
_items.AddRange(other._items)
End If
End Sub
Public Function Terminate() As SqlDecimal
If _items.Count <> 0 Then
Dim result As Decimal
_items = _items.OrderBy(Function(i) i).ToList()
If _items.Count Mod 2 = 0 Then
result = ((_items((_items.Count / 2) - 1)) + (_items(_items.Count / 2))) / 2#
Else
result = _items((_items.Count - 1) / 2)
End If
Return New SqlDecimal(result)
Else
Return New SqlDecimal()
End If
End Function
Public Sub Read(r As BinaryReader) Implements IBinarySerialize.Read
'deserialize it from a string
Dim list = r.ReadString()
_items = New List(Of Decimal)
For Each value In list.Split(","c)
Dim number As Decimal
If Decimal.TryParse(value, number) Then
_items.Add(number)
End If
Next
End Sub
Public Sub Write(w As BinaryWriter) Implements IBinarySerialize.Write
'serialize the list to a string
Dim list = ""
For Each item In _items
If list <> "" Then
list += ","
End If
list += item.ToString()
Next
w.Write(list)
End Sub
End Class
Then compile it and copy the DLL and PDB file to your SQL Server machine and run the following command in SQL Server:
CREATE ASSEMBLY CustomAggregate FROM '{path to your DLL}'
WITH PERMISSION_SET=SAFE;
GO
CREATE AGGREGATE Median(#value decimal(9, 3))
RETURNS decimal(9, 3)
EXTERNAL NAME [CustomAggregate].[{namespace of your DLL}.Median];
GO
You can then write a query to calculate the median like this:
SELECT dbo.Median(Field) FROM Table

Answer 9

I just came across this page while looking for a set based solution to median. After looking at some of the solutions here, I came up with the following. Hope is helps/works.
DECLARE #test TABLE(
i int identity(1,1),
id int,
score float
)
INSERT INTO #test (id,score) VALUES (1,10)
INSERT INTO #test (id,score) VALUES (1,11)
INSERT INTO #test (id,score) VALUES (1,15)
INSERT INTO #test (id,score) VALUES (1,19)
INSERT INTO #test (id,score) VALUES (1,20)
INSERT INTO #test (id,score) VALUES (2,20)
INSERT INTO #test (id,score) VALUES (2,21)
INSERT INTO #test (id,score) VALUES (2,25)
INSERT INTO #test (id,score) VALUES (2,29)
INSERT INTO #test (id,score) VALUES (2,30)
INSERT INTO #test (id,score) VALUES (3,20)
INSERT INTO #test (id,score) VALUES (3,21)
INSERT INTO #test (id,score) VALUES (3,25)
INSERT INTO #test (id,score) VALUES (3,29)
DECLARE #counts TABLE(
id int,
cnt int
)
INSERT INTO #counts (
id,
cnt
)
SELECT
id,
COUNT(*)
FROM
#test
GROUP BY
id
SELECT
drv.id,
drv.start,
AVG(t.score)
FROM
(
SELECT
MIN(t.i)-1 AS start,
t.id
FROM
#test t
GROUP BY
t.id
) drv
INNER JOIN #test t ON drv.id = t.id
INNER JOIN #counts c ON t.id = c.id
WHERE
t.i = ((c.cnt+1)/2)+drv.start
OR (
t.i = (((c.cnt+1)%2) * ((c.cnt+2)/2))+drv.start
AND ((c.cnt+1)%2) * ((c.cnt+2)/2) <> 0
)
GROUP BY
drv.id,
drv.start

Answer 10

The following query returns the median from a list of values in one column. It cannot be used as or along with an aggregate function, but you can still use it as a sub-query with a WHERE clause in the inner select.
SQL Server 2005+:
SELECT TOP 1 value from
(
SELECT TOP 50 PERCENT value
FROM table_name
ORDER BY value
)for_median
ORDER BY value DESC

Answer 11

Although Justin grant's solution appears solid I found that when you have a number of duplicate values within a given partition key the row numbers for the ASC duplicate values end up out of sequence so they do not properly align.
Here is a fragment from my result:
KEY VALUE ROWA ROWD
13 2 22 182
13 1 6 183
13 1 7 184
13 1 8 185
13 1 9 186
13 1 10 187
13 1 11 188
13 1 12 189
13 0 1 190
13 0 2 191
13 0 3 192
13 0 4 193
13 0 5 194
I used Justin's code as the basis for this solution. Although not as efficient given the use of multiple derived tables it does resolve the row ordering problem I encountered. Any improvements would be welcome as I am not that experienced in T-SQL.
SELECT PKEY, cast(AVG(VALUE)as decimal(5,2)) as MEDIANVALUE
FROM
(
SELECT PKEY,VALUE,ROWA,ROWD,
'FLAG' = (CASE WHEN ROWA IN (ROWD,ROWD-1,ROWD+1) THEN 1 ELSE 0 END)
FROM
(
SELECT
PKEY,
cast(VALUE as decimal(5,2)) as VALUE,
ROWA,
ROW_NUMBER() OVER (PARTITION BY PKEY ORDER BY ROWA DESC) as ROWD
FROM
(
SELECT
PKEY,
VALUE,
ROW_NUMBER() OVER (PARTITION BY PKEY ORDER BY VALUE ASC,PKEY ASC ) as ROWA
FROM [MTEST]
)T1
)T2
)T3
WHERE FLAG = '1'
GROUP BY PKEY
ORDER BY PKEY

Answer 12

In a UDF, write:
Select Top 1 medianSortColumn from Table T
Where (Select Count(*) from Table
Where MedianSortColumn <
(Select Count(*) From Table) / 2)
Order By medianSortColumn

Answer 13

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

Answer 14

Using a single statement - One way is to use ROW_NUMBER(), COUNT() window function and filter the sub-query. Here is to find the median salary:
SELECT AVG(e_salary)
FROM
(SELECT
ROW_NUMBER() OVER(ORDER BY e_salary) as row_no,
e_salary,
(COUNT(*) OVER()+1)*0.5 AS row_half
FROM Employee) t
WHERE row_no IN (FLOOR(row_half),CEILING(row_half))
I have seen similar solutions over the net using FLOOR and CEILING but tried to use a single statement. (edited)

Answer 15

Median Finding
This is the simplest method to find the median of an attribute.
Select round(S.salary,4) median from employee S
where (select count(salary) from station
where salary < S.salary ) = (select count(salary) from station
where salary > S.salary)

Answer 16

See other solutions for median calculation in SQL here:
"Simple way to calculate median with MySQL" (the solutions are mostly vendor-independent).

Answer 17

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

Answer 18

For a continuous variable/measure 'col1' from 'table1'
select col1
from
(select top 50 percent col1,
ROW_NUMBER() OVER(ORDER BY col1 ASC) AS Rowa,
ROW_NUMBER() OVER(ORDER BY col1 DESC) AS Rowd
from table1 ) tmp
where tmp.Rowa = tmp.Rowd

Answer 19

Frequently, we may need to calculate Median not just for the whole table, but for aggregates with respect to some ID. In other words, calculate median for each ID in our table, where each ID has many records. (based on the solution edited by #gdoron: good performance and works in many SQL)
SELECT our_id, AVG(1.0 * our_val) as Median
FROM
( SELECT our_id, our_val,
COUNT(*) OVER (PARTITION BY our_id) AS cnt,
ROW_NUMBER() OVER (PARTITION BY our_id ORDER BY our_val) AS rnk
FROM our_table
) AS x
WHERE rnk IN ((cnt + 1)/2, (cnt + 2)/2) GROUP BY our_id;
Hope it helps.

Answer 20

For large scale datasets, you can try this GIST:
https://gist.github.com/chrisknoll/1b38761ce8c5016ec5b2
It works by aggregating the distinct values you would find in your set (such as ages, or year of birth, etc.), and uses SQL window functions to locate any percentile position you specify in the query.

