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
通常情况下,我们不仅需要为整个表计算Median,还需要为与某个ID相关的聚合计算Median。换句话说,计算表中每个ID的中位数,其中每个ID有许多记录。(基于@gdoron编辑的解决方案:性能良好,适用于许多SQL)
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;
希望能有所帮助。
其他回答
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
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
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)
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)
我最初的回答是:
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
这将使您一举获得中位数和四分位范围。如果你真的只想要一行作为中值,那么取消注释where子句。
当你把它放入解释计划时,60%的工作是对数据进行排序,这在计算像这样的位置依赖统计数据时是不可避免的。
我修改了答案,以遵循Robert Ševčík-Robajz在下面的评论中提出的优秀建议:
;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
当您有偶数个数据项时,这应该计算正确的中位数和百分比值。同样,如果您只想要中位数而不是整个百分位数分布,请取消最后的where子句的注释。