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
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
其他回答
在Jeff Atwood的答案的基础上,它是用GROUP BY和一个相关的子查询来获得每个组的中位数。
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
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/
以下解决方案在这些假设下有效:
无重复值 没有取消
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) ;
我想自己想出一个解决办法,但我的大脑绊倒了。我觉得很管用,但别让我早上解释。: 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)
如果你使用的是SQL 2005或更好的版本,这是一个很好的,简单的中位数计算表中的单列:
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