根据MSDN, Median在Transact-SQL中不能作为聚合函数使用。但是,我想知道是否可以创建此功能(使用create Aggregate函数、用户定义函数或其他方法)。

最好的方法(如果可能的话)是什么——允许在聚合查询中计算中值(假设是数值数据类型)?


当前回答

在SQL Server 2012中,您应该使用PERCENTILE_CONT:

SELECT SalesOrderID, OrderQty,
    PERCENTILE_CONT(0.5) 
        WITHIN GROUP (ORDER BY OrderQty)
        OVER (PARTITION BY SalesOrderID) AS MedianCont
FROM Sales.SalesOrderDetail
WHERE SalesOrderID IN (43670, 43669, 43667, 43663)
ORDER BY SalesOrderID DESC

参见:http://blog.sqlauthority.com/2011/11/20/sql-server-introduction-to-percentile_cont-analytic-functions-introduced-in-sql-server-2012/

其他回答

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。

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

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)  

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)

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