Finding the Median value using SQL

Here is a simple method of finding the median value using SQL. The first will be a simple example that finds the median number of units sold for a given book (ISBN) over a period of time. There will be 3 sections of code (that will be combined using CTEs). First we need to find the median number, that is for a given book, how many times was it sold. The sales_data table contains 3 columns: ISBN, SaleDate, UnitsSold

select ISBN,
       ceiling(count(*)/2.0) as median
  from sales_data
 group by ISBN

The use of ceiling in the above equation accounts for the fact that if we have only 1 sale for an ISBN, then 1/2=0, whereas ceiling(1/2.0)=1. The need for the median to start at 1 and not 0 will become evident in the text below.

select ISBN,
       ROW_NUMBER() OVER(Partition by ISBN Order by Units_Sold) as r_num
  from sales_data

Since we want to know the median of UnitsSold we must Order by UnitsSold. Using the rownumber() function we get a set of numbers, starting at 1 (here is the need for ceiling above) and continuing until we get a new ISBN. Once we get a new ISBN, the numbering restarts at 1. Let’s say that our test data looks like this:

ISBN Units_Sold r_num

9780000000001 10 1

9780000000001 13 2

9780000000001 20 3

Our median will be ceiling(3/2.0)=2. That means the median value for Units_Sold for ISBN=9780000000001 is 13. The full code is below. Please note I create this as a view so there is no need to be a ; before the first with.

Create view MedianUnitsSold
with base
select ISBN,
       ceiling(count(*)/2.0) as median
  from sales_data
 group by ISBN
select ISBN,
       ROW_NUMBER() OVER(Partition by ISBN Order by Units_Sold) as r_num
  from sales_data
select p.ISBN,
  from population p
  join base b on (b.ISBN=p.ISBN
       and b.median=p.r_num)

As always, I hope this post has been able to help you solve a problem that you’ve had or provided you with some new insight. Feel free to post comments below.

Auto-increment without the need of a table.

I came across a situation where I need unique numbers for both a large and small set of data. The standard approach is to create a table with an identity column and do your inserts. For a table with tens of thousands of records, this might be the best approach, but there is another one: ROW_NUMBER().

The key to using the ROW_NUMBER() function is to provide a constant so you will get a contiguous sequence.  Let's look at sample data.


Let's look at the function:

ROW_NUMBER() OVER (Partition by <group> order by <columns>)

I can first construct the function as follows

 select Format,

However this will give us the following results:

id Format
 1   AB
 1   AL
 1   CD
 1   CE
 1   GG
 1   LI

The reason is your PARTITION BY creates a unique number for each format. To solve this problem you need to put a value that will never change over the life of the select. I use 1.

 select Format,

The new result set is:

id Format
 1   AB
 2   AL
 3   CD
 4   CE
 5   GG
 6   LI

Hope this helps improve your sql. As always, leave comments below.





Grouping data by 15 minute intervals

After a rather long course change (I was doing little work with SQL Server and work with other RDBMS systems), I'm back working on Microsoft SQL Server and a few other Microsoft products. Here is an interesting problem presented to me a few days ago.

A colleague of mine need timed data grouped by 15 minute intervals. I found this to be an interesting problem and didn't see much online, so thought I'd post this quick tidbit for others to find. The technique can be used to group by any interval with a little tweaking.

The field in question is a datetime field. The code needed to group data in 15 min intervals is this:


Let's break it down. I take the minutes of the datetime field and divide it by 15, which is my interval. That will give me a number between 0 and 3, because it is an integer/integer. I then multiple that by 15. The distribution is as follows:

  • (00-14)/15 = 0*15=0
  • (15-29)/15= 1*15 = 15
  • (30-44)/15=2*15=30
  • (45-59)=3*15=45

Datepart(mi,"2018-5-9 11:48:33")/15) = 48/15 = 3*15 = 45. Now add in the the date and hours.

left(convert(varchar,"2018-5-9 11:48:33",120),14)+RIGHT('00' + CONVERT(VARCHAR(2), (datepart(mi,"2018-5-9 11:48:33")/15)*15), 2) = "2018-5-9 11:45:00"

If you wanted to do a 10 min interval, this: datepart(mi,C001_TimeEnd)/15)*15 becomes this datepart(mi,C001_TimeEnd)/10)*10.


Using Row_Number to pick the correct entry.

There have been numerous occasions where I have to pick a one of two records out of a given result set, but how I determine which one has some unique rules.  I have two examples of this and I show how I used Row_Number() to solve this.

The first set of data has these sets of rules:

  • For any given ISBN13, give me the 'Acquisition Editor' if they exist and if they do not exist, provide the 'Editor', known as [Role].
  • If there are multiple persons for the chosen [Role], pick the one with the lowest [SortOrder].
  • If [SortOrder] is null, pick the first name alphabetically.

Here is the solution:

select ISBN13,
  from (select ISBN13,
               ROW_NUMBER() OVER(Partition by ISBN13 order by [Role],isnull(cast(SortOrder as varchar(40)),DisplayName)) as Row
          from dbo.ProductContact
          where C006_Role in ('Acquisition Editor','Editor')) d
where Row=1

The second set of data had the following rules:

  • For a given ISBN13, provide the sum of the  'First Printing', which can be under either PrintNumber '0000' or '0001'. 
  • PrintNumber '0000' is always chosen over '0001' if it exists.

Here is the solution:

select isbn13,
   from (SELECT isbn13,
                Sum(Print_Run) as First_Run,                
                ROW_NUMBER() OVER(PARTITION by isbn13 order by Printing_Number) as Row
           FROM dbo.PPB
          where Printing_Number in ('0000','0001')
          group by isbn13,Printing_Number) d
  where Row=1