Simple way to calculate median with MySQL

What's the simplest (and hopefully not too slow) way to calculate the median with MySQL? I've used AVG(x) for finding the mean, but I'm having a hard time finding a simple way of calculating the median. For now, I'm returning all the rows to PHP, doing a sort, and then picking the middle row, but surely there must be some simple way of doing it in a single MySQL query.

Example data:

id | val
--------
 1    4
 2    7
 3    2
 4    2
 5    9
 6    8
 7    3

Sorting on val gives 2 2 3 4 7 8 9, so the median should be 4, versus SELECT AVG(val) which == 5.

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.

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;

A comment on this page in the MySQL documentation has the following suggestion:

-- (mostly) High Performance scaling MEDIAN function per group
-- Median defined in http://en.wikipedia.org/wiki/Median
--
-- by Peter Hlavac
-- 06.11.2008
--
-- Example Table:

DROP table if exists table_median;
CREATE TABLE table_median (id INTEGER(11),val INTEGER(11));
COMMIT;


INSERT INTO table_median (id, val) VALUES
(1, 7), (1, 4), (1, 5), (1, 1), (1, 8), (1, 3), (1, 6),
(2, 4),
(3, 5), (3, 2),
(4, 5), (4, 12), (4, 1), (4, 7);



-- Calculating the MEDIAN
SELECT @a := 0;
SELECT
id,
AVG(val) AS MEDIAN
FROM (
SELECT
id,
val
FROM (
SELECT
-- Create an index n for every id
@a := (@a + 1) mod o.c AS shifted_n,
IF(@a mod o.c=0, o.c, @a) AS n,
o.id,
o.val,
-- the number of elements for every id
o.c
FROM (
SELECT
t_o.id,
val,
c
FROM
table_median t_o INNER JOIN
(SELECT
id,
COUNT(1) AS c
FROM
table_median
GROUP BY
id
) t2
ON (t2.id = t_o.id)
ORDER BY
t_o.id,val
) o
) a
WHERE
IF(
-- if there is an even number of elements
-- take the lower and the upper median
-- and use AVG(lower,upper)
c MOD 2 = 0,
n = c DIV 2 OR n = (c DIV 2)+1,

-- if its an odd number of elements
-- take the first if its only one element
-- or take the one in the middle
IF(
c = 1,
n = 1,
n = c DIV 2 + 1
)
)
) a
GROUP BY
id;

-- Explanation:
-- The Statement creates a helper table like
--
-- n id val count
-- ----------------
-- 1, 1, 1, 7
-- 2, 1, 3, 7
-- 3, 1, 4, 7
-- 4, 1, 5, 7
-- 5, 1, 6, 7
-- 6, 1, 7, 7
-- 7, 1, 8, 7
--
-- 1, 2, 4, 1

-- 1, 3, 2, 2
-- 2, 3, 5, 2
--
-- 1, 4, 1, 4
-- 2, 4, 5, 4
-- 3, 4, 7, 4
-- 4, 4, 12, 4


-- from there we can select the n-th element on the position: count div 2 + 1 

I have this below code which I found on HackerRank and it is pretty simple and works in each and every case.

SELECT M.MEDIAN_COL FROM MEDIAN_TABLE M WHERE  
  (SELECT COUNT(MEDIAN_COL) FROM MEDIAN_TABLE WHERE MEDIAN_COL < M.MEDIAN_COL ) = 
  (SELECT COUNT(MEDIAN_COL) FROM MEDIAN_TABLE WHERE MEDIAN_COL > M.MEDIAN_COL );

If MySQL has ROW_NUMBER, then the MEDIAN is (be inspired by this SQL Server query):

WITH Numbered AS 
(
SELECT *, COUNT(*) OVER () AS Cnt,
    ROW_NUMBER() OVER (ORDER BY val) AS RowNum
FROM yourtable
)
SELECT id, val
FROM Numbered
WHERE RowNum IN ((Cnt+1)/2, (Cnt+2)/2)
;

The IN is used in case you have an even number of entries.

If you want to find the median per group, then just PARTITION BY group in your OVER clauses.

Rob

Most of the solutions above work only for one field of the table, you might need to get the median (50th percentile) for many fields on the query.

I use this:

SELECT CAST(SUBSTRING_INDEX(SUBSTRING_INDEX(
 GROUP_CONCAT(field_name ORDER BY field_name SEPARATOR ','),
  ',', 50/100 * COUNT(*) + 1), ',', -1) AS DECIMAL) AS `Median`
FROM table_name;

You can replace the "50" in example above to any percentile, is very efficient.

