Simple way to calculate median with MySQL

Question

In MariaDB / MySQL:

SELECT AVG(dd.val) as median_val
FROM (
SELECT d.val, @rownum:=@rownum+1 as `row_number`, @total_rows:=@rownum
  FROM data d, (SELECT @rownum:=0) r
  WHERE d.val is NOT NULL
  -- put some where clause here
  ORDER BY d.val
) as dd
WHERE dd.row_number IN ( FLOOR((@total_rows+1)/2), FLOOR((@total_rows+2)/2) );

Steve Cohen points out, that after the first pass, @rownum will contain the total number of rows. This can be used to determine the median, so no second pass or join is needed.

Also AVG(dd.val) and dd.row_number IN(...) is used to correctly produce a median when there are an even number of records. Reasoning:

SELECT FLOOR((3+1)/2),FLOOR((3+2)/2); -- when total_rows is 3, avg rows 2 and 2
SELECT FLOOR((4+1)/2),FLOOR((4+2)/2); -- when total_rows is 4, avg rows 2 and 3

Finally, MariaDB 10.3.3+ contains a MEDIAN function

I just found another answer online in the comments:

For medians in almost any SQL:

SELECT x.val from data x, data y
GROUP BY x.val
HAVING SUM(SIGN(1-SIGN(y.val-x.val))) = (COUNT(*)+1)/2

Make sure your columns are well indexed and the index is used for filtering and sorting. Verify with the explain plans.

select count(*) from table --find the number of rows

Calculate the "median" row number. Maybe use: median_row = floor(count / 2).

Then pick it out of the list:

select val from table order by val asc limit median_row,1

This should return you one row with just the value you want.

Jacob

I found the accepted solution didn't work on my MySQL install, returning an empty set, but this query worked for me in all situations that I tested it on:

SELECT x.val from data x, data y
GROUP BY x.val
HAVING SUM(SIGN(1-SIGN(y.val-x.val)))/COUNT(*) > .5
LIMIT 1

Unfortunately, neither TheJacobTaylor's nor velcrow's answers return accurate results for current versions of MySQL.

Velcro's answer from above is close, but it does not calculate correctly for result sets with an even number of rows. Medians are defined as either 1) the middle number on odd numbered sets, or 2) the average of the two middle numbers on even number sets.

So, here's velcro's solution patched to handle both odd and even number sets:

SELECT AVG(middle_values) AS 'median' FROM (
  SELECT t1.median_column AS 'middle_values' FROM
    (
      SELECT @row:=@row+1 as `row`, x.median_column
      FROM median_table AS x, (SELECT @row:=0) AS r
      WHERE 1
      -- put some where clause here
      ORDER BY x.median_column
    ) AS t1,
    (
      SELECT COUNT(*) as 'count'
      FROM median_table x
      WHERE 1
      -- put same where clause here
    ) AS t2
    -- the following condition will return 1 record for odd number sets, or 2 records for even number sets.
    WHERE t1.row >= t2.count/2 and t1.row <= ((t2.count/2) +1)) AS t3;

To use this, follow these 3 easy steps:

1. Replace "median_table" (2 occurrences) in the above code with the name of your table
2. Replace "median_column" (3 occurrences) with the column name you'd like to find a median for
3. If you have a WHERE condition, replace "WHERE 1" (2 occurrences) with your where condition

I propose a faster way.

Get the row count:

SELECT CEIL(COUNT(*)/2) FROM data;

Then take the middle value in a sorted subquery:

SELECT max(val) FROM (SELECT val FROM data ORDER BY val limit @middlevalue) x;

I tested this with a 5x10e6 dataset of random numbers and it will find the median in under 10 seconds.

Install and use this mysql statistical functions: http://www.xarg.org/2012/07/statistical-functions-in-mysql/

After that, calculate median is easy:

SELECT median(val) FROM data;