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

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)

这是我能想到的求中位数的最优解。示例中的名称基于Justin示例。确保表有索引 销售。SalesOrderHeader以索引列CustomerId和TotalDue的顺序存在。

SELECT
 sohCount.CustomerId,
 AVG(sohMid.TotalDue) as TotalDueMedian
FROM 
(SELECT 
  soh.CustomerId,
  COUNT(*) as NumberOfRows
FROM 
  Sales.SalesOrderHeader soh 
GROUP BY soh.CustomerId) As sohCount
CROSS APPLY 
    (Select 
       soh.TotalDue
    FROM 
    Sales.SalesOrderHeader soh 
    WHERE soh.CustomerId = sohCount.CustomerId 
    ORDER BY soh.TotalDue
    OFFSET sohCount.NumberOfRows / 2 - ((sohCount.NumberOfRows + 1) % 2) ROWS 
    FETCH NEXT 1 + ((sohCount.NumberOfRows + 1) % 2) ROWS ONLY
    ) As sohMid
GROUP BY sohCount.CustomerId

更新

我有点不确定哪种方法性能最好，所以我比较了我的方法Justin Grants和Jeff Atwoods，在一个批量中运行基于这三种方法的查询，每个查询的批量成本为:

没有指数:

我的30% Justin Grants 13% Jeff Atwoods 58%

还有index

我的3%。 Justin Grants 10% Jeff Atwoods 87%

I tried to see how well the queries scale if you have index by creating more data from around 14 000 rows by a factor of 2 up to 512 which means in the end around 7,2 millions rows. Note I made sure CustomeId field where unique for each time I did a single copy, so the proportion of rows compared to unique instance of CustomerId was kept constant. While I was doing this I ran executions where I rebuilt index afterwards, and I noticed the results stabilized at around a factor of 128 with the data I had to these values:

我的3%。 贾斯汀·格兰特5% Jeff Atwoods 92%

我想知道，在保持惟一CustomerId不变的情况下，扩展行数会如何影响性能，因此我设置了一个新的测试，在其中执行了上述操作。现在，批成本比率并没有稳定下来，而是不断分化，每个CustomerId平均大约有20行，最后每个这样唯一的Id大约有10000行。数字如下:

我的4% 贾斯汀60% 杰夫斯35%

通过比较结果，我确保我正确地实现了每个方法。 我的结论是，只要索引存在，我使用的方法通常更快。还要注意，本文针对这个特定问题推荐使用这种方法https://www.microsoftpressstore.com/articles/article.aspx?p=2314819&seqNum=5

进一步提高对该查询的后续调用的性能的一种方法是在辅助表中持久化计数信息。您甚至可以通过一个触发器来维护它，该触发器更新并保存有关依赖于CustomerId的SalesOrderHeader行计数的信息，当然您也可以简单地存储中值。

在我的解决方案表中是一个只有分数列的学生表，我正在计算分数的中位数，这个解决方案是基于SQL server 2019的

with total_c as ( --Total_c CTE counts total number of rows in a table
    select count(*) as n from student
),
even as ( --Even CTE extract two middle rows if the number of rows are even
    select marks from student 
    order by marks 
    offset (select n from total_c)/2 -1 rows
    fetch next 2 rows only
),
odd as ( --Odd CTE extract middle row if the number of rows are odd
    select marks from student 
    order by marks 
    offset (select n + 1 from total_c)/2 -1 rows
    fetch next 1 rows only
    )
--Case statement helps to select odd or even CTE based on number of rows
select                                                        
case when n%2 = 0 then (select avg(cast(marks as float)) from even)
    else (select marks from odd)
end as med_marks
from total_c

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!!