中找到

这是查找属性中值的最简单方法。

Select round(S.salary,4) median from employee S 
where (select count(salary) from station 
where salary < S.salary ) = (select count(salary) from station
where salary > S.salary)

以下是我的解决方案:

with tempa as

 (

    select value,row_number() over (order by value) as Rn,/* Assigning a 
                                                           row_number */
           count(value) over () as Cnt /*Taking total count of the values */
    from numbers
    where value is not null /* Excluding the null values */
 ),

tempb as

  (

    /* Since we don't know whether the number of rows is odd or even, we shall 
     consider both the scenarios */

    select round(cnt/2) as Ref from tempa where mod(cnt,2)=1
    union all
    select round(cnt/2) a Ref from tempa where mod(cnt,2)=0
     union all
    select round(cnt/2) + 1 as Ref from tempa where mod(cnt,2)=0
   )
  select avg(value) as Median_Value

  from tempa where rn in

    ( select Ref from tempb);

在UDF中，写:

 Select Top 1 medianSortColumn from Table T
  Where (Select Count(*) from Table
         Where MedianSortColumn <
           (Select Count(*) From Table) / 2)
  Order By medianSortColumn

试试下面的逻辑来找出中位数:

考虑一个包含以下数字的表格: 1、1、2、3、4、5所示

中位数是2.5

with tempa as 
(
    select num,count(num) over() as Cnt,
        row_number() over (order by num) as Rnum
    from temp),
tempb as
    (
        select round(cnt/2) as ref_value
        from tempa where mod(cnt,2)<>0
        union all
        select round(cnt/2) from tempa where mod(cnt,2)=0
        union all
        select round(cnt/2+1)
        from tempa where mod(cnt,2)=0
    )
select avg(num) from tempa
where rnum in (select * from tempb);
    

查看SQL中位数计算的其他解决方案: “用MySQL计算中位数的简单方法”(解决方案大多与供应商无关)。

MS SQL Server 2012(及以后版本)有PERCENTILE_DISC函数，计算排序值的特定百分比。PERCENTILE_DISC(0.5)将计算中位数- https://msdn.microsoft.com/en-us/library/hh231327.aspx