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);
    

我最初的回答是:

select  max(my_column) as [my_column], quartile
from    (select my_column, ntile(4) over (order by my_column) as [quartile]
         from   my_table) i
--where quartile = 2
group by quartile

这将使您一举获得中位数和四分位范围。如果你真的只想要一行作为中值，那么取消注释where子句。

当你把它放入解释计划时，60%的工作是对数据进行排序，这在计算像这样的位置依赖统计数据时是不可避免的。

我修改了答案，以遵循Robert Ševčík-Robajz在下面的评论中提出的优秀建议:

;with PartitionedData as
  (select my_column, ntile(10) over (order by my_column) as [percentile]
   from   my_table),
MinimaAndMaxima as
  (select  min(my_column) as [low], max(my_column) as [high], percentile
   from    PartitionedData
   group by percentile)
select
  case
    when b.percentile = 10 then cast(b.high as decimal(18,2))
    else cast((a.low + b.high)  as decimal(18,2)) / 2
  end as [value], --b.high, a.low,
  b.percentile
from    MinimaAndMaxima a
  join  MinimaAndMaxima b on (a.percentile -1 = b.percentile) or (a.percentile = 10 and b.percentile = 10)
--where b.percentile = 5

当您有偶数个数据项时，这应该计算正确的中位数和百分比值。同样，如果您只想要中位数而不是整个百分位数分布，请取消最后的where子句的注释。

我想自己想出一个解决办法，但我的大脑绊倒了。我觉得很管用，但别让我早上解释。: P

DECLARE @table AS TABLE
(
    Number int not null
);

insert into @table select 2;
insert into @table select 4;
insert into @table select 9;
insert into @table select 15;
insert into @table select 22;
insert into @table select 26;
insert into @table select 37;
insert into @table select 49;

DECLARE @Count AS INT
SELECT @Count = COUNT(*) FROM @table;

WITH MyResults(RowNo, Number) AS
(
    SELECT RowNo, Number FROM
        (SELECT ROW_NUMBER() OVER (ORDER BY Number) AS RowNo, Number FROM @table) AS Foo
)
SELECT AVG(Number) FROM MyResults WHERE RowNo = (@Count+1)/2 OR RowNo = ((@Count+1)%2) * ((@Count+2)/2)

下面的查询从一列中的值列表返回中位数。它不能作为聚合函数使用，也不能与聚合函数一起使用，但仍然可以在内部select中使用带有WHERE子句的子查询。

SQL Server 2005+:

SELECT TOP 1 value from
(
    SELECT TOP 50 PERCENT value 
    FROM table_name 
    ORDER BY  value
)for_median
ORDER BY value DESC

对于像我这样正在学习基础知识的新手来说，我个人觉得这个例子更容易理解，因为它更容易理解到底发生了什么以及中值来自哪里……

select
 ( max(a.[Value1]) + min(a.[Value1]) ) / 2 as [Median Value1]
,( max(a.[Value2]) + min(a.[Value2]) ) / 2 as [Median Value2]

from (select
    datediff(dd,startdate,enddate) as [Value1]
    ,xxxxxxxxxxxxxx as [Value2]
     from dbo.table1
     )a

不过，对上面的一些代码绝对敬畏!!

with t1 as (select *, row_number() over(order by ordqty) as rn,
count(*) over() as rc from ord_line)
select rn,* from t1 where rn in((rc+1)/2, (rc+2)/2);

它将计算偶数和奇数的中位数。

Ord_line是一个表 Ordqty是一个列