Answer 21

To get median value of salary from employee table
with cte as (select salary, ROW_NUMBER() over (order by salary asc) as num from employees)
select avg(salary) from cte where num in ((select (count(*)+1)/2 from employees), (select (count(*)+2)/2 from employees));

Answer 22

I wanted to work out a solution by myself, but my brain tripped and fell on the way. I think it works, but don't ask me to explain it in the morning. :P
DECLARE #table AS TABLE
(
Number int not null
);
insert into #table select 2;
insert into #table select 4;
insert into #table select 9;
insert into #table select 15;
insert into #table select 22;
insert into #table select 26;
insert into #table select 37;
insert into #table select 49;
DECLARE #Count AS INT
SELECT #Count = COUNT(*) FROM #table;
WITH MyResults(RowNo, Number) AS
(
SELECT RowNo, Number FROM
(SELECT ROW_NUMBER() OVER (ORDER BY Number) AS RowNo, Number FROM #table) AS Foo
)
SELECT AVG(Number) FROM MyResults WHERE RowNo = (#Count+1)/2 OR RowNo = ((#Count+1)%2) * ((#Count+2)/2)

Answer 23

--Create Temp Table to Store Results in
DECLARE #results AS TABLE
(
[Month] datetime not null
,[Median] int not null
);
--This variable will determine the date
DECLARE #IntDate as int
set #IntDate = -13
WHILE (#IntDate < 0)
BEGIN
--Create Temp Table
DECLARE #table AS TABLE
(
[Rank] int not null
,[Days Open] int not null
);
--Insert records into Temp Table
insert into #table
SELECT
rank() OVER (ORDER BY DATEADD(mm, DATEDIFF(mm, 0, DATEADD(ss, SVR.close_date, '1970')), 0), DATEDIFF(day,DATEADD(ss, SVR.open_date, '1970'),DATEADD(ss, SVR.close_date, '1970')),[SVR].[ref_num]) as [Rank]
,DATEDIFF(day,DATEADD(ss, SVR.open_date, '1970'),DATEADD(ss, SVR.close_date, '1970')) as [Days Open]
FROM
mdbrpt.dbo.View_Request SVR
LEFT OUTER JOIN dbo.dtv_apps_systems vapp
on SVR.category = vapp.persid
LEFT OUTER JOIN dbo.prob_ctg pctg
on SVR.category = pctg.persid
Left Outer Join [mdbrpt].[dbo].[rootcause] as [Root Cause]
on [SVR].[rootcause]=[Root Cause].[id]
Left Outer Join [mdbrpt].[dbo].[cr_stat] as [Status]
on [SVR].[status]=[Status].[code]
LEFT OUTER JOIN [mdbrpt].[dbo].[net_res] as [net]
on [net].[id]=SVR.[affected_rc]
WHERE
SVR.Type IN ('P')
AND
SVR.close_date IS NOT NULL
AND
[Status].[SYM] = 'Closed'
AND
SVR.parent is null
AND
[Root Cause].[sym] in ( 'RC - Application','RC - Hardware', 'RC - Operational', 'RC - Unknown')
AND
(
[vapp].[appl_name] in ('3PI','Billing Rpts/Files','Collabrent','Reports','STMS','STMS 2','Telco','Comergent','OOM','C3-BAU','C3-DD','DIRECTV','DIRECTV Sales','DIRECTV Self Care','Dealer Website','EI Servlet','Enterprise Integration','ET','ICAN','ODS','SB-SCM','SeeBeyond','Digital Dashboard','IVR','OMS','Order Services','Retail Services','OSCAR','SAP','CTI','RIO','RIO Call Center','RIO Field Services','FSS-RIO3','TAOS','TCS')
OR
pctg.sym in ('Systems.Release Health Dashboard.Problem','DTV QA Test.Enterprise Release.Deferred Defect Log')
AND
[Net].[nr_desc] in ('3PI','Billing Rpts/Files','Collabrent','Reports','STMS','STMS 2','Telco','Comergent','OOM','C3-BAU','C3-DD','DIRECTV','DIRECTV Sales','DIRECTV Self Care','Dealer Website','EI Servlet','Enterprise Integration','ET','ICAN','ODS','SB-SCM','SeeBeyond','Digital Dashboard','IVR','OMS','Order Services','Retail Services','OSCAR','SAP','CTI','RIO','RIO Call Center','RIO Field Services','FSS-RIO3','TAOS','TCS')
)
AND
DATEADD(mm, DATEDIFF(mm, 0, DATEADD(ss, SVR.close_date, '1970')), 0) = DATEADD(mm, DATEDIFF(mm,0,DATEADD(mm,#IntDate,getdate())), 0)
ORDER BY [Days Open]
DECLARE #Count AS INT
SELECT #Count = COUNT(*) FROM #table;
WITH MyResults(RowNo, [Days Open]) AS
(
SELECT RowNo, [Days Open] FROM
(SELECT ROW_NUMBER() OVER (ORDER BY [Days Open]) AS RowNo, [Days Open] FROM #table) AS Foo
)
insert into #results
SELECT
DATEADD(mm, DATEDIFF(mm,0,DATEADD(mm,#IntDate,getdate())), 0) as [Month]
,AVG([Days Open])as [Median] FROM MyResults WHERE RowNo = (#Count+1)/2 OR RowNo = ((#Count+1)%2) * ((#Count+2)/2)
set #IntDate = #IntDate+1
DELETE FROM #table
END
select *
from #results
order by [Month]