Just make sure you have enough memory for the GROUP_CONCAT, you can change it with:

SET group_concat_max_len = 10485760; #10MB max length

More details: http://web.performancerasta.com/metrics-tips-calculating-95th-99th-or-any-percentile-with-single-mysql-query/

You could use the user-defined function that's found here.

Building off of velcro's answer, for those of you having to do a median off of something that is grouped by another parameter:

SELECT grp_field, t1.val FROM (
SELECT grp_field, @rownum:=IF(@s = grp_field, @rownum + 1, 0) AS SELECT grp_field, t1.val FROM (
SELECT grp_field, @rownum:=IF(@s = grp_field, @rownum + 1, 0) AS row_number,
@s:=IF(@s = grp_field, @s, grp_field) AS sec, d.val
FROM data d,  (SELECT @rownum:=0, @s:=0) r
ORDER BY grp_field, d.val
) as t1 JOIN (
SELECT grp_field, count(*) as total_rows
FROM data d
GROUP BY grp_field
) as t2
ON t1.grp_field = t2.grp_field
WHERE t1.row_number=floor(total_rows/2)+1;
,
@s:=IF(@s = grp_field, @s, grp_field) AS sec, d.val
FROM data d,  (SELECT @rownum:=0, @s:=0) r
ORDER BY grp_field, d.val
) as t1 JOIN (
SELECT grp_field, count(*) as total_rows
FROM data d
GROUP BY grp_field
) as t2
ON t1.grp_field = t2.grp_field
WHERE t1.row_number=floor(total_rows/2)+1;

Takes care about an odd value count - gives the avg of the two values in the middle in that case.

SELECT AVG(val) FROM
  ( SELECT x.id, x.val from data x, data y
      GROUP BY x.id, x.val
      HAVING SUM(SIGN(1-SIGN(IF(y.val-x.val=0 AND x.id != y.id, SIGN(x.id-y.id), y.val-x.val)))) IN (ROUND((COUNT(*))/2), ROUND((COUNT(*)+1)/2))
  ) sq

My code, efficient without tables or additional variables:

SELECT
((SUBSTRING_INDEX(SUBSTRING_INDEX(group_concat(val order by val), ',', floor(1+((count(val)-1) / 2))), ',', -1))
+
(SUBSTRING_INDEX(SUBSTRING_INDEX(group_concat(val order by val), ',', ceiling(1+((count(val)-1) / 2))), ',', -1)))/2
as median
FROM table;

Single query to archive the perfect median:

SELECT 
COUNT(*) as total_rows, 
IF(count(*)%2 = 1, CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*)), ',', -1) AS DECIMAL), ROUND((CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*) + 1), ',', -1) AS DECIMAL) + CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*)), ',', -1) AS DECIMAL)) / 2)) as median, 
AVG(val) as average 
FROM 
data

Optionally, you could also do this in a stored procedure:

DROP PROCEDURE IF EXISTS median;
DELIMITER //
CREATE PROCEDURE median (table_name VARCHAR(255), column_name VARCHAR(255), where_clause VARCHAR(255))
BEGIN
  -- Set default parameters
  IF where_clause IS NULL OR where_clause = '' THEN
    SET where_clause = 1;
  END IF;

  -- Prepare statement
  SET @sql = CONCAT(
    "SELECT AVG(middle_values) AS 'median' FROM (
      SELECT t1.", column_name, " AS 'middle_values' FROM
        (
          SELECT @row:=@row+1 as `row`, x.", column_name, "
          FROM ", table_name," AS x, (SELECT @row:=0) AS r
          WHERE ", where_clause, " ORDER BY x.", column_name, "
        ) AS t1,
        (
          SELECT COUNT(*) as 'count'
          FROM ", table_name, " x
          WHERE ", where_clause, "
        ) 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
    ");

  -- Execute statement
  PREPARE stmt FROM @sql;
  EXECUTE stmt;
END//
DELIMITER ;


-- Sample usage:
-- median(table_name, column_name, where_condition);
CALL median('products', 'price', NULL);

My solution presented below works in just one query without creation of table, variable or even sub-query. Plus, it allows you to get median for each group in group-by queries (this is what i needed !):

SELECT `columnA`, 
SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(`columnB` ORDER BY `columnB`), ',', CEILING((COUNT(`columnB`)/2))), ',', -1) medianOfColumnB
FROM `tableC`
-- some where clause if you want
GROUP BY `columnA`;

It works because of a smart use of group_concat and substring_index.

