功能的计算中位数在SQL服务器
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20-09-2019 - |
解决方案
有很多方法可以做到这一点,有显着不同的绩效。这里有一个特别优化的解决方案,从 中位数,ROW_NUMBERs和性能.这是一个特别的最佳解决方案,当涉及到实际I/o执行过程中产生的–它看起来代价更高于其他解决方案,但它实际上是要快得多。
该网页还包含讨论的其他解决方案并且性能测试的细节。注意使用的一个独特的列作为一个disambiguator的情况下,有多个行相同价值的中位列。
如同所有的数据库性能的方案,总是试图测试解决方案的真实数据上的真正的硬件–你永远不知道什么时候的变化到SQL服务器的优化或特点在环境将使常迅速的解决方案速度较慢。
SELECT
CustomerId,
AVG(TotalDue)
FROM
(
SELECT
CustomerId,
TotalDue,
-- SalesOrderId in the ORDER BY is a disambiguator to break ties
ROW_NUMBER() OVER (
PARTITION BY CustomerId
ORDER BY TotalDue ASC, SalesOrderId ASC) AS RowAsc,
ROW_NUMBER() OVER (
PARTITION BY CustomerId
ORDER BY TotalDue DESC, SalesOrderId DESC) AS RowDesc
FROM Sales.SalesOrderHeader SOH
) x
WHERE
RowAsc IN (RowDesc, RowDesc - 1, RowDesc + 1)
GROUP BY CustomerId
ORDER BY CustomerId;
其他提示
如果你使用SQL 2005或更高,这是在一个表中单列一个不错的,简单的上下的中位数计算:
SELECT
(
(SELECT MAX(Score) FROM
(SELECT TOP 50 PERCENT Score FROM Posts ORDER BY Score) AS BottomHalf)
+
(SELECT MIN(Score) FROM
(SELECT TOP 50 PERCENT Score FROM Posts ORDER BY Score DESC) AS TopHalf)
) / 2 AS Median
在SQL服务器2012年你应该使用 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
我原来的快速回答是:
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
这会给你一举中位数和四分位范围。如果你真的只想要一排就是中位数则取消对where子句。
当你坚持到这一个解释计划,工作的60%的分选位置算出的统计依赖这样时,这是不可避免的的数据。
我已经修改了答案跟随罗伯特Ševčík-Robajz在下面的意见的很好的建议:
;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
当你有一个甚至多个数据项这应该计算出正确的位数和百分率值。此外,取消最后where子句如果只想位数,而不是整个百分位分布。
甚至更好:
SELECT @Median = AVG(1.0 * val)
FROM
(
SELECT o.val, rn = ROW_NUMBER() OVER (ORDER BY o.val), c.c
FROM dbo.EvenRows AS o
CROSS JOIN (SELECT c = COUNT(*) FROM dbo.EvenRows) AS c
) AS x
WHERE rn IN ((c + 1)/2, (c + 2)/2);
从主自己,伊茨克奔甘的!