Answer 24

This works with SQL 2000:
DECLARE #testTable TABLE
(
VALUE INT
)
--INSERT INTO #testTable -- Even Test
--SELECT 3 UNION ALL
--SELECT 5 UNION ALL
--SELECT 7 UNION ALL
--SELECT 12 UNION ALL
--SELECT 13 UNION ALL
--SELECT 14 UNION ALL
--SELECT 21 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 29 UNION ALL
--SELECT 40 UNION ALL
--SELECT 56
--
--INSERT INTO #testTable -- Odd Test
--SELECT 3 UNION ALL
--SELECT 5 UNION ALL
--SELECT 7 UNION ALL
--SELECT 12 UNION ALL
--SELECT 13 UNION ALL
--SELECT 14 UNION ALL
--SELECT 21 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 29 UNION ALL
--SELECT 39 UNION ALL
--SELECT 40 UNION ALL
--SELECT 56
DECLARE #RowAsc TABLE
(
ID INT IDENTITY,
Amount INT
)
INSERT INTO #RowAsc
SELECT VALUE
FROM #testTable
ORDER BY VALUE ASC
SELECT AVG(amount)
FROM #RowAsc ra
WHERE ra.id IN
(
SELECT ID
FROM #RowAsc
WHERE ra.id -
(
SELECT MAX(id) / 2.0
FROM #RowAsc
) BETWEEN 0 AND 1
)

Answer 25

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!!!

Answer 26

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

Answer 27

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(R)
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) ;

Answer 28

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

Answer 29

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;

Answer 30

For your question, Jeff Atwood had already given the simple and effective solution. But, if you are looking for some alternative approach to calculate the median, below SQL code will help you.
create table employees(salary int);
insert into employees values(8); insert into employees values(23); insert into employees values(45); insert into employees values(123); insert into employees values(93); insert into employees values(2342); insert into employees values(2238);
select * from employees;
declare #odd_even int; declare #cnt int; declare #middle_no int;
set #cnt=(select count(*) from employees); set #middle_no=(#cnt/2)+1; select #odd_even=case when (#cnt%2=0) THEN -1 ELse 0 END ;
select AVG(tbl.salary) from (select salary,ROW_NUMBER() over (order by salary) as rno from employees group by salary) tbl where tbl.rno=#middle_no or tbl.rno=#middle_no+#odd_even;
If you are looking to calculate median in MySQL, this github link will be useful.