But, to allow big group_concat, you have to set group_concat_max_len to a higher value (1024 char by default). You can set it like that (for current sql session) :

SET SESSION group_concat_max_len = 10000; 
-- up to 4294967295 in 32-bits platform.

More infos for group_concat_max_len: https://dev.mysql.com/doc/refman/5.1/en/server-system-variables.html#sysvar_group_concat_max_len

Another riff on Velcrow's answer, but uses a single intermediate table and takes advantage of the variable used for row numbering to get the count, rather than performing an extra query to calculate it. Also starts the count so that the first row is row 0 to allow simply using Floor and Ceil to select the median row(s).

SELECT Avg(tmp.val) as median_val
    FROM (SELECT inTab.val, @rows := @rows + 1 as rowNum
              FROM data as inTab,  (SELECT @rows := -1) as init
              -- Replace with better where clause or delete
              WHERE 2 > 1
              ORDER BY inTab.val) as tmp
    WHERE tmp.rowNum in (Floor(@rows / 2), Ceil(@rows / 2));

Knowing exact row count you can use this query:

SELECT <value> AS VAL FROM <table> ORDER BY VAL LIMIT 1 OFFSET <half>

Where <half> = ceiling(<size> / 2.0) - 1

SELECT 
    SUBSTRING_INDEX(
        SUBSTRING_INDEX(
            GROUP_CONCAT(field ORDER BY field),
            ',',
            ((
                ROUND(
                    LENGTH(GROUP_CONCAT(field)) - 
                    LENGTH(
                        REPLACE(
                            GROUP_CONCAT(field),
                            ',',
                            ''
                        )
                    )
                ) / 2) + 1
            )),
            ',',
            -1
        )
FROM
    table

The above seems to work for me.

I used a two query approach:

• first one to get count, min, max and avg
• second one (prepared statement) with a "LIMIT @count/2, 1" and "ORDER BY .." clauses to get the median value

These are wrapped in a function defn, so all values can be returned from one call.

If your ranges are static and your data does not change often, it might be more efficient to precompute/store these values and use the stored values instead of querying from scratch every time.

as i just needed a median AND percentile solution, I made a simple and quite flexible function based on the findings in this thread. I know that I am happy myself if I find "readymade" functions that are easy to include in my projects, so I decided to quickly share:

function mysql_percentile($table, $column, $where, $percentile = 0.5) {

	$sql = "
			SELECT `t1`.`".$column."` as `percentile` FROM (
			SELECT @rownum:=@rownum+1 as `row_number`, `d`.`".$column."`
			  FROM `".$table."` `d`,  (SELECT @rownum:=0) `r`
			  ".$where."
			  ORDER BY `d`.`".$column."`
			) as `t1`, 
			(
			  SELECT count(*) as `total_rows`
			  FROM `".$table."` `d`
			  ".$where."
			) as `t2`
			WHERE 1
			AND `t1`.`row_number`=floor(`total_rows` * ".$percentile.")+1;
		";

	$result = sql($sql, 1);

	if (!empty($result)) {
		return $result['percentile'];		
	} else {
		return 0;
	}

}

Usage is very easy, example from my current project:

...
$table = DBPRE."zip_".$slug;
$column = 'seconds';
$where = "WHERE `reached` = '1' AND `time` >= '".$start_time."'";

	$reaching['median'] = mysql_percentile($table, $column, $where, 0.5);
	$reaching['percentile25'] = mysql_percentile($table, $column, $where, 0.25);
	$reaching['percentile75'] = mysql_percentile($table, $column, $where, 0.75);
...

Here is my way . Of course, you could put it into a procedure :-)

SET @median_counter = (SELECT FLOOR(COUNT(*)/2) - 1 AS `median_counter` FROM `data`);

SET @median = CONCAT('SELECT `val` FROM `data` ORDER BY `val` LIMIT ', @median_counter, ', 1');

PREPARE median FROM @median;

EXECUTE median;

You could avoid the variable @median_counter, if you substitude it:

SET @median = CONCAT( 'SELECT `val` FROM `data` ORDER BY `val` LIMIT ',
                      (SELECT FLOOR(COUNT(*)/2) - 1 AS `median_counter` FROM `data`),
                      ', 1'
                    );

PREPARE median FROM @median;

EXECUTE median;

Based on @bob's answer, this generalizes the query to have the ability to return multiple medians, grouped by some criteria.

Think, e.g., median sale price for used cars in a car lot, grouped by year-month.

