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。

使用一条语句——一种方法是使用ROW_NUMBER()， COUNT()窗口函数并过滤子查询。下面是薪资中位数:

 SELECT AVG(e_salary) 
 FROM                                                             
    (SELECT 
      ROW_NUMBER() OVER(ORDER BY e_salary) as row_no, 
      e_salary,
      (COUNT(*) OVER()+1)*0.5 AS row_half
     FROM Employee) t
 WHERE row_no IN (FLOOR(row_half),CEILING(row_half))

我在网上看到过类似的解决方案，使用地板和天花板，但尝试使用单一的语句。(编辑)

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);
    

对于连续变量/测量'table1'中的'col1'

select col1  
from
    (select top 50 percent col1, 
    ROW_NUMBER() OVER(ORDER BY col1 ASC) AS Rowa,
    ROW_NUMBER() OVER(ORDER BY col1 DESC) AS Rowd
    from table1 ) tmp
where tmp.Rowa = tmp.Rowd

DECLARE @Obs int
DECLARE @RowAsc table
(
ID      INT IDENTITY,
Observation  FLOAT
)
INSERT INTO @RowAsc
SELECT Observations FROM MyTable
ORDER BY 1 
SELECT @Obs=COUNT(*)/2 FROM @RowAsc
SELECT Observation AS Median FROM @RowAsc WHERE ID=@Obs

对于大规模数据集，您可以尝试以下GIST:

https://gist.github.com/chrisknoll/1b38761ce8c5016ec5b2

它通过聚合您在集合中找到的不同值(例如年龄或出生年份等)来工作，并使用SQL窗口函数来定位您在查询中指定的任何百分比位置。