Function to Calculate Median in SQL Server
Sql ServerTsqlAggregate FunctionsMedianSql Server Problem Overview
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?
Sql Server Solutions
Solution 1 - Sql Server
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
Solution 2 - Sql Server
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;
Solution 3 - Sql Server
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
Solution 4 - Sql Server
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.
Solution 5 - Sql Server
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!
Solution 6 - Sql Server
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
Solution 7 - Sql Server
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)
Solution 8 - Sql Server
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
Solution 9 - Sql Server
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
Solution 10 - Sql Server
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
Solution 11 - Sql Server
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
Solution 12 - Sql Server
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
Solution 13 - Sql Server
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
Solution 14 - Sql Server
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)
Solution 15 - Sql Server
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)
Solution 16 - Sql Server
See other solutions for median calculation in SQL here: "Simple way to calculate median with MySQL" (the solutions are mostly vendor-independent).
Solution 17 - Sql Server
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.
Solution 18 - Sql Server
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
Solution 19 - Sql Server
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
Solution 20 - Sql Server
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.
Solution 21 - Sql Server
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));
Solution 22 - Sql Server
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)
Solution 23 - Sql Server
--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]
Solution 24 - Sql Server
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
)
Solution 25 - 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!!!
Solution 26 - Sql Server
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
Solution 27 - Sql Server
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) ;
Solution 28 - Sql Server
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
Solution 29 - Sql Server
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;
Solution 30 - Sql Server
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.
Solution 31 - Sql Server
This is the most optimal solution for finding medians that I can think of. The names in the example is based on Justin example. Make sure an index for table Sales.SalesOrderHeader exists with index columns CustomerId and TotalDue in that order.
SELECT
sohCount.CustomerId,
AVG(sohMid.TotalDue) as TotalDueMedian
FROM
(SELECT
soh.CustomerId,
COUNT(*) as NumberOfRows
FROM
Sales.SalesOrderHeader soh
GROUP BY soh.CustomerId) As sohCount
CROSS APPLY
(Select
soh.TotalDue
FROM
Sales.SalesOrderHeader soh
WHERE soh.CustomerId = sohCount.CustomerId
ORDER BY soh.TotalDue
OFFSET sohCount.NumberOfRows / 2 - ((sohCount.NumberOfRows + 1) % 2) ROWS
FETCH NEXT 1 + ((sohCount.NumberOfRows + 1) % 2) ROWS ONLY
) As sohMid
GROUP BY sohCount.CustomerId
UPDATE
I was a bit unsure about which method has best performance, so I did a comparison between my method Justin Grants and Jeff Atwoods by running query based on all three methods in one batch and the batch cost of each query were:
Without index:
- Mine 30%
- Justin Grants 13%
- Jeff Atwoods 58%
And with index
- Mine 3%.
- Justin Grants 10%
- Jeff Atwoods 87%
I tried to see how well the queries scale if you have index by creating more data from around 14 000 rows by a factor of 2 up to 512 which means in the end around 7,2 millions rows. Note I made sure CustomeId field where unique for each time I did a single copy, so the proportion of rows compared to unique instance of CustomerId was kept constant. While I was doing this I ran executions where I rebuilt index afterwards, and I noticed the results stabilized at around a factor of 128 with the data I had to these values:
- Mine 3%.
- Justin Grants 5%
- Jeff Atwoods 92%
I wondered how the performance could have been affected by scaling number of of rows but keeping unique CustomerId constant, so I setup a new test where I did just this. Now instead of stabilizing, the batch cost ratio kept diverging, also instead of about 20 rows per CustomerId per average I had in the end around 10000 rows per such unique Id. The numbers where:
- Mine 4%
- Justins 60%
- Jeffs 35%
I made sure I implemented each method correct by comparing the results. My conclusion is the method I used is generally faster as long as index exists. Also noticed that this method is what's recommended for this particular problem in this article https://www.microsoftpressstore.com/articles/article.aspx?p=2314819&seqNum=5
A way to even further improve performance of subsequent calls to this query even further is to persist the count information in an auxiliary table. You could even maintain it by having a trigger that update and holds information regarding the count of SalesOrderHeader rows dependant on CustomerId, of course you can then simple store the median as well.
