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
这是我能想到的求中位数的最优解。示例中的名称基于Justin示例。确保表有索引 销售。SalesOrderHeader以索引列CustomerId和TotalDue的顺序存在。
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性能最好,所以我比较了我的方法Justin Grants和Jeff Atwoods,在一个批量中运行基于这三种方法的查询,每个查询的批量成本为:
没有指数:
我的30% Justin Grants 13% Jeff Atwoods 58%
还有index
我的3%。 Justin Grants 10% Jeff Atwoods 87%
I tried to see how well the queries scale if you have index by creating more data from around 14 000 rows by a factor of 2 up to 512 which means in the end around 7,2 millions rows. Note I made sure CustomeId field where unique for each time I did a single copy, so the proportion of rows compared to unique instance of CustomerId was kept constant. While I was doing this I ran executions where I rebuilt index afterwards, and I noticed the results stabilized at around a factor of 128 with the data I had to these values:
我的3%。 贾斯汀·格兰特5% Jeff Atwoods 92%
我想知道,在保持惟一CustomerId不变的情况下,扩展行数会如何影响性能,因此我设置了一个新的测试,在其中执行了上述操作。现在,批成本比率并没有稳定下来,而是不断分化,每个CustomerId平均大约有20行,最后每个这样唯一的Id大约有10000行。数字如下:
我的4% 贾斯汀60% 杰夫斯35%
通过比较结果,我确保我正确地实现了每个方法。 我的结论是,只要索引存在,我使用的方法通常更快。还要注意,本文针对这个特定问题推荐使用这种方法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在辅助表中持久化计数信息。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行计数的信息,当然您也可以简单地存储中值。
其他回答
这是我能想到的求中位数的最优解。示例中的名称基于Justin示例。确保表有索引 销售。SalesOrderHeader以索引列CustomerId和TotalDue的顺序存在。
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性能最好,所以我比较了我的方法Justin Grants和Jeff Atwoods,在一个批量中运行基于这三种方法的查询,每个查询的批量成本为:
没有指数:
我的30% Justin Grants 13% Jeff Atwoods 58%
还有index
我的3%。 Justin Grants 10% Jeff Atwoods 87%
I tried to see how well the queries scale if you have index by creating more data from around 14 000 rows by a factor of 2 up to 512 which means in the end around 7,2 millions rows. Note I made sure CustomeId field where unique for each time I did a single copy, so the proportion of rows compared to unique instance of CustomerId was kept constant. While I was doing this I ran executions where I rebuilt index afterwards, and I noticed the results stabilized at around a factor of 128 with the data I had to these values:
我的3%。 贾斯汀·格兰特5% Jeff Atwoods 92%
我想知道,在保持惟一CustomerId不变的情况下,扩展行数会如何影响性能,因此我设置了一个新的测试,在其中执行了上述操作。现在,批成本比率并没有稳定下来,而是不断分化,每个CustomerId平均大约有20行,最后每个这样唯一的Id大约有10000行。数字如下:
我的4% 贾斯汀60% 杰夫斯35%
通过比较结果,我确保我正确地实现了每个方法。 我的结论是,只要索引存在,我使用的方法通常更快。还要注意,本文针对这个特定问题推荐使用这种方法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在辅助表中持久化计数信息。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行计数的信息,当然您也可以简单地存储中值。
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)
对于像我这样正在学习基础知识的新手来说,我个人觉得这个例子更容易理解,因为它更容易理解到底发生了什么以及中值来自哪里……
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
不过,对上面的一些代码绝对敬畏!!
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。
虽然Justin grant的解决方案看起来很可靠,但我发现当您在给定的分区键中有许多重复值时,ASC重复值的行号最终会不按顺序排列,因此它们不能正确对齐。
以下是我的研究结果的一个片段:
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
我使用Justin的代码作为这个解决方案的基础。尽管考虑到使用多个派生表效率不高,但它确实解决了我遇到的行排序问题。任何改进都会受到欢迎,因为我在T-SQL方面不是那么有经验。
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