这段代码有点长，但很容易理解

medii是有列val的表，它有数据集， Smedi是一个cte，它将列idx作为行号，val作为medi表中的'val'，该表是升序排序的。 这是基本的数学，如果行号是奇数，那么它的中值来自smedi。 当它是偶数时，它是中间两个值的平均值。

with smedi(idx,vals) as(
                select ROW_NUMBER() over(order by val),val from medi
                )
select (case
            when (select count(*) from medi)%2!=0 then (select vals from smedi where (((select count(*) from medi)/2))=idx)
            else (select avg(vals) from smedi where idx in ((select count(*)/2 from medi),(select (count(*)/2)+1 from medi)))
            end)

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。

--Create Temp Table to Store Results in
DECLARE @results AS TABLE 
(
    [Month] datetime not null
 ,[Median] int not null
);

--This variable will determine the date
DECLARE @IntDate as int 
set @IntDate = -13


WHILE (@IntDate < 0) 
BEGIN

--Create Temp Table
DECLARE @table AS TABLE 
(
    [Rank] int not null
 ,[Days Open] int not null
);

--Insert records into Temp Table
insert into @table 

SELECT 
    rank() OVER (ORDER BY DATEADD(mm, DATEDIFF(mm, 0, DATEADD(ss, SVR.close_date, '1970')), 0), DATEDIFF(day,DATEADD(ss, SVR.open_date, '1970'),DATEADD(ss, SVR.close_date, '1970')),[SVR].[ref_num]) as [Rank]
 ,DATEDIFF(day,DATEADD(ss, SVR.open_date, '1970'),DATEADD(ss, SVR.close_date, '1970')) as [Days Open]
FROM
 mdbrpt.dbo.View_Request SVR
 LEFT OUTER JOIN dbo.dtv_apps_systems vapp 
 on SVR.category = vapp.persid
 LEFT OUTER JOIN dbo.prob_ctg pctg 
 on SVR.category = pctg.persid
 Left Outer Join [mdbrpt].[dbo].[rootcause] as [Root Cause] 
 on [SVR].[rootcause]=[Root Cause].[id]
 Left Outer Join [mdbrpt].[dbo].[cr_stat] as [Status]
 on [SVR].[status]=[Status].[code]
 LEFT OUTER JOIN [mdbrpt].[dbo].[net_res] as [net] 
 on [net].[id]=SVR.[affected_rc]
WHERE
 SVR.Type IN ('P') 
 AND
 SVR.close_date IS NOT NULL 
 AND
 [Status].[SYM] = 'Closed'
 AND
 SVR.parent is null
 AND
 [Root Cause].[sym] in ( 'RC - Application','RC - Hardware', 'RC - Operational', 'RC - Unknown')
 AND
 (
  [vapp].[appl_name] in ('3PI','Billing Rpts/Files','Collabrent','Reports','STMS','STMS 2','Telco','Comergent','OOM','C3-BAU','C3-DD','DIRECTV','DIRECTV Sales','DIRECTV Self Care','Dealer Website','EI Servlet','Enterprise Integration','ET','ICAN','ODS','SB-SCM','SeeBeyond','Digital Dashboard','IVR','OMS','Order Services','Retail Services','OSCAR','SAP','CTI','RIO','RIO Call Center','RIO Field Services','FSS-RIO3','TAOS','TCS')
 OR
  pctg.sym in ('Systems.Release Health Dashboard.Problem','DTV QA Test.Enterprise Release.Deferred Defect Log')
 AND  
  [Net].[nr_desc] in ('3PI','Billing Rpts/Files','Collabrent','Reports','STMS','STMS 2','Telco','Comergent','OOM','C3-BAU','C3-DD','DIRECTV','DIRECTV Sales','DIRECTV Self Care','Dealer Website','EI Servlet','Enterprise Integration','ET','ICAN','ODS','SB-SCM','SeeBeyond','Digital Dashboard','IVR','OMS','Order Services','Retail Services','OSCAR','SAP','CTI','RIO','RIO Call Center','RIO Field Services','FSS-RIO3','TAOS','TCS')
 )
 AND
 DATEADD(mm, DATEDIFF(mm, 0, DATEADD(ss, SVR.close_date, '1970')), 0) = DATEADD(mm, DATEDIFF(mm,0,DATEADD(mm,@IntDate,getdate())), 0)
ORDER BY [Days Open]



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

WITH MyResults(RowNo, [Days Open]) AS
(
    SELECT RowNo, [Days Open] FROM
        (SELECT ROW_NUMBER() OVER (ORDER BY [Days Open]) AS RowNo, [Days Open] FROM @table) AS Foo
)


insert into @results
SELECT 
 DATEADD(mm, DATEDIFF(mm,0,DATEADD(mm,@IntDate,getdate())), 0) as [Month]
 ,AVG([Days Open])as [Median] FROM MyResults WHERE RowNo = (@Count+1)/2 OR RowNo = ((@Count+1)%2) * ((@Count+2)/2) 


set @IntDate = @IntDate+1
DELETE FROM @table
END

select *
from @results
order by [Month]

查看SQL中位数计算的其他解决方案: “用MySQL计算中位数的简单方法”(解决方案大多与供应商无关)。

MS SQL Server 2012(及以后版本)有PERCENTILE_DISC函数，计算排序值的特定百分比。PERCENTILE_DISC(0.5)将计算中位数- https://msdn.microsoft.com/en-us/library/hh231327.aspx

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