Question

在StackOverflow和其他网站中找到了一些解决方案，以找到中位数，但是所有这些都可与以下主体一起使用-中位数是该行的行数的一半更少，另一半更大。但是对于偶数行，中位数是两个中间值的平均值。如何在MySQL中计算？

Answer 1

给出以下模式

CREATE TABLE numbers (
  i INT AUTO_INCREMENT PRIMARY KEY,
  n INT,
  INDEX(n)
);

我们想找到n列的中位数。

ROW_NUMBER（）和CTE（需要MySQL 8或MariaDB 10.2）

with sorted as (
    select t.n, row_number() over (order by t.n) as rn
    from numbers t
), cnt as (
    select count(*) as c from numbers
)
select avg(s.n) as median
from sorted s
cross join cnt
where s.rn between floor((c+1)/2) and ceil((c+1)/2);

性能：可以（对于10万行140ms）

具有AUTO_INCREMENT的临时表

drop temporary table if exists tmp;
create temporary table tmp(
    rn int auto_increment primary key,
    n int
) engine=memory;

insert into tmp(n) 
    select n
    from numbers
    order by n;

select avg(n) as median
from tmp
cross join (select count(*) as c from numbers) cnt
where rn between floor((c+1)/2) and ceil((c+1)/2);

性能：可以（对于10万行，为110ms）

具有计算的LIMIT / OFFSET的准备好的语句

set @c = (select count(*) from numbers);
set @limit = 2 - (@c % 2);
set @offset = (@c+1) div 2 - 1;

prepare stmt from '
    select avg(n) as median
    from (
        select n
        from numbers
        limit ? offset ?
    ) sub
';

execute stmt using @limit, @offset;

性能：最佳（10万行50ms）

交叉加入

使用主键

select avg(n) as median
from (
  select t.n
  from numbers t
  cross join numbers t2
  group by t.i
  having greatest(sum(t2.n < t.n), sum(t2.n > t.n))
      <= (select count(*) from numbers) / 2
) sub

没有主键

select avg(n) as median
from (
    select t.n
    from numbers t
    cross join numbers t2
    group by t.n
    having greatest(sum(t2.n < t.n), sum(t2.n > t.n)) / sqrt(SUM(t2.n = t.n))
        <= (select count(*) from numbers)/2
) sub

性能：最差-O（n²）（每1K行500ms）

MySQL如何从偶数行中找到中位数

1 个答案: