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

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) 

其他回答

我发现接受的解决方案在我的MySQL安装上不起作用,返回一个空集,但这个查询在我测试的所有情况下都适用:

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
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。

这种方法似乎包括偶数和奇数计数,没有子查询。

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