SQL Server has traditionally shied away from providing native solutions to some of the more common statistical questions, such as calculating a median. According to WikiPedia, "median is described as the numerical value separating the higher half of a sample, a population, or a probability distribution, from the lower half. The median of a finite list of numbers can be found by arranging all the observations from lowest value to highest value and picking the middle one. If there is an even number of observations, then there is no single middle value; the median is then usually defined to be the mean of the two middle values."

In terms of a SQL Server query, the key thing you'll take away from that is that you need to "arrange" (sort) all of the values. Sorting in SQL Server is typically a pretty expensive operation if there isn't a supporting index, and adding an index to support an operation which probably isn't requested that often may not be worthwhile.

Let's examine how we have typically solved this problem in previous versions of SQL Server. First let's create a very simple table so that we can eyeball that our logic is correct and deriving an accurate median. We can test the following two tables, one with an even number of rows, and the other with an odd number of rows:

CREATE TABLE dbo.EvenRows ( id INT PRIMARY KEY, val INT ); CREATE TABLE dbo.OddRows ( id INT PRIMARY KEY, val INT ); INSERT dbo.EvenRows(id,val) SELECT 1, 6 UNION ALL SELECT 2, 11 UNION ALL SELECT 3, 4 UNION ALL SELECT 4, 4 UNION ALL SELECT 5, 15 UNION ALL SELECT 6, 14 UNION ALL SELECT 7, 4 UNION ALL SELECT 8, 9; INSERT dbo.OddRows(id,val) SELECT 1, 6 UNION ALL SELECT 2, 11 UNION ALL SELECT 3, 4 UNION ALL SELECT 4, 4 UNION ALL SELECT 5, 15 UNION ALL SELECT 6, 14 UNION ALL SELECT 7, 4; DECLARE @Median DECIMAL(12, 2);

Just from casual observance, we can see that the median for the table with odd rows should be 6, and for the even table it should be 7.5 ((6+9)/2). So now let's see some solutions that have been used over the years:

## SQL Server 2000

In SQL Server 2000, we were constrained to a very limited T-SQL dialect. I'm investigating these options for comparison because some people out there are still running SQL Server 2000, and others may have upgraded but, since their median calculations were written "back in the day," the code might still look like this today.

### 2000_A – max of one half, min of the other

This approach takes the highest value from the first 50 percent, the lowest value from the last 50 percent, then divides them by two. This works for even or odd rows because, in the even case, the two values are the two middle rows, and in the odd case, the two values are actually from the same row.

SELECT @Median = ( (SELECT MAX(val) FROM (SELECT TOP 50 PERCENT val FROM dbo.EvenRows ORDER BY val, id) AS t) + (SELECT MIN(val) FROM (SELECT TOP 50 PERCENT val FROM dbo.EvenRows ORDER BY val DESC, id DESC) AS b) ) / 2.0;

### 2000_B – #temp table

This example first creates a #temp table, and using the same type of math as above, determines the two "middle" rows with assistance from a contiguous `IDENTITY`

column ordered by the val column. (The order of assignment of `IDENTITY`

values can only be relied upon because of the `MAXDOP`

setting.)

CREATE TABLE #x ( i INT IDENTITY(1,1), val DECIMAL(12, 2) ); CREATE CLUSTERED INDEX v ON #x(val); INSERT #x(val) SELECT val FROM dbo.EvenRows ORDER BY val OPTION (MAXDOP 1); SELECT @Median = AVG(val) FROM #x AS x WHERE EXISTS ( SELECT 1 FROM #x WHERE x.i - (SELECT MAX(i) / 2.0 FROM #x) IN (0, 0.5, 1) );

## SQL Server 2005, 2008, 2008 R2

SQL Server 2005 introduced some interesting new window functions, such as `ROW_NUMBER()`

, which can help solve statistical problems like median a little easier than we could in SQL Server 2000. These approaches all work in SQL Server 2005 and above:

### 2005_A – dueling row numbers

This example uses `ROW_NUMBER()`

to walk up and down the values once in each direction, then finds the "middle" one or two rows based on that calculation. This is quite similar to the first example above, with easier syntax:

SELECT @Median = AVG(1.0 * val) FROM ( SELECT val, ra = ROW_NUMBER() OVER (ORDER BY val, id), rd = ROW_NUMBER() OVER (ORDER BY val DESC, id DESC) FROM dbo.EvenRows ) AS x WHERE ra BETWEEN rd - 1 AND rd + 1;

