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
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) ;
其他回答
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。
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)
试试下面的逻辑来找出中位数:
考虑一个包含以下数字的表格: 1、1、2、3、4、5所示
中位数是2.5
with tempa as
(
select num,count(num) over() as Cnt,
row_number() over (order by num) as Rnum
from temp),
tempb as
(
select round(cnt/2) as ref_value
from tempa where mod(cnt,2)<>0
union all
select round(cnt/2) from tempa where mod(cnt,2)=0
union all
select round(cnt/2+1)
from tempa where mod(cnt,2)=0
)
select avg(num) from tempa
where rnum in (select * from tempb);
这段代码有点长,但很容易理解
medii是有列val的表,它有数据集, Smedi是一个cte,它将列idx作为行号,val作为medi表中的'val',该表是升序排序的。 这是基本的数学,如果行号是奇数,那么它的中值来自smedi。 当它是偶数时,它是中间两个值的平均值。
with smedi(idx,vals) as(
select ROW_NUMBER() over(order by val),val from medi
)
select (case
when (select count(*) from medi)%2!=0 then (select vals from smedi where (((select count(*) from medi)/2))=idx)
else (select avg(vals) from smedi where idx in ((select count(*)/2 from medi),(select (count(*)/2)+1 from medi)))
end)