用MySQL计算中位数最简单(希望不会太慢)的方法是什么?我已经使用AVG(x)来寻找平均值，但我很难找到一个简单的方法来计算中位数。现在，我将所有的行返回到PHP，进行排序，然后选择中间的行，但是肯定有一些简单的方法可以在一个MySQL查询中完成它。
示例数据:
id | val
--------
1 4
2 7
3 2
4 2
5 9
6 8
7 3
对val排序得到2 2 3 4 7 8 9，因此中位数应该是4，而SELECT AVG(val) == 5。
当前回答
我刚刚在网上的评论中找到了另一个答案:
对于几乎所有SQL中的中位数: SELECT x.val from data x, data y GROUP BY x.val 总和(符号(1-SIGN (y.val-x.val))) = (COUNT (*) + 1) / 2
确保列有良好的索引，并且索引用于筛选和排序。与解释计划核对。
select count(*) from table --find the number of rows
计算“中值”行号。可能使用:median_row = floor(count / 2)。
然后把它从列表中挑出来:
select val from table order by val asc limit median_row,1
这将返回您想要的值的一行。
其他回答
MySQL从8.0版本开始支持窗口函数，您可以使用ROW_NUMBER或DENSE_RANK(不要使用RANK，因为它将相同的RANK分配给相同的值，就像在体育排名):
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));
最简单和快速的方法来计算中位数在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)
基于@bob的回答，这将查询泛化为能够返回多个中位数，并按某些标准分组。
想想，例如，一个车场二手车的中位数销售价格，按年-月分组。
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;
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;
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)
