简单、快速、准确

SELECT x.Amount 
FROM   (SELECT amount, 
               Count(1) OVER (partition BY 'A')        AS TotalRows, 
               Row_number() OVER (ORDER BY Amount ASC) AS AmountOrder 
        FROM   facttransaction ft) x 
WHERE  x.AmountOrder = Round(x.TotalRows / 2.0, 0)  

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。

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);
    

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);

虽然Justin grant的解决方案看起来很可靠，但我发现当您在给定的分区键中有许多重复值时，ASC重复值的行号最终会不按顺序排列，因此它们不能正确对齐。

以下是我的研究结果的一个片段:

KEY VALUE ROWA ROWD  

13  2     22   182
13  1     6    183
13  1     7    184
13  1     8    185
13  1     9    186
13  1     10   187
13  1     11   188
13  1     12   189
13  0     1    190
13  0     2    191
13  0     3    192
13  0     4    193
13  0     5    194

我使用Justin的代码作为这个解决方案的基础。尽管考虑到使用多个派生表效率不高，但它确实解决了我遇到的行排序问题。任何改进都会受到欢迎，因为我在T-SQL方面不是那么有经验。

SELECT PKEY, cast(AVG(VALUE)as decimal(5,2)) as MEDIANVALUE
FROM
(
  SELECT PKEY,VALUE,ROWA,ROWD,
  'FLAG' = (CASE WHEN ROWA IN (ROWD,ROWD-1,ROWD+1) THEN 1 ELSE 0 END)
  FROM
  (
    SELECT
    PKEY,
    cast(VALUE as decimal(5,2)) as VALUE,
    ROWA,
    ROW_NUMBER() OVER (PARTITION BY PKEY ORDER BY ROWA DESC) as ROWD 

    FROM
    (
      SELECT
      PKEY, 
      VALUE,
      ROW_NUMBER() OVER (PARTITION BY PKEY ORDER BY VALUE ASC,PKEY ASC ) as ROWA 
      FROM [MTEST]
    )T1
  )T2
)T3
WHERE FLAG = '1'
GROUP BY PKEY
ORDER BY PKEY

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。