Using MEDIAN along side MAX, MIN, and AVG functions in MySQL

Question

I have the following MySQL query which is working perfectly:

select 
    count(*) as `# of Data points`, 
    name, 
    max((QNTY_Sell/QNTYDelivered)*1000) as `MAX Thousand Price`,
    min((QNTY_Sell/QNTYDelivered)*1000) as `MIN Thousand Price`,
    avg((QNTY_Sell/QNTYDelivered)*1000) as `MEAN Thousand Price` 
from 
    table_name 
where 
    year(date) >= 2012 and 
    name like "%the_name%" and 
    QNTYDelivered > 0 and 
    QNTY_Sell > 0 
group by name 
order by name;

Now I wish to also add a result column that gives me the MEDIAN of the data for each line. Under SELECT this would look like this in a perfect world:

median((QNTY_Sell/QNTYDelivered)*1000) as `MEDIAN Thousand Price`

Searching Google for a MySQL median function brought me to this answer, which seems ok if you are interested in the median of a data set for an entire table: Simple way to calculate median with MySQL

The difference here is that I am grouping the data in my table by the name column, and want to get the median for each line of the data grouped by this column.

Does anyone know a nifty trick to allow me to do this?

Thanks!

Answer 1

You can calculate the median with GROUP BY in MySQL even though there is no median function built in.

Consider the table:

Acrington   200.00
Acrington   200.00
Acrington   300.00
Acrington   400.00
Bulingdon   200.00
Bulingdon   300.00
Bulingdon   400.00
Bulingdon   500.00
Cardington  100.00
Cardington  149.00
Cardington  151.00
Cardington  300.00
Cardington  300.00

For each row you can count the number of similar items that are less. You can also count how many values are less than or equal:

name        v       <   <=
Acrington   200.00  0   2
Acrington   200.00  0   2
Acrington   300.00  2   3
Acrington   400.00  3   4
Bulingdon   200.00  0   1
Bulingdon   300.00  1   2
Bulingdon   400.00  2   3
Bulingdon   500.00  3   4
Cardington  100.00  0   1
Cardington  149.00  1   2
Cardington  151.00  2   3
Cardington  300.00  3   5
Cardington  300.00  3   5

With query

SELECT name,v, (SELECT COUNT(1) FROM sale WHERE v<o.v AND name=o.name) as ls
             , (SELECT COUNT(1) FROM sale WHERE v<=o.v AND name=o.name) as lse
  FROM sale o

The median value will occur when the less-than-or-equal count is half the number of items

• Acrington has 4 items. Half of this is 2 which is in the range 0..2 (corresponding to 200.00) and also in the range 2..3 (corresponding to 300.00)

• Bullingdon also has 4 items. 2 is in the range 1..2 (value 300.00) and 2..3 (value 400.00)

• Cardington has 5 items. The value 2.5 is between 2 and 3 which corresponds to Cardington 151.

The median value is the mean of the min and max values returned by:

SELECT cs.name,v
   FROM
   (SELECT name,v, (SELECT COUNT(1) FROM sale WHERE v<o.v AND name=o.name) as ls
                 , (SELECT COUNT(1) FROM sale WHERE v<=o.v AND name=o.name) as lse
      FROM sale o) cs JOIN
   (SELECT name,COUNT(1)*.5 as cn
      FROM sale
      GROUP BY name) cc ON cs.name=cc.name
 WHERE cn between ls and lse

Which gives:

Acrington   200.00
Acrington   200.00
Acrington   300.00
Bulingdon   300.00
Bulingdon   400.00
Cardington  151.00

Finally we can get the median:

SELECT name,(MAX(v)+MIN(v))/2 FROM
(SELECT cs.name,v
   FROM
   (SELECT name,v, (SELECT COUNT(1) FROM sale WHERE v<o.v AND name=o.name) as ls
                 , (SELECT COUNT(1) FROM sale WHERE v<=o.v AND name=o.name) as lse
      FROM sale o) cs JOIN
   (SELECT name,COUNT(1)*.5 as cn
      FROM sale
     GROUP BY name) cc ON cs.name=cc.name
 WHERE cn between ls and lse
 ) AS medians
GROUP BY name

Giving

Acrington   250.000000
Bulingdon   350.000000
Cardington  151.000000