### 2005_B – row number + count

This one is quite similar to the above, using a single calculation of `ROW_NUMBER()`

and then using the total `COUNT()`

to find the "middle" one or two rows:

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

### 2005_C – variation on row number + count

Fellow MVP Itzik Ben-Gan showed me this method, which achieves the same answer as the above two methods, but in a very slightly different way:

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);

## SQL Server 2012

In SQL Server 2012, we have new windowing capabilities in T-SQL that allow statistical calculations like median to be expressed more directly. To calculate the median for a set of values, we can use `PERCENTILE_CONT()`

. We can also use the new "paging" extension to the `ORDER BY`

clause (`OFFSET / FETCH`

).

### 2012_A – new distribution functionality

This solution uses a very straightforward calculation using distribution (if you don't want the average between the two middle values in the case of an even number of rows).

SELECT @Median = PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY val) OVER () FROM dbo.EvenRows;

### 2012_B – paging trick

This example implements a clever use of `OFFSET / FETCH`

(and not exactly one for which it was intended) – we simply move to the row that is one before half the count, then take the next one or two rows depending on whether the count was odd or even. Thanks to Itzik Ben-Gan for pointing out this approach.

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;

## But which one performs better?

We've verified that the above methods all produce the expected results on our little table, and we know that the SQL Server 2012 version has the cleanest and most logical syntax. But which one should you be using in your busy production environment? We can build a much bigger table from system metadata, making sure we have plenty of duplicate values. This script will produce a table with 10,000,000 non-unique integers:

USE tempdb; GO CREATE TABLE dbo.obj(id INT IDENTITY(1,1), val INT); CREATE CLUSTERED INDEX x ON dbo.obj(val, id); INSERT dbo.obj(val) SELECT TOP (10000000) o.[object_id] FROM sys.all_columns AS c CROSS JOIN sys.all_objects AS o CROSS JOIN sys.all_objects AS o2 WHERE o.[object_id] > 0 ORDER BY c.[object_id];

On my system the median for this table should be 146,099,561. I can calculate this pretty quickly without a manual spot check of 10,000,000 rows by using the following query:

SELECT val FROM ( SELECT val, rn = ROW_NUMBER() OVER (ORDER BY val) FROM dbo.obj ) AS x WHERE rn IN (4999999, 5000000, 5000001);

Results:

val rn ---- ---- 146099561 4999999 146099561 5000000 146099561 5000001

So now we can create a stored procedure for each method, verify that each one produces the correct output, and then measure performance metrics such as duration, CPU and reads. We'll perform all of these steps with the existing table, and also with a copy of the table that does not benefit from the clustered index (we'll drop it and re-create the table as a heap).

I've created seven procedures implementing the query methods above. For brevity I won't list them here, but each one is named `dbo.Median_<version>`

, e.g. `dbo.Median_2000_A`

, `dbo.Median_2000_B`

, etc. corresponding to the approaches described above. If we run these seven procedures using the free SQL Sentry Plan Explorer, here is what we observe in terms of duration, CPU and reads (note that we run DBCC FREEPROCCACHE and DBCC DROPCLEANBUFFERS in between executions):

And these metrics don't change much at all if we operate against a heap instead. The biggest percentage change was the method that still ended up being the fastest: the paging trick using OFFSET / FETCH:

Here is a graphical representation of the results. To make it more clear, I highlighted the slowest performer in red and the fastest approach in green.

I was surprised to see that, in both cases, `PERCENTILE_CONT()`

– which was designed for this type of calculation – is actually worse than all of the other earlier solutions. I guess it just goes to show that while sometimes newer syntax might make our coding easier, it doesn't always guarantee that performance will improve. I was also surprised to see `OFFSET / FETCH`

prove to be so useful in scenarios that usually wouldn't seem to fit its purpose – pagination.

In any case, I hope I have demonstrated which approach you should use, depending on your version of SQL Server (and that the choice should be the same whether or not you have a supporting index for the calculation).

Aaron, interesting article. A few comments:

A) 2012_B can be re-written for SQL Server 2005 or 2008 by doing something like the following. (I adapted the algorithm directly from Itzik's query — perhaps it can be simplified.)

B) It would be nice to see a followup article comparing the relative merits of different techniques for partitioned median, rather than a median on the entire set. (e.g. median for each product, or for each salesperson, or customer, or whatever).

