Calculating the Median with Mysql

Question

I\'m having trouble with calculating the median of a list of values, not the average.

I found this article Simple way to calculate median with MySQL

It has a

Answer 1

val is your time column, x and y are two references to the data table (you can write data AS x, data AS y).

EDIT: To avoid computing your sums twice, you can store the intermediate results.

CREATE TEMPORARY TABLE average_user_total_time 
      (SELECT SUM(time) AS time_taken 
            FROM scores 
            WHERE created_at >= '2010-10-10' 
                    and created_at <= '2010-11-11' 
            GROUP BY user_id);

Then you can compute median over these values which are in a named table.

EDIT: Temporary table won't work here. You could try using a regular table with "MEMORY" table type. Or just have your subquery that computes the values for the median twice in your query. Apart from this, I don't see another solution. This doesn't mean there isn't a better way, maybe somebody else will come with an idea.

Answer 2

If you have a table R with a column named A, and you want the median of A, you can do as follows:

SELECT A FROM R R1
WHERE ( SELECT COUNT(A) FROM R R2 WHERE R2.A < R1.A ) = ( SELECT COUNT(A) FROM R R3 WHERE R3.A > R1.A )

Note: This will only work if there are no duplicated values in A. Also, null values are not allowed.

Answer 3

First try to understand what the median is: it is the middle value in the sorted list of values.

Once you understand that, the approach is two steps:

1. sort the values in either order
2. pick the middle value (if not an odd number of values, pick the average of the two middle values)

Example:

Median of 0 1 3 7 9 10: 5 (because (7+3)/2=5)
Median of 0 1 3 7 9 10 11: 7 (because 7 is the middle value)

So, to sort dates you need a numerical value; you can get their time stamp (as seconds elapsed from epoch) and use the definition of median.

Answer 4

Simplest ways me and my friend have found out... ENJOY!!

SELECT count(*) INTO @c from station;
select ROUND((@c+1)/2) into @final; 
SELECT round(lat_n,4) from station a where @final-1=(select count(lat_n) from station b where b.lat_n > a.lat_n);

Answer 5

Here is a solution that is easy to understand. Just replace Your_Column and Your_Table as per your requirement.

SET @r = 0;

SELECT AVG(Your_Column)
FROM (SELECT (@r := @r + 1) AS r, Your_Column FROM Your_Table ORDER BY Your_Column) Temp
WHERE
    r = (SELECT CEIL(COUNT(*) / 2) FROM Your_Table) OR
    r = (SELECT FLOOR((COUNT(*) / 2) + 1) FROM Your_Table)

Originally adopted from this thread.

Answer 6

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.

This will find an arbitrary percentile by replacing the COUNT(*)/2 with COUNT(*)*n where n is the percentile (.5 for median, .75 for 75th percentile, etc).