MS SQL Server 2012的(和更高)具有计算用于排序的值的特定百分位数的PERCENTILE_DISC功能。 PERCENTILE_DISC(0.5)将计算中值 - https://msdn.microsoft.com/ EN-US /库/ hh231327.aspx
简单,快速,准确的
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)
如果您要使用在SQL Server中创建聚合函数,这是如何做到这一点。这样做,这样具有能够写出干净查询的好处。注意此该过程可以适用于很容易计算的百分数值。
创建新的Visual Studio项目,并设置目标框架到.NET 3.5(这是用于SQL 2008,它可以是SQL 2012不同)。然后,创建一个类文件,并把在下面的代码,或c#当量:
Imports Microsoft.SqlServer.Server
Imports System.Data.SqlTypes
Imports System.IO
<Serializable>
<SqlUserDefinedAggregate(Format.UserDefined, IsInvariantToNulls:=True, IsInvariantToDuplicates:=False, _
IsInvariantToOrder:=True, MaxByteSize:=-1, IsNullIfEmpty:=True)>
Public Class Median
Implements IBinarySerialize
Private _items As List(Of Decimal)
Public Sub Init()
_items = New List(Of Decimal)()
End Sub
Public Sub Accumulate(value As SqlDecimal)
If Not value.IsNull Then
_items.Add(value.Value)
End If
End Sub
Public Sub Merge(other As Median)
If other._items IsNot Nothing Then
_items.AddRange(other._items)
End If
End Sub
Public Function Terminate() As SqlDecimal
If _items.Count <> 0 Then
Dim result As Decimal
_items = _items.OrderBy(Function(i) i).ToList()
If _items.Count Mod 2 = 0 Then
result = ((_items((_items.Count / 2) - 1)) + (_items(_items.Count / 2))) / 2@
Else
result = _items((_items.Count - 1) / 2)
End If
Return New SqlDecimal(result)
Else
Return New SqlDecimal()
End If
End Function
Public Sub Read(r As BinaryReader) Implements IBinarySerialize.Read
'deserialize it from a string
Dim list = r.ReadString()
_items = New List(Of Decimal)
For Each value In list.Split(","c)
Dim number As Decimal
If Decimal.TryParse(value, number) Then
_items.Add(number)
End If
Next
End Sub
Public Sub Write(w As BinaryWriter) Implements IBinarySerialize.Write
'serialize the list to a string
Dim list = ""
For Each item In _items
If list <> "" Then
list += ","
End If
list += item.ToString()
Next
w.Write(list)
End Sub
End Class
然后编译它,DLL和PDB文件复制到您的SQL Server计算机和SQL Server运行下面的命令:
CREATE ASSEMBLY CustomAggregate FROM '{path to your DLL}'
WITH PERMISSION_SET=SAFE;
GO
CREATE AGGREGATE Median(@value decimal(9, 3))
RETURNS decimal(9, 3)
EXTERNAL NAME [CustomAggregate].[{namespace of your DLL}.Median];
GO
然后,您可以编写一个查询来计算这样的中位数: SELECT dbo.Median(场)FROM表
我只是碰到这个页面就在寻找一个基于集合的解决方案中位数。寻找一些解决方案在这里后,我想出了以下内容。希望是帮助/作品。
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
下面的查询返回从值在一列中的列表的中值即可。它不能被用作或连同聚合函数,但仍然可以使用它作为一个子查询与内选择WHERE子句。
<强> SQL Server 2005的+:强>
SELECT TOP 1 value from
(
SELECT TOP 50 PERCENT value
FROM table_name
ORDER BY value
)for_median
ORDER BY value DESC
虽然贾斯汀·格兰特的解决方案似乎很扎实,我发现,当你有一个给定的分区键的ASC重复值的行号结束了序列内的号码重复值,因此他们没有正确对齐。
下面是从我的结果的片段:
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
我用Justin的代码作为这个解决方案的基础。虽然效率不高,考虑到使用多个派生表的它确实解决我遇到的行排序问题。因为我不是在T-SQL经历过任何的改进将受到欢迎。
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
以上Justin的例子是非常好的。不过,该主键必须要非常清楚地说明。我已经看到了在野外的代码没有钥匙,结果都是不好的。
我得到了PERCENTILE_CONT的抱怨是,它不会让你从数据集中的实际值。 得到一个“中间”,即从数据集的实际值使用PERCENTILE_DISC。
SELECT SalesOrderID, OrderQty,
PERCENTILE_DISC(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
在一个UDF,写:
Select Top 1 medianSortColumn from Table T
Where (Select Count(*) from Table
Where MedianSortColumn <
(Select Count(*) From Table) / 2)
Order By medianSortColumn
请参阅SQL中位数算出这里其他的解决方案: “简单的方法来计算中位数与MySQL ”(解决方案大多是独立于供应商)。
有关的连续可变/测量从 '表1' 'COL1'
select col1
from
(select top 50 percent col1,
ROW_NUMBER() OVER(ORDER BY col1 ASC) AS Rowa,
ROW_NUMBER() OVER(ORDER BY col1 DESC) AS Rowd
from table1 ) tmp
where tmp.Rowa = tmp.Rowd
我想通过自己制定出一个解决方案,但我的大脑绊倒了在路上了。我的认为的它的工作原理,但不要问我来解释它是在早晨。 :P
DECLARE @table AS TABLE
(
Number int not null
);
insert into @table select 2;
insert into @table select 4;
insert into @table select 9;
insert into @table select 15;
insert into @table select 22;
insert into @table select 26;
insert into @table select 37;
insert into @table select 49;
DECLARE @Count AS INT
SELECT @Count = COUNT(*) FROM @table;
WITH MyResults(RowNo, Number) AS
(
SELECT RowNo, Number FROM
(SELECT ROW_NUMBER() OVER (ORDER BY Number) AS RowNo, Number FROM @table) AS Foo
)
SELECT AVG(Number) FROM MyResults WHERE RowNo = (@Count+1)/2 OR RowNo = ((@Count+1)%2) * ((@Count+2)/2)
--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]
此可与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
)
对于像我这样的新手谁是学习最基础的,我个人觉得这个例子更容易理解,因为它更容易明白到底发生了什么,并在那里中间值是从...来
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
在上面虽然一些代码的绝对畏惧!!!