C) When inserting data into your sample tables you might consider using the TABLOCK hint; on my end this makes the insert about 3x faster, which is nice when we're talking about performance :-)

Is there a mistake with 2005_C – variation on row number + count (possibly elsewhere…)?

I was testing it with a toy example in Excel to verify and I think the where clause should be : WHERE rn IN ((c )/2, (c + 1)/2),(c + 2)/2);

Hi John, I don't think so, can you explain to me what result set would require the ((c)/2) expression to be included? in the IN() check? I played with it a bit tonight and couldn't contrive a case where the query I posted returned incorrect results. I'm more than happy to look at your sample data, what the query returns, and what you think it should return – you can e-mail me at abertrand AT sqlsentry DOT net. Sorry, I'm old-school "security through obscurity."

I think we are both correct (in a sense) .

Since SQL will round down these numbers in the where clause ,my example I made in excel to test your logic is not quite right.

If SQL did not round then my code would be correct or if you convert my code to FLOATs it should work.

I learned something!

Ah, I see, I should state explicitly somewhere that this is meant to find the median of a set of *whole numbers* not float values. :-)

I ran over the same data set 2005_A and 2005_B.

I think there is something not quite right about the where clause with 2005_A. I modified it to calculate within a group (median per id). It had a lower count than 2005_B which was spot on.

I'm not sure what the issue is.

Looks like they have the same counts on the derived tables until the where clause is applied.

Here's my code.

I found the issue with 2005_A. I have ID's with ties. For example, 10 entries but there are 8 zeros. Those ties fall out due to how the where clause is working. Long story, short, I'm sticking with 2005_B.

2005_C modified to handle multiple Id's was 9x as long time-wise.

I took Adam Machanic's suggestion and decided to find out how some Median solutions would perform with partitioned sets:

https://www.simple-talk.com/sql/t-sql-programming/calculating-the-median-value-within-a-partitioned-set-using-t-sql/

I hope you guys don't mind me building on your work.

Calculate the median depend group

`SELECT @Median = AVG(1.0 * Val)`

FROM

(

SELECT val,

c = COUNT(*) OVER (Partition by EventsID),

rn = ROW_NUMBER() OVER (Partition by EventsID ORDER BY val)

FROM dbo.EvenRows

) AS x

WHERE rn IN ((c + 1)/2, (c + 2)/2)

group by EventsID

Not at all Dwain; hope you don't mind me building on yours (and Peso's). :-)

http://www.sqlperformance.com/2014/02/t-sql-queries/grouped-median

Aaron,

Of course I don't mind. The "science of SQL" is like any other type of science. Where would we be if we couldn't build on the work of earlier researchers? Constructing our homes with stone knives and bear skins most likely.

Keep up the great work!

Hi Aaron,

If performance is king and you don't mind leaving your successors some tricky code to whine about, I think it's possible to do a lot faster than 2012_B for a heap with this approach:

If stats don't exist on this column, create them (0s or 380ms on my server)

You already know where I'm going with this – without needing to be accurate the stats give you up to 200 RANGE_HI_KEYs , with the median likely to be near the middle one.

Dump the histogram into a temp table (8ms)

Scan the dbo.obj table a few times with queries like these to establish the range that truly contains the median:

select count(1) from dbo.obj where val < 677577452 –RANGE_HI_KEY no. 100 (400ms)

select count(1) from dbo.obj where 677577452 <= val and val < 693577509 –RANGE_HI_KEY no. 100 and 101 (250ms)

select count(1) from dbo.obj where 693577509 <= val and val < 709577566 –RANGE_HI_KEY no. 101 and 102 (250ms)

If you're lucky two queries are sufficient, but in my example I needed 3. You don't need to see how badly I coded the while loop with variables in place of literals.

These counts also give us the total number of rows lower than the range, so now we can run a much smaller sort for our offset fetch:

SELECT AVG(1.0 * val)

FROM (

SELECT val FROM dbo.Obj where 693577509 <= val and val < 709577566

ORDER BY val

OFFSET (51202) / 2 ROWS

FETCH NEXT 2 ROWS ONLY

) AS x;

That's another 265ms on my laptop making a total of 1550 ms vs. 4650 ms when I reproduce the vanilla 2012_B. If I already had statistics, and found the interval at the first attempt (i.e. only 2 of those 3 queries), this would be under 1000ms.

The performance comparison is of course more dramatic for 2005.

I didn't explain that too well, so I've written up a fuller description here:

https://simoncbirch.wordpress.com/2016/02/06/median-strip/

Great Article! Thanks.