SQL Server中是否有任何线性回归函数？
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SQL Server 2005/2008中是否有任何线性回归函数，类似于Linear Regression functions in Oracle？
答案
据我所知，没有。写一个很简单。以下为y = Alpha + Beta * x + epsilon提供恒定的alpha和斜率beta：
-- test data (GroupIDs 1, 2 normal regressions, 3, 4 = no variance)
WITH some_table(GroupID, x, y) AS
( SELECT 1, 1, 1 UNION SELECT 1, 2, 2 UNION SELECT 1, 3, 1.3
UNION SELECT 1, 4, 3.75 UNION SELECT 1, 5, 2.25 UNION SELECT 2, 95, 85
UNION SELECT 2, 85, 95 UNION SELECT 2, 80, 70 UNION SELECT 2, 70, 65
UNION SELECT 2, 60, 70 UNION SELECT 3, 1, 2 UNION SELECT 3, 1, 3
UNION SELECT 4, 1, 2 UNION SELECT 4, 2, 2),
-- linear regression query
/*WITH*/ mean_estimates AS
( SELECT GroupID
,AVG(x * 1.) AS xmean
,AVG(y * 1.) AS ymean
FROM some_table
GROUP BY GroupID
),
stdev_estimates AS
( SELECT pd.GroupID
-- T-SQL STDEV() implementation is not numerically stable
,CASE SUM(SQUARE(x - xmean)) WHEN 0 THEN 1
ELSE SQRT(SUM(SQUARE(x - xmean)) / (COUNT(*) - 1)) END AS xstdev
, SQRT(SUM(SQUARE(y - ymean)) / (COUNT(*) - 1)) AS ystdev
FROM some_table pd
INNER JOIN mean_estimates pm ON pm.GroupID = pd.GroupID
GROUP BY pd.GroupID, pm.xmean, pm.ymean
),
standardized_data AS -- increases numerical stability
( SELECT pd.GroupID
,(x - xmean) / xstdev AS xstd
,CASE ystdev WHEN 0 THEN 0 ELSE (y - ymean) / ystdev END AS ystd
FROM some_table pd
INNER JOIN stdev_estimates ps ON ps.GroupID = pd.GroupID
INNER JOIN mean_estimates pm ON pm.GroupID = pd.GroupID
),
standardized_beta_estimates AS
( SELECT GroupID
,CASE WHEN SUM(xstd * xstd) = 0 THEN 0
ELSE SUM(xstd * ystd) / (COUNT(*) - 1) END AS betastd
FROM standardized_data pd
GROUP BY GroupID
)
SELECT pb.GroupID
,ymean - xmean * betastd * ystdev / xstdev AS Alpha
,betastd * ystdev / xstdev AS Beta
FROM standardized_beta_estimates pb
INNER JOIN stdev_estimates ps ON ps.GroupID = pb.GroupID
INNER JOIN mean_estimates pm ON pm.GroupID = pb.GroupID
这里GroupID用于显示如何按源数据表中的某个值进行分组。如果您只想要表中所有数据的统计信息（不是特定的子组），则可以删除它和连接。为了清楚起见，我使用了WITH声明。作为替代方案，您可以使用子查询。请注意表中使用的数据类型的精度，因为如果精度相对于数据不够高，数值稳定性会迅速恶化。
编辑:(在评论中回答彼得关于R2等其他统计数据的问题）
您可以使用相同的技术轻松计算其他统计信息。这是一个具有R2，相关性和样本协方差的版本：
-- test data (GroupIDs 1, 2 normal regressions, 3, 4 = no variance)
WITH some_table(GroupID, x, y) AS
( SELECT 1, 1, 1 UNION SELECT 1, 2, 2 UNION SELECT 1, 3, 1.3
UNION SELECT 1, 4, 3.75 UNION SELECT 1, 5, 2.25 UNION SELECT 2, 95, 85
UNION SELECT 2, 85, 95 UNION SELECT 2, 80, 70 UNION SELECT 2, 70, 65
UNION SELECT 2, 60, 70 UNION SELECT 3, 1, 2 UNION SELECT 3, 1, 3
UNION SELECT 4, 1, 2 UNION SELECT 4, 2, 2),
-- linear regression query
/*WITH*/ mean_estimates AS
( SELECT GroupID
,AVG(x * 1.) AS xmean
,AVG(y * 1.) AS ymean
FROM some_table pd
GROUP BY GroupID
),
stdev_estimates AS
( SELECT pd.GroupID
-- T-SQL STDEV() implementation is not numerically stable
,CASE SUM(SQUARE(x - xmean)) WHEN 0 THEN 1
ELSE SQRT(SUM(SQUARE(x - xmean)) / (COUNT(*) - 1)) END AS xstdev
, SQRT(SUM(SQUARE(y - ymean)) / (COUNT(*) - 1)) AS ystdev
FROM some_table pd
INNER JOIN mean_estimates pm ON pm.GroupID = pd.GroupID
GROUP BY pd.GroupID, pm.xmean, pm.ymean
),
standardized_data AS -- increases numerical stability
( SELECT pd.GroupID
,(x - xmean) / xstdev AS xstd
,CASE ystdev WHEN 0 THEN 0 ELSE (y - ymean) / ystdev END AS ystd
FROM some_table pd
INNER JOIN stdev_estimates ps ON ps.GroupID = pd.GroupID
INNER JOIN mean_estimates pm ON pm.GroupID = pd.GroupID
),
standardized_beta_estimates AS
( SELECT GroupID
,CASE WHEN SUM(xstd * xstd) = 0 THEN 0
ELSE SUM(xstd * ystd) / (COUNT(*) - 1) END AS betastd
FROM standardized_data
GROUP BY GroupID
)
SELECT pb.GroupID
,ymean - xmean * betastd * ystdev / xstdev AS Alpha
,betastd * ystdev / xstdev AS Beta
,CASE ystdev WHEN 0 THEN 1 ELSE betastd * betastd END AS R2
,betastd AS Correl
,betastd * xstdev * ystdev AS Covar
FROM standardized_beta_estimates pb
INNER JOIN stdev_estimates ps ON ps.GroupID = pb.GroupID
INNER JOIN mean_estimates pm ON pm.GroupID = pb.GroupID
编辑2通过标准化数据（而不是仅居中）和通过用STDEV替换numerical stability issues来提高数值稳定性。对我来说，目前的实施似乎是稳定性和复杂性之间的最佳平衡。我可以通过用数值稳定的在线算法替换我的标准偏差来提高稳定性，但这会使实现变得非常复杂（并且减慢它）。类似地，使用例如Kahan（-Babuška-Neumaier）对SUM和AVG的补偿似乎在有限的测试中表现得更好，但使查询更加复杂。只要我不知道T-SQL如何实现SUM和AVG（例如它可能已经使用成对求和），我不能保证这些修改总能提高准确性。
另一答案
这是一种基于blog post on Linear Regression in T-SQL的替代方法，它使用以下等式：
博客中的SQL建议虽然使用了游标。这是我用过的forum answer的美化版本：
table
-----
X (numeric)
Y (numeric)
/**
* m = (nSxy - SxSy) / (nSxx - SxSx)
* b = Ay - (Ax * m)
* N.B. S = Sum, A = Mean
*/
DECLARE @n INT
SELECT @n = COUNT(*) FROM table
SELECT (@n * SUM(X*Y) - SUM(X) * SUM(Y)) / (@n * SUM(X*X) - SUM(X) * SUM(X)) AS M,
AVG(Y) - AVG(X) *
(@n * SUM(X*Y) - SUM(X) * SUM(Y)) / (@n * SUM(X*X) - SUM(X) * SUM(X)) AS B
FROM table
另一答案
我实际上使用Gram-Schmidt正交化编写了一个SQL例程。它以及其他机器学习和预测程序可在sqldatamine.blogspot.com获得
根据Brad Larson的建议，我在这里添加了代码，而不仅仅是将用户引导到我的博客。这会产生与Excel中的linest函数相同的结果。我的主要资料来源是Hastie，Tibshirni和Friedman的“统计学习元素”（2008）。
--Create a table of data
create table #rawdata (id int,area float, rooms float, odd float, price float)
insert into #rawdata select 1, 2201,3,1,400
insert into #rawdata select 2, 1600,3,0,330
insert into #rawdata select 3, 2400,3,1,369
insert into #rawdata select 4, 1416,2,1,232
insert into #rawdata select 5, 3000,4,0,540
--Insert the data into x & y vectors
select id xid, 0 xn,1 xv into #x from #rawdata
union all
select id, 1,rooms from #rawdata
union all
select id, 2,area from #rawdata
union all
select id, 3,odd from #rawdata
select id yid, 0 yn, price yv into #y from #rawdata
--create a residuals table and insert the intercept (1)
create table #z (zid int, zn int, zv float)
insert into #z select id , 0 zn,1 zv from #rawdata
--create a table for the orthoganal (#c) & regression(#b) parameters
create table #c(cxn int, czn int, cv float)
create table #b(bn int, bv float)
--@p is the number of independent variables including the intercept (@p = 0)
declare @p int
set @p = 1
--Loop through each independent variable and estimate the orthagonal parameter (#c)
-- then estimate the residuals and insert into the residuals table (#z)
while @p <= (select max(xn) from #x)
begin
insert into #c
select xn cxn, zn czn, sum(xv*zv)/sum(zv*zv) cv
from #x join #z on xid = zid where zn = @p-1 and xn>zn group by x以上是关于SQL Server中是否有任何线性回归函数？的主要内容，如果未能解决你的问题，请参考以下文章