这很简单,只要我能想出一个答案。我的数据运行良好。如果要排除某些值只需添加一个where子句内选择。
SELECT TOP 1
ValueField AS MedianValue
FROM
(SELECT TOP(SELECT COUNT(1)/2 FROM tTABLE)
ValueField
FROM
tTABLE
ORDER BY
ValueField) A
ORDER BY
ValueField DESC
以下解决方案在这些假设下有效:
- 没有重复值
- 无 NULL
代码:
IF OBJECT_ID('dbo.R', 'U') IS NOT NULL
DROP TABLE dbo.R
CREATE TABLE R (
A FLOAT NOT NULL);
INSERT INTO R VALUES (1);
INSERT INTO R VALUES (2);
INSERT INTO R VALUES (3);
INSERT INTO R VALUES (4);
INSERT INTO R VALUES (5);
INSERT INTO R VALUES (6);
-- Returns Median(R)
select SUM(A) / CAST(COUNT(A) AS FLOAT)
from R R1
where ((select count(A) from R R2 where R1.A > R2.A) =
(select count(A) from R R2 where R1.A < R2.A)) OR
((select count(A) from R R2 where R1.A > R2.A) + 1 =
(select count(A) from R R2 where R1.A < R2.A)) OR
((select count(A) from R R2 where R1.A > R2.A) =
(select count(A) from R R2 where R1.A < R2.A) + 1) ;
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
我尝试用几个备选方案,但由于我的数据记录已经重复值时,ROW_NUMBER版本好像是不是我的选择。因此,这里所述查询我使用(具有一个NTILE版):
SELECT distinct
CustomerId,
(
MAX(CASE WHEN Percent50_Asc=1 THEN TotalDue END) OVER (PARTITION BY CustomerId) +
MIN(CASE WHEN Percent50_desc=1 THEN TotalDue END) OVER (PARTITION BY CustomerId)
)/2 MEDIAN
FROM
(
SELECT
CustomerId,
TotalDue,
NTILE(2) OVER (
PARTITION BY CustomerId
ORDER BY TotalDue ASC) AS Percent50_Asc,
NTILE(2) OVER (
PARTITION BY CustomerId
ORDER BY TotalDue DESC) AS Percent50_desc
FROM Sales.SalesOrderHeader SOH
) x
ORDER BY CustomerId;
杰夫·阿特伍德的回答大厦上面在这里它与GROUP BY和相关子查询来获得各组的中位数。
SELECT TestID,
(
(SELECT MAX(Score) FROM
(SELECT TOP 50 PERCENT Score FROM Posts WHERE TestID = Posts_parent.TestID ORDER BY Score) AS BottomHalf)
+
(SELECT MIN(Score) FROM
(SELECT TOP 50 PERCENT Score FROM Posts WHERE TestID = Posts_parent.TestID ORDER BY Score DESC) AS TopHalf)
) / 2 AS MedianScore,
AVG(Score) AS AvgScore, MIN(Score) AS MinScore, MAX(Score) AS MaxScore
FROM Posts_parent
GROUP BY Posts_parent.TestID
通常,我们可能需要计算中值不只是为整个表,但用于相对于一些ID聚集体。换句话说,计算中位数在我们的表中的每个ID,每个ID具有多条记录。 (基于由@gdoron编辑的溶液:良好的性能,并且在许多SQL作品)
SELECT our_id, AVG(1.0 * our_val) as Median
FROM
( SELECT our_id, our_val,
