让我们创建一个名为numbers的示例表

这个答案是针对mysql数据库的

在postgres Sql中，它简单地使用per_cont函数

创建表数字( num INT, 频率整数 ）;

在数字表中插入值

插入数字 (7) 0 (1, 1), (2、3), (1) 3 (9,1), (1, 1), (2、3), (1) 3 (9,1);

——select * from numbers

作为递归num_frequency (num,frequency, i) （ 选择num,频率,1 从数字 UNION ALL 选择num,频率,i + 1 从num_frequency num_frequency的地方。I < num_frequency.frequency ）

select * (max(当numbers=lower_limit时，则num else null end)/2 +max(当数字=upper_limit时，则num else null end)/2)作为中位数 从( select *, total_number % 2, 情况下 当total_number%2=0时，total_number/2 Else (total_number+1)/2 end as lower_limit， 情况下 当total_number%2=0时，total_number/2+1 其他(total_number + 1) / 2 结束为upper_limit

从( Select *，max(numbers) over() as total_number from ( Select num,row_number() over(按num排序) 作为num_frequency中的数字 b) b) b)

我有一个包含大约10亿行的数据库，我们需要它来确定集合中的年龄中位数。对十亿行进行排序是困难的，但如果你将可以找到的不同值(年龄范围从0到100)聚合在一起，你可以对这个列表进行排序，并使用一些算术魔术来找到你想要的任何百分位数，如下所示:

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
;

这个查询取决于你的db支持窗口函数(包括ROWS UNBOUNDED precede)，但如果你没有，这是一个简单的事情，将aggData CTE与自身连接，并将所有先前的总数聚合到' cumulative '列，用于确定哪个值包含指定的预分词。上面的示例计算p10、p25、p50(中位数)、p75和p90。

屁股的

您可以使用窗口函数row_number()来回答查询以找到介质

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);

你可以使用这里的用户定义函数。

我发现这个答案非常有用——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));

不幸的是，无论是TheJacobTaylor还是velcrow的答案都不会返回当前版本MySQL的准确结果。

从上面来看，魔术贴的答案是接近的，但它不能正确计算具有偶数行数的结果集。中位数定义为1)奇数集上的中间数，或2)偶数集上两个中间数的平均值。

所以，这里是魔术贴的解决方案修补处理奇数和偶数集:

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;

要使用它，请遵循以下3个简单步骤:

将上面代码中的“median_table”(出现2次)替换为您的表名 将“median_column”(3次)替换为您希望为其查找中位数的列名 如果你有一个WHERE条件，用WHERE条件替换“WHERE 1”(2次)