SQL Server 2005/2008中是否有任何线性回归函数，类似于Linear Regression functions in Oracle？
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。
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
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 xn, zn
insert into #z
select zid, xn,xv- sum(cv*zv)
from #x join #z on xid = zid join #c on czn = zn and cxn = xn where xn = @p and zn<xn group by zid, xn,xv
set @p = @p +1
end
--Loop through each independent variable and estimate the regression parameter by regressing the orthoganal
-- resiuduals on the dependent variable y
while @p>=0
begin
insert into #b
select zn, sum(yv*zv)/ sum(zv*zv)
from #z join
(select yid, yv-isnull(sum(bv*xv),0) yv from #x join #y on xid = yid left join #b on xn=bn group by yid, yv) y
on zid = yid where zn = @p group by zn
set @p = @p-1
end
--The regression parameters
select * from #b
--Actual vs. fit with error
select yid, yv, fit, yv-fit err from #y join
(select xid, sum(xv*bv) fit from #x join #b on xn = bn group by xid) f
on yid = xid
--R Squared
select 1-sum(power(err,2))/sum(power(yv,2)) from
(select yid, yv, fit, yv-fit err from #y join
(select xid, sum(xv*bv) fit from #x join #b on xn = bn group by xid) f
on yid = xid) d
SQL Server中没有线性回归函数。但是要计算数据点x，y对之间的简单线性回归（Y'= bX + A） - 包括相关系数，确定系数（R ^ 2）和标准误差估计（标准偏差）的计算，请执行下列操作：
对于表regression_data与数字列x和y：
declare @total_points int
declare @intercept DECIMAL(38, 10)
declare @slope DECIMAL(38, 10)
declare @r_squared DECIMAL(38, 10)
declare @standard_estimate_error DECIMAL(38, 10)
declare @correlation_coefficient DECIMAL(38, 10)
declare @average_x DECIMAL(38, 10)
declare @average_y DECIMAL(38, 10)
declare @sumX DECIMAL(38, 10)
declare @sumY DECIMAL(38, 10)
declare @sumXX DECIMAL(38, 10)
declare @sumYY DECIMAL(38, 10)
declare @sumXY DECIMAL(38, 10)
declare @Sxx DECIMAL(38, 10)
declare @Syy DECIMAL(38, 10)
declare @Sxy DECIMAL(38, 10)
Select
@total_points = count(*),
@average_x = avg(x),
@average_y = avg(y),
@sumX = sum(x),
@sumY = sum(y),
@sumXX = sum(x*x),
@sumYY = sum(y*y),
@sumXY = sum(x*y)
from regression_data
set @Sxx = @sumXX - (@sumX * @sumX) / @total_points
set @Syy = @sumYY - (@sumY * @sumY) / @total_points
set @Sxy = @sumXY - (@sumX * @sumY) / @total_points
set @correlation_coefficient = @Sxy / SQRT(@Sxx * @Syy)
set @slope = (@total_points * @sumXY - @sumX * @sumY) / (@total_points * @sumXX - power(@sumX,2))
set @intercept = @average_y - (@total_points * @sumXY - @sumX * @sumY) / (@total_points * @sumXX - power(@sumX,2)) * @average_x
set @r_squared = (@intercept * @sumY + @slope * @sumXY - power(@sumY,2) / @total_points) / (@sumYY - power(@sumY,2) / @total_points)
-- calculate standard_estimate_error (standard deviation)
Select
@standard_estimate_error = sqrt(sum(power(y - (@slope * x + @intercept),2)) / @total_points)
From regression_data
要添加到@ icc97答案，我已经包含了斜率和截距的加权版本。如果值都是常量，则斜率将为NULL（使用适当的设置SET ARITHABORT OFF; SET ANSI_WARNINGS OFF;），并且需要通过coalesce（）替换0。