SELECT 
	period, 
	AVG(middle_values) AS 'median' 
FROM (
	SELECT t1.sale_price AS 'middle_values', t1.row_num, t1.period, t2.count
	FROM (
		SELECT 
			@last_period:=@period AS 'last_period',
			@period:=DATE_FORMAT(sale_date, '%Y-%m') AS 'period',
			IF (@period<>@last_period, @row:=1, @row:=@row+1) as `row_num`, 
			x.sale_price
	      FROM listings AS x, (SELECT @row:=0) AS r
	      WHERE 1
	      	-- where criteria goes here
	      ORDER BY DATE_FORMAT(sale_date, '%Y%m'), x.sale_price
		) AS t1
	LEFT JOIN (  
	      SELECT COUNT(*) as 'count', DATE_FORMAT(sale_date, '%Y-%m') AS 'period'
	      FROM listings x
	      WHERE 1
			-- same where criteria goes here
	      GROUP BY DATE_FORMAT(sale_date, '%Y%m')
		) AS t2
		ON t1.period = t2.period
	) AS t3
WHERE 
	row_num >= (count/2) 
	AND row_num <= ((count/2) + 1)
GROUP BY t3.period
ORDER BY t3.period;

Often, we may need to calculate Median not just for the whole table, but for aggregates with respect to our ID. In other words, calculate median for each ID in our table, where each ID has many records. (good performance and works in many SQL + fixes problem of even and odds, more about performance of different Median-methods https://sqlperformance.com/2012/08/t-sql-queries/median )

SELECT our_id, AVG(1.0 * our_val) as Median
FROM
( SELECT our_id, our_val, 
  COUNT(*) OVER (PARTITION BY our_id) AS cnt,
  ROW_NUMBER() OVER (PARTITION BY our_id ORDER BY our_val) AS rn
  FROM our_table
) AS x
WHERE rn IN ((cnt + 1)/2, (cnt + 2)/2) GROUP BY our_id;

Hope it helps

MySQL has supported window functions since version 8.0, you can use ROW_NUMBER or DENSE_RANK (DO NOT use RANK as it assigns the same rank to same values, like in sports ranking):

SELECT AVG(t1.val) AS median_val
  FROM (SELECT val, 
               ROW_NUMBER() OVER(ORDER BY val) AS rownum
          FROM data) t1,
       (SELECT COUNT(*) AS num_records FROM data) t2
 WHERE t1.row_num IN
       (FLOOR((t2.num_records + 1) / 2), 
        FLOOR((t2.num_records + 2) / 2));

A simple way to calculate Median in MySQL

set @ct := (select count(1) from station);
set @row := 0;

select avg(a.val) as median from 
(select * from  table order by val) a
where (select @row := @row + 1)
between @ct/2.0 and @ct/2.0 +1;

Simple Solution For ORACLE:

SELECT ROUND(MEDIAN(Lat_N), 4) FROM Station;

Easy Solution to Understand For MySQL:

select case MOD(count(lat_n),2) 
when 1 then (select round(S.LAT_N,4) from station S where (select count(Lat_N) from station where Lat_N < S.LAT_N ) = (select count(Lat_N) from station where Lat_N > S.LAT_N))
else (select round(AVG(S.LAT_N),4) from station S where 1 = (select count(Lat_N) from station where Lat_N < S.LAT_N ) - (select count(Lat_N) from station where Lat_N > S.LAT_N))
end from station;

Explanation

STATION is table name. LAT_N is the column name having numeric value

Suppose there are 101 records(odd number) in station table. This means that the median is 51st record if the tabled sorted either asc or desc.

In above query for every S.LAT_N of S table I am creating two tables. One for number of LAT_N values less than S.LAT_N and another for number of LAT_N values greater than S.LAT_N. Later I am comparing these two tables and if they are matched then I am selecting that S.LAT_N value. When I check for 51st records there are 50 values less than 51st record and there 50 records greater than 51st record. As you see, there are 50 records in both tables. So this is our answer. For every other record there are different number of records in two tables created for comparison. So, only 51st record meets the condition.

Now suppose there are 100 records(even number) in station table. This means that the median is average of 50th and 51st records if the tabled sorted either asc or desc.

Same as odd logic I am creating two tables. One for number of LAT_N values less than S.LAT_N and another for number of LAT_N values greater than S.LAT_N. Later I am comparing these two tables and if their difference is equal to 1 then I am selecting that S.LAT_N value and find the average. When I check for 50th records there are 49 values less than 50th record and there 51 records greater than 50th record. As you see, there is difference of 1 record in both tables. So this(50th record) is our 1st record for average. Similarly, When I check for 51st records there are 50 values less than 51st record and there 49 records greater than 51st record. As you see, there is difference of 1 record in both tables. So this(51st record) is our 2nd record for average. For every other record there are different number of records in two tables created for comparison. So, only 50th and 51st records meet the condition.

I am using the below table for the solution in MySQL:

CREATE TABLE transactions (
  transaction_id int , user_id int , merchant_name varchar(255), transaction_date date , amount int
);

INSERT INTO transactions (transaction_id, user_id, merchant_name, transaction_date, amount)  
VALUES (1, 1 ,'abc', '2015-08-17', 100),(2, 2, 'ced', '2015-2-17', 100),(3, 1, 'def', '2015-2-16', 121),
(4, 1 ,'ced', '2015-3-17', 110),(5, 1, 'ced', '2015-3-17', 150),(6, 2 ,'abc', '2015-4-17', 130), 
(7, 3 ,'ced', '2015-12-17', 10),(8, 3 ,'abc', '2015-8-17', 100),(9, 2 ,'abc', '2015-12-17', 140),(10, 1,'abc', '2015-9-17', 100),
(11, 1 ,'abc', '2015-08-17', 121),(12, 2 ,'ced', '2015-12-23', 130),(13, 1 ,'def', '2015-12-23', 13),(3, 4, 'abc', '2015-2-16', 120),(3, 4, 'def', '2015-2-16', 121),(3, 4, 'ced', '2015-2-16', 121);

Calculating Median for 'amount' column:

WITH Numbered AS 
(
SELECT *, COUNT(*) OVER () AS TotatRecords,
    ROW_NUMBER() OVER (ORDER BY amount) AS RowNum
FROM transactions
)
SELECT Avg(amount)
FROM Numbered
WHERE RowNum IN ( FLOOR((TotatRecords+1)/2), FLOOR((TotatRecords+2)/2) )
;

TotalRecords = 16 and Median = 120.5000

This query will work for both the conditions i.e. Even and Odd records.

After reading all previous ones they didn't match with my actual requirement so I implemented my own one which doesn't need any procedure or complicate statements, just I GROUP_CONCAT all values from the column I wanted to obtain the MEDIAN and applying a COUNT DIV BY 2 I extract the value in from the middle of the list like the following query does :

(POS is the name of the column I want to get its median)

(query) SELECT
SUBSTRING_INDEX ( 
   SUBSTRING_INDEX ( 
       GROUP_CONCAT(pos ORDER BY CAST(pos AS SIGNED INTEGER) desc SEPARATOR ';') 
    , ';', COUNT(*)/2 ) 
, ';', -1 ) AS `pos_med`
FROM table_name
GROUP BY any_criterial

I hope this could be useful for someone in the way many of other comments were for me from this website.

I have a database containing about 1 billion rows that we require to determine the median age in the set. Sorting a billion rows is hard, but if you aggregate the distinct values that can be found (ages range from 0 to 100), you can sort THIS list, and use some arithmetic magic to find any percentile you want as follows:

with rawData(count_value) as
(
  	select p.YEAR_OF_BIRTH
		from dbo.PERSON p
),
overallStats (avg_value, stdev_value, min_value, max_value, total) as
(
  select avg(1.0 * count_value) as avg_value,
    stdev(count_value) as stdev_value,
    min(count_value) as min_value,
    max(count_value) as max_value,
    count(*) as total
  from rawData
),
aggData (count_value, total, accumulated) as
(
  select count_value, 
  	count(*) as total, 
		SUM(count(*)) OVER (ORDER BY count_value ROWS UNBOUNDED PRECEDING) as accumulated
  FROM rawData
  group by count_value
)
select o.total as count_value,
  o.min_value,
	o.max_value,
	o.avg_value,
	o.stdev_value,
	MIN(case when d.accumulated >= .50 * o.total then count_value else o.max_value end) as median_value,
	MIN(case when d.accumulated >= .10 * o.total then count_value else o.max_value end) as p10_value,
	MIN(case when d.accumulated >= .25 * o.total then count_value else o.max_value end) as p25_value,
	MIN(case when d.accumulated >= .75 * o.total then count_value else o.max_value end) as p75_value,
	MIN(case when d.accumulated >= .90 * o.total then count_value else o.max_value end) as p90_value
from aggData d
cross apply overallStats o
GROUP BY o.total, o.min_value, o.max_value, o.avg_value, o.stdev_value
;

This query depends on your db supporting window functions (including ROWS UNBOUNDED PRECEDING) but if you do not have that it is a simple matter to join aggData CTE with itself and aggregate all prior totals into the 'accumulated' column which is used to determine which value contains the specified precentile. The above sample calcuates p10, p25, p50 (median), p75, and p90.

-Chris

Taken from: http://mdb-blog.blogspot.com/2015/06/mysql-find-median-nth-element-without.html

I would suggest another way, without join, but working with strings

i did not checked it with tables with large data, but small/medium tables it works just fine.

The good thing here, that it works also by GROUPING so it can return the median for several items.

here is test code for test table:

DROP TABLE test.test_median
CREATE TABLE test.test_median AS
SELECT 'book' AS grp, 4 AS val UNION ALL
SELECT 'book', 7 UNION ALL
SELECT 'book', 2 UNION ALL
SELECT 'book', 2 UNION ALL
SELECT 'book', 9 UNION ALL
SELECT 'book', 8 UNION ALL
SELECT 'book', 3 UNION ALL

SELECT 'note', 11 UNION ALL

SELECT 'bike', 22 UNION ALL
SELECT 'bike', 26 

and the code for finding the median for each group:

SELECT grp,
		 SUBSTRING_INDEX( SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val), ',', COUNT(*)/2 ), ',', -1) as the_median,
		 GROUP_CONCAT(val ORDER BY val) as all_vals_for_debug
FROM test.test_median
GROUP BY grp

Output:

grp | the_median| all_vals_for_debug
bike| 22        | 22,26
book| 4         | 2,2,3,4,7,8,9
note| 11        | 11

In some cases median gets calculated as follows :

The "median" is the "middle" value in the list of numbers when they are ordered by value. For even count sets, median is average of the two middle values. I've created a simple code for that :

$midValue = 0;
$rowCount = "SELECT count(*) as count {$from} {$where}";

$even = FALSE;
$offset = 1;
$medianRow = floor($rowCount / 2);
if ($rowCount % 2 == 0 && !empty($medianRow)) {
  $even = TRUE;
  $offset++;
  $medianRow--;
}

$medianValue = "SELECT column as median 
               {$fromClause} {$whereClause} 
               ORDER BY median 
               LIMIT {$medianRow},{$offset}";

$medianValDAO = db_query($medianValue);
while ($medianValDAO->fetch()) {
  if ($even) {
    $midValue = $midValue + $medianValDAO->median;
  }
  else {
    $median = $medianValDAO->median;
  }
}
if ($even) {
  $median = $midValue / 2;
}
return $median;

The $median returned would be the required result :-)

Medians grouped by dimension:

SELECT your_dimension, avg(t1.val) as median_val FROM (
SELECT @rownum:=@rownum+1 AS `row_number`,
   IF(@dim <> d.your_dimension, @rownum := 0, NULL),
   @dim := d.your_dimension AS your_dimension,
   d.val
   FROM data d,  (SELECT @rownum:=0) r, (SELECT @dim := 'something_unreal') d
  WHERE 1
  -- put some where clause here
  ORDER BY d.your_dimension, d.val
) as t1
INNER JOIN  
(
  SELECT d.your_dimension,
    count(*) as total_rows
  FROM data d
  WHERE 1
  -- put same where clause here
  GROUP BY d.your_dimension
) as t2 USING(your_dimension)
WHERE 1
AND t1.row_number in ( floor((total_rows+1)/2), floor((total_rows+2)/2) )

GROUP BY your_dimension;

This way seems include both even and odd count without subquery.

SELECT AVG(t1.x)
FROM table t1, table t2
GROUP BY t1.x
HAVING SUM(SIGN(t1.x - t2.x)) = 0

These methods select from the same table twice. If the source data are coming from an expensive query, this is a way to avoid running it twice:

select KEY_FIELD, AVG(VALUE_FIELD) MEDIAN_VALUE
from (
    select KEY_FIELD, VALUE_FIELD, RANKF
    , @rownumr := IF(@prevrowidr=KEY_FIELD,@rownumr+1,1) RANKR
	, @prevrowidr := KEY_FIELD
    FROM (
		SELECT KEY_FIELD, VALUE_FIELD, RANKF
        FROM (
			SELECT KEY_FIELD, VALUE_FIELD 
			, @rownumf := IF(@prevrowidf=KEY_FIELD,@rownumf+1,1) RANKF
			, @prevrowidf := KEY_FIELD     
			FROM (
				SELECT KEY_FIELD, VALUE_FIELD 
				FROM (
					-- some expensive query
				)	B
				ORDER BY  KEY_FIELD, VALUE_FIELD
			) C
			, (SELECT @rownumf := 1) t_rownum
			, (SELECT @prevrowidf := '*') t_previd
        ) D
		ORDER BY  KEY_FIELD, RANKF DESC
	) E
	, (SELECT @rownumr := 1) t_rownum
	, (SELECT @prevrowidr := '*') t_previd
) F
WHERE RANKF-RANKR BETWEEN -1 and 1
GROUP BY KEY_FIELD