COUNT(*) OVER (PARTITION BY our_id) AS cnt,
ROW_NUMBER() OVER (PARTITION BY our_id ORDER BY our_val) AS rnk
FROM our_table
) AS x
WHERE rnk IN ((cnt + 1)/2, (cnt + 2)/2) GROUP BY our_id;
希望它能帮助。
对于你的问题,杰夫阿特伍德已经给出的简单和有效的解决方案。但是,如果你正在寻找一些替代方法计算中位数,下面的SQL代码会帮助你。
create table employees(salary int);
insert into employees values(8); insert into employees values(23); insert into employees values(45); insert into employees values(123); insert into employees values(93); insert into employees values(2342); insert into employees values(2238);
select * from employees;
declare @odd_even int; declare @cnt int; declare @middle_no int;
set @cnt=(select count(*) from employees); set @middle_no=(@cnt/2)+1; select @odd_even=case when (@cnt%2=0) THEN -1 ELse 0 END ;
select AVG(tbl.salary) from (select salary,ROW_NUMBER() over (order by salary) as rno from employees group by salary) tbl where tbl.rno=@middle_no or tbl.rno=@middle_no+@odd_even;
如果您正在寻找在MySQL中位数计算,这 GitHub的链接一>将是有用的。
这是我能想到的寻找中位数的最佳解决方案。示例中的名称基于 Justin 示例。确保表销售的索引。Salesorderheader存在索引列的customerId和TotalDue按此顺序。
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具有最佳性能,因此我通过在一批中运行基于所有三种方法的查询来比较我的方法 Justin Grants 和 Jeff Atwoods,每个查询的批量成本为:
没有索引:
- 我的 30%
- 贾斯汀·格兰特 13%
- 杰夫·阿特伍兹 58%
并带有索引
- 我的3%。
- 贾斯汀·格兰特 10%
- 杰夫·阿特伍兹 87%
我试图通过从大约 14000 行创建更多数据(乘以 2 倍到 512 行,这意味着最终大约 720 万行)来查看查询的扩展程度如何。请注意,我确保每次执行单个副本时,CustomeId 字段都是唯一的,因此与 CustomerId 的唯一实例相比,行的比例保持不变。当我这样做时,我运行了执行程序,之后重建了索引,我注意到结果稳定在 128 倍左右,数据为这些值:
- 我的3%。
- 贾斯汀·格兰特 5%
- 杰夫·阿特伍兹 92%
我想知道扩展行数但保持唯一的 CustomerId 不变会如何影响性能,因此我设置了一个新的测试,我就是这样做的。现在,批量成本比率非但没有稳定下来,反而不断发散,而且每个 CustomerId 平均不是 20 行,最终每个这样的唯一 ID 大约有 10000 行。数字其中:
- 我的 4%
- 贾斯汀 60%
- 杰夫斯 35%
通过比较结果,我确保我正确地实施了每种方法。我的结论是,只要索引存在,我使用的方法通常会更快。另请注意,此方法是本文中针对此特定问题推荐的方法 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将计数信息保留在辅助表中。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 行数的信息,当然您也可以简单地存储中位数。
对于大型数据集,你可以试试这个要点是:
https://gist.github.com/chrisknoll/1b38761ce8c5016ec5b2
其工作原理是在您所设定的聚集不同的值,你会发现(如年龄,或出生等一年),并使用SQL窗口功能来定位你在查询中指定的任何百分位置。
<强>平均查找强>
这是找到一个属性的值的最简单的方法。
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)