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)?
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
--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]
其他回答
我只是在寻找一个基于集的中位数的解决方案时偶然发现了这一页。在研究了一些解决方案之后,我想到了以下几点。希望是有用的。
DECLARE @test TABLE(
i int identity(1,1),
id int,
score float
)
INSERT INTO @test (id,score) VALUES (1,10)
INSERT INTO @test (id,score) VALUES (1,11)
INSERT INTO @test (id,score) VALUES (1,15)
INSERT INTO @test (id,score) VALUES (1,19)
INSERT INTO @test (id,score) VALUES (1,20)
INSERT INTO @test (id,score) VALUES (2,20)
INSERT INTO @test (id,score) VALUES (2,21)
INSERT INTO @test (id,score) VALUES (2,25)
INSERT INTO @test (id,score) VALUES (2,29)
INSERT INTO @test (id,score) VALUES (2,30)
INSERT INTO @test (id,score) VALUES (3,20)
INSERT INTO @test (id,score) VALUES (3,21)
INSERT INTO @test (id,score) VALUES (3,25)
INSERT INTO @test (id,score) VALUES (3,29)
DECLARE @counts TABLE(
id int,
cnt int
)
INSERT INTO @counts (
id,
cnt
)
SELECT
id,
COUNT(*)
FROM
@test
GROUP BY
id
SELECT
drv.id,
drv.start,
AVG(t.score)
FROM
(
SELECT
MIN(t.i)-1 AS start,
t.id
FROM
@test t
GROUP BY
t.id
) drv
INNER JOIN @test t ON drv.id = t.id
INNER JOIN @counts c ON t.id = c.id
WHERE
t.i = ((c.cnt+1)/2)+drv.start
OR (
t.i = (((c.cnt+1)%2) * ((c.cnt+2)/2))+drv.start
AND ((c.cnt+1)%2) * ((c.cnt+2)/2) <> 0
)
GROUP BY
drv.id,
drv.start
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)
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)。
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!!
这适用于SQL 2000:
DECLARE @testTable TABLE
(
VALUE INT
)
--INSERT INTO @testTable -- Even Test
--SELECT 3 UNION ALL
--SELECT 5 UNION ALL
--SELECT 7 UNION ALL
--SELECT 12 UNION ALL
--SELECT 13 UNION ALL
--SELECT 14 UNION ALL
--SELECT 21 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 29 UNION ALL
--SELECT 40 UNION ALL
--SELECT 56
--
--INSERT INTO @testTable -- Odd Test
--SELECT 3 UNION ALL
--SELECT 5 UNION ALL
--SELECT 7 UNION ALL
--SELECT 12 UNION ALL
--SELECT 13 UNION ALL
--SELECT 14 UNION ALL
--SELECT 21 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 29 UNION ALL
--SELECT 39 UNION ALL
--SELECT 40 UNION ALL
--SELECT 56
DECLARE @RowAsc TABLE
(
ID INT IDENTITY,
Amount INT
)
INSERT INTO @RowAsc
SELECT VALUE
FROM @testTable
ORDER BY VALUE ASC
SELECT AVG(amount)
FROM @RowAsc ra
WHERE ra.id IN
(
SELECT ID
FROM @RowAsc
WHERE ra.id -
(
SELECT MAX(id) / 2.0
FROM @RowAsc
) BETWEEN 0 AND 1
)