这是用SQL编写的解决方案：
with d as (select segment,w,x,y from somedatasource)
select segment,
avg(y) - avg(x) *
((count(*) * sum(x*y)) - (sum(x)*sum(y)))/
((count(*) * sum(x*x)) - (Sum(x)*Sum(x))) as intercept,
((count(*) * sum(x*y)) - (sum(x)*sum(y)))/
((count(*) * sum(x*x)) - (sum(x)*sum(x))) AS slope,
avg(y) - ((avg(x*y) - avg(x)*avg(y))/var_samp(X)) * avg(x) as interceptUnstable,
(avg(x*y) - avg(x)*avg(y))/var_samp(X) as slopeUnstable,
(Avg(x * y) - Avg(x) * Avg(y)) / (stddev_pop(x) * stddev_pop(y)) as correlationUnstable,
(sum(y*w)/sum(w)) - (sum(w*x)/sum(w)) *
((sum(w)*sum(x*y*w)) - (sum(x*w)*sum(y*w)))/
((sum(w)*sum(x*x*w)) - (sum(x*w)*sum(x*w))) as wIntercept,
((sum(w)*sum(x*y*w)) - (sum(x*w)*sum(y*w)))/
((sum(w)*sum(x*x*w)) - (sum(x*w)*sum(x*w))) as wSlope,
(count(*) * sum(x * y) - sum(x) * sum(y)) / (sqrt(count(*) * sum(x * x) - sum(x) * sum(x))
* sqrt(count(*) * sum(y * y) - sum(y) * sum(y))) as correlation,
count(*) as n
from d where x is not null and y is not null group by segment
其中w是重量。我对R进行了双重检查以确认结果。可能需要将数据从somedatasource转换为浮点数。我包含了不稳定的版本来警告你不要那些。 （特别感谢斯蒂芬在另一个答案。）
请记住，相关性是数据点x和y的相关性，而不是预测的相关性。
我已经翻译了Excel中功能预测中使用的线性回归函数，并创建了一个返回a，b和预测的SQL函数。您可以在excel帮助中查看完整的teorical解释，以获得FORECAST功能。您需要创建表数据类型XYFloatType的所有内容：
CREATE TYPE [dbo].[XYFloatType]
AS TABLE(
[X] FLOAT,
[Y] FLOAT)
然后编写以下函数：
/*
-- =============================================
-- Author: Me :)
-- Create date: Today :)
-- Description: (Copied Excel help):
--Calculates, or predicts, a future value by using existing values.
The predicted value is a y-value for a given x-value.
The known values are existing x-values and y-values, and the new value is predicted by using linear regression.
You can use this function to predict future sales, inventory requirements, or consumer trends.
-- =============================================
*/
CREATE FUNCTION dbo.FN_GetLinearRegressionForcast
(@PtXYData as XYFloatType READONLY ,@PnFuturePointint)
RETURNS @ABDData TABLE( a FLOAT, b FLOAT, Forecast FLOAT)
AS
BEGIN
DECLARE @LnAvX Float
,@LnAvY Float
,@LnB Float
,@LnA Float
,@LnForeCast Float
Select @LnAvX = AVG([X])
,@LnAvY = AVG([Y])
FROM @PtXYData;
SELECT @LnB = SUM ( ([X]-@LnAvX)*([Y]-@LnAvY) ) / SUM (POWER([X]-@LnAvX,2))
FROM @PtXYData;
SET @LnA = @LnAvY - @LnB * @LnAvX;
SET @LnForeCast = @LnA + @LnB * @PnFuturePoint;
INSERT INTO @ABDData ([A],[B],[Forecast]) VALUES (@LnA,@LnB,@LnForeCast)
RETURN
END
/*
your tests:
(I used the same values that are in the excel help)
DECLARE @t XYFloatType
INSERT @t VALUES(20,6),(28,7),(31,9),(38,15),(40,21) -- x and y values
SELECT *, A+B*30 [Prueba]FROM dbo.FN_GetLinearRegressionForcast@t,30);
*/
这里是一个函数，它采用一种类型的表类型：table（Y float，X double），它被称为XYDoubleType，并假设我们的线性函数的形式为AX + B.它返回A和B一个Table列以防万一你想把它放在一个连接或什么的
CREATE FUNCTION FN_GetABForData(
@XYData as XYDoubleType READONLY
) RETURNS @ABData TABLE(
A FLOAT,
B FLOAT,
Rsquare FLOAT )
AS
BEGIN
DECLARE @sx FLOAT, @sy FLOAