我没有将这个解决方案的性能与这里发布的其他答案进行比较，但我发现这个解决方案是最容易理解的，并且涵盖了计算中位数的全部数学公式。换句话说，这个解决方案对于偶数和奇数数据集足够健壮:

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>

MySQL文档中这一页的注释有以下建议:

-- (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 

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)

我使用下表的解决方案在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);

计算“金额”列的中位数:

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

此查询将适用于两种情况，即偶数和奇数记录。

这是我的办法。当然，你可以把它放到一个过程中:-)

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;

你可以避免变量@median_counter，如果你替换它:

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;

在某些情况下，中位数的计算如下:

“中位数”是数字列表中按值排序时的“中间”值。对于偶数集，中位数是两个中间值的平均值。 我为此创建了一个简单的代码:

$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;

返回的$中位数将是所需的结果:-)