sql求中位数_sql中位数函数

中位数是指有序数列中，位于 中间位置的数的值
若为奇数，则中间数开始位置=结束位置
若为偶数，则中位数结束位置-开始位置=1
即
求解公司员工薪水的中位数

select  com_id, floor((count(salary)+1)/2) as start,
floor((count(salary)+2)/2) as end
from employee group by com_id order by com_id
1
2
3

1. 窗口函数：根据中位数的位置信息进行求解

• 分奇偶条件判断

select com_id,salary
from
	(
	select com_id,
	salary,
	row_number() over(partition by com_id order by salary desc) as rnk,
	count(salary) over(partition by com_id) as num
	from employee 
	) t1
where t1.rnk in (floor(num/2)+1, if(mod(num,2)=0,floor(num/2),floor(num/2)+1)
order by com_id

1
2
3
4
5
6
7
8
9
10
11
12

• 中位数条件由排序和总和计算

select com_id, salary
from
(
	select com_id,salary,
	row_number() over(partition by com_id order by salary Desc) rnk,
	count(salary) over(partition by com_id) as num
	from employee
) t1
where abs(t1.rnk - (t1.num+1)/2) < 1
order by com_id
1
2
3
4
5
6
7
8
9
10

注意：不可在一次查询中对窗口函数的结果进行操作
因为查询的顺序为：from->where->group by->having->select->order by

2. 中位数，正排倒排都是中位数

select com_id,salary
from
(
	select *,
	row_number() over(partition by com_id order by salary) as rnk1,
	row_number() over(partition by com_id order by salary desc) as rnk2
	from employee
) t1
where rnk1=rnk2 or abs(rnk1-rnk2)=1
order by com_id

1
2
3
4
5
6
7
8
9
10
11

报错：BIGINT UNSIGNED value is out of range

两种方式修改：

直接修改设置SET sql_mode=‘NO_UNSIGNED_SUBTRACTION’

或者修改代码

select com_id,salary
from
(
	select *,
	row_number() over(partition by com_id order by salary) as rnk1,
	row_number() over(partition by com_id order by salary desc) as rnk2
	from employee
) t1
where t1.rnk1 = t1.rnk2 or abs(cast(t1.rnk1 as signed)-cast(t1.rnk2 as signed)) = 1
1
2
3
4
5
6
7
8
9

ref:SQL 求中位数