create table med(id integer);
insert into med(id) values(1);
insert into med(id) values(2);
insert into med(id) values(3);
insert into med(id) values(4);
insert into med(id) values(5);
insert into med(id) values(6);

select (MIN(count)+MAX(count))/2 from 
(select case when (select count(*) from 
med A where A.id<B.id)=(select count(*)/2 from med) OR 
(select count(*) from med A where A.id>B.id)=(select count(*)/2 
from med) then cast(B.id as float)end as count from med B) C;

 ?column? 
----------
  3.5
(1 row)

OR

select cast(avg(id) as float) from 
(select t1.id from med t1 JOIN med t2 on t1.id!= t2.id 
group by t1.id having ABS(SUM(SIGN(t1.id-t2.id)))=1) A;

The following SQL Code will help you to calculate the median in MySQL using user defined variables.

create table employees(salary int);

insert into employees values(8);
insert into employees values(23);
insert into employees values(45);
insert into employees values(123);
insert into employees values(93);
insert into employees values(2342);
insert into employees values(2238);

select * from employees;

Select salary from employees  order by salary;

set @rowid=0;
set @cnt=(select count(*) from employees);
set @middle_no=ceil(@cnt/2);
set @odd_even=null;

select AVG(salary) from 
(select salary,@rowid:=@rowid+1 as rid, (CASE WHEN(mod(@cnt,2)=0) THEN @odd_even:=1 ELSE @odd_even:=0 END) as odd_even_status  from employees  order by salary) as tbl where tbl.rid=@middle_no or tbl.rid=(@middle_no+@odd_even);

If you are looking for detailed explanation, please refer this blog.

I found this answer very helpful - https://www.eversql.com/how-to-calculate-median-value-in-mysql-using-a-simple-sql-query/

SET @rowindex := -1;

SELECT
   AVG(g.grade)
FROM
   (SELECT @rowindex:=@rowindex + 1 AS rowindex,
       grades.grade AS grade
    FROM grades
    ORDER BY grades.grade) AS g
WHERE
g.rowindex IN (FLOOR(@rowindex / 2) , CEIL(@rowindex / 2));

The below query will work perfect for both even or odd number of rows. In the subquery, we are finding the value(s) which has same number of rows before and after it. In case of odd rows the having clause will evaluate to 0 (same number of rows before and after cancels out the sign).

Similarly, for even rows the having clause evaluates to 1 for two rows (the center 2 rows) because they will (collectively) have same number of rows before and after.

In the outer query, we will avg out either the single value (in case of odd rows) or (2 values in case of even rows).

select avg(val) as median
from
(
    select d1.val
    from data d1 cross join data d2
    group by d1.val
    having abs(sum(sign(d1.val-d2.val))) in (0,1)
) sub

Note: In case your table has duplicate values, the above having clause should be changed to the below condition. In this case, there could be values outside of the original possibilities of 0,1. The below condition will make this condition dynamic and work in case of duplicates too.

having sum(case when d1.val=d2.val then 1 else 0 end)>=
abs(sum(sign(d1.val-d2.val)))

Try something like :

SELECT  
CAST (AVG(val) AS DECIMAL(10,4))
FROM
(
    SELECT 
    val,
    ROW_NUMBER() OVER( ORDER BY val ) -1 AS rn,
    COUNT(1) OVER () -1 AS cnt
    FROM STATION
) as tmp
WHERE rn IN (FLOOR(cnt/2),CEILING (cnt/2))

**

> Note : The reason for -1 is to make it zero indexed..i.e row number > now starts from 0 instead of 1

**

I have not compared the performance of this solution to the rest of the answers posted here, but I found this to be the most straight-forward to understand, and covers the full extent of the mathematical formula for calculating a median. In other words, this solution will be robust enough for even- and odd-numbered data sets:

SELECT CASE 
-- odd-numbered data sets:
WHEN MOD(COUNT(*), 2) = 1 THEN (SELECT median.<value> AS median
FROM
(SELECT t1.<value>
  FROM (SELECT <value>, 
               ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
          FROM <data>) t1,
       (SELECT COUNT(*) AS num_records FROM <data>) t2
 WHERE t1.rownum =(t2.num_records) / 2) as median)
-- even-numbered data sets:
ELSE (select (low_bound.<value> + up_bound.<value>) / 2 AS median
FROM
(SELECT t1.<value>
  FROM (SELECT <value>, 
               ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
          FROM <data>) t1,
       (SELECT COUNT(*) AS num_records FROM <data>) t2
 WHERE t1.rownum =(t2.num_records - 1) / 2) as low_bound,
 (SELECT t1.<value>
  FROM (SELECT <value>, 
               ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
          FROM station) t1,
       (SELECT COUNT(*) AS num_records FROM data) t2
 WHERE t1.rownum =(t2.num_records + 1) / 2) as up_bound)
END
FROM <data>

The most simple and fast way to calculate median in mysql.

select x.col
from   (select lat_n, 
               count(1) over (partition by 'A')        as total_rows, 
               row_number() over (order by col asc) as rank_Order 
        from   station ft) x 
where  x.rank_Order = round(x.total_rows / 2.0, 0) 

If this is MySQL, there are window functions now and you can just do it this way (assuming you want to round up to nearest integer - otherwise just replace ROUND with CEIL or FLOOR or what have you). The following solution works for tables regardless of whether they have an even number of rows or an odd number of rows:

WITH CTE AS (
	SELECT val,
			ROW_NUMBER() OVER (ORDER BY val ASC) AS rn,
			COUNT(*) OVER () AS total_count
	FROM data
)
SELECT ROUND(AVG(val)) AS median
FROM CTE
WHERE
	rn BETWEEN
	total_count / 2.0 AND
	total_count / 2.0 + 1;

I think some of the more recent answers on this thread were already getting at this approach, but it also seemed like people were overthinking it, so consider this an improved version. Regardless of SQL flavor, there is no reason anyone should be writing a huge paragraph of code with multiple subqueries just to get the median in 2021. However, please note that the above query only works if you're asked to find the median for a continuous series. Of course, regardless of row number, sometimes people do make a distinction between what is referred to as the Discrete Median and what is referred to as the Interpolated Median for a continuous series.

If you're asked to find the median for a discrete series and the table has an even number of rows, the above solution will not work for you, and you should revert to using one of the other solutions, like TheJacobTaylor's.

The second solution below is a slightly modified version of TheJacobTaylor's, where I explicitly state CROSS JOIN. This will work for tables that have an odd number of rows too, regardless of whether you're asked to find the median for a continuous or discrete series, but I would specifically use this when asked to find the median of a discrete series. Otherwise, use the first solution. That way, you'll never have to think about whether the data contains an 'even' or 'odd' number of data points.

SELECT x.val AS median
FROM data x
CROSS JOIN data y
GROUP BY x.val
HAVING SUM(SIGN(1 - SIGN(y.val - x.val))) = (COUNT(*) + 1) / 2;

Finally, you can easily do this in PostgreSQL using built-in functions. Here is a nice explanation, along with an effective summary on discrete vs interpolated medians.

https://leafo.net/guides/postgresql-calculating-percentile.html#calculating-the-median

For a table station and column lat_n, here is MySQL code to get the median:

set @rows := (select count(1) from station);
set @v1 := 0;
set @sql1 := concat('select lat_n into @v1 from station order by lat_n asc limit 1 offset ', ceil(@rows/2) - 1);
prepare statement1 from @sql1;
execute statement1;
set @v2 := 0;
set @sql2 := concat('select lat_n into @v2 from station order by lat_n asc limit 1 offset ', ceil((@rows + 1)/2) - 1);
prepare statement2 from @sql2;
execute statement2;
select (@v1 + @v2)/2;

You can use window function row_number() to answer the query to find medium

select val 
from (select val, row_number() over (order by val) as rownumber, x.cnt 
from data, (select count(*) as cnt from data) x) abc
where rownumber=ceil(cnt/2);

set @r = 0;

select  
    case when mod(c,2)=0 then round(sum(lat_N),4)
    else round(sum(lat_N)/2,4) 
    end as Med  
from 
    (select lat_N, @r := @r+1, @r as id from station order by lat_N) A
    cross join
    (select (count(1)+1)/2 as c from station) B
where id >= floor(c) and id <=ceil(c)