Solution 32 - Sql Server
Below is My solution:
with tempa as
(
select value,row_number() over (order by value) as Rn,/* Assigning a
row_number */
count(value) over () as Cnt /*Taking total count of the values */
from numbers
where value is not null /* Excluding the null values */
),
tempb as
(
/* Since we don't know whether the number of rows is odd or even, we shall
consider both the scenarios */
select round(cnt/2) as Ref from tempa where mod(cnt,2)=1
union all
select round(cnt/2) a Ref from tempa where mod(cnt,2)=0
union all
select round(cnt/2) + 1 as Ref from tempa where mod(cnt,2)=0
)
select avg(value) as Median_Value
from tempa where rn in
( select Ref from tempb);
Solution 33 - Sql Server
Using the COUNT aggregate, you can first count how many rows there are and store in a variable called @cnt. Then you can compute parameters for the OFFSET-FETCH filter to specify, based on qty ordering, how many rows to skip (offset value) and how many to filter (fetch value).
The number of rows to skip is (@cnt - 1) / 2. It’s clear that for an odd count this calculation is correct because you first subtract 1 for the single middle value, before you divide by 2.
This also works correctly for an even count because the division used in the expression is integer division; so, when subtracting 1 from an even count, you’re left with an odd value.
When dividing that odd value by 2, the fraction part of the result (.5) is truncated. The number of rows to fetch is 2 - (@cnt % 2). The idea is that when the count is odd the result of the modulo operation is 1, and you need to fetch 1 row. When the count is even the result of the modulo operation is 0, and you need to fetch 2 rows. By subtracting the 1 or 0 result of the modulo operation from 2, you get the desired 1 or 2, respectively. Finally, to compute the median quantity, take the one or two result quantities, and apply an average after converting the input integer value to a numeric one as follows:
DECLARE @cnt AS INT = (SELECT COUNT(*) FROM [Sales].[production].[stocks]);
SELECT AVG(1.0 * quantity) AS median
FROM ( SELECT quantity
FROM [Sales].[production].[stocks]
ORDER BY quantity
OFFSET (@cnt - 1) / 2 ROWS FETCH NEXT 2 - @cnt % 2 ROWS ONLY ) AS D;
Solution 34 - Sql Server
In my solution table is a student table having marks column only and I am calculating median of marks and this solution is based on SQL server 2019
with total_c as ( --Total_c CTE counts total number of rows in a table
select count(*) as n from student
),
even as ( --Even CTE extract two middle rows if the number of rows are even
select marks from student
order by marks
offset (select n from total_c)/2 -1 rows
fetch next 2 rows only
),
odd as ( --Odd CTE extract middle row if the number of rows are odd
select marks from student
order by marks
offset (select n + 1 from total_c)/2 -1 rows
fetch next 1 rows only
)
--Case statement helps to select odd or even CTE based on number of rows
select
case when n%2 = 0 then (select avg(cast(marks as float)) from even)
else (select marks from odd)
end as med_marks
from total_c
Solution 35 - Sql Server
This code is a bit lengthy, but easy to understand
medi is table which has column 'val' which has dataset, smedi is a cte which has column idx as rownumber and vals as 'val' from medi table which is sorted in ascending order. Then its basic math, if row number is odd then its middle value from smedi. and when its even its average of middle two values.
with smedi(idx,vals) as(
select ROW_NUMBER() over(order by val),val from medi
)
select (case
when (select count(*) from medi)%2!=0 then (select vals from smedi where (((select count(*) from medi)/2))=idx)
else (select avg(vals) from smedi where idx in ((select count(*)/2 from medi),(select (count(*)/2)+1 from medi)))
end)
Solution 36 - Sql Server
Try the below logic to find out the median:
Consider a table with the below numbers: 1,1,2,3,4,5
THE MEDIAN is 2.5
with tempa as
(
select num,count(num) over() as Cnt,
row_number() over (order by num) as Rnum
from temp),
tempb as
(
select round(cnt/2) as ref_value
from tempa where mod(cnt,2)<>0
union all
select round(cnt/2) from tempa where mod(cnt,2)=0
union all
select round(cnt/2+1)
from tempa where mod(cnt,2)=0
)
select avg(num) from tempa
where rnum in (select * from tempb);