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
以下是我的解决方案:
with tempa as
(
select value,row_number() over (order by value) as Rn,/* Assigning a
row_number */
count(value) over () as Cnt /*Taking total count of the values */
from numbers
where value is not null /* Excluding the null values */
),
tempb as
(
/* Since we don't know whether the number of rows is odd or even, we shall
consider both the scenarios */
select round(cnt/2) as Ref from tempa where mod(cnt,2)=1
union all
select round(cnt/2) a Ref from tempa where mod(cnt,2)=0
union all
select round(cnt/2) + 1 as Ref from tempa where mod(cnt,2)=0
)
select avg(value) as Median_Value
from tempa where rn in
( select Ref from tempb);
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)
更好的是:
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);
来自主人,伊兹克·本甘!
我尝试了几种替代方案,但由于我的数据记录有重复的值,ROW_NUMBER版本似乎不是我的选择。这里是我使用的查询(NTILE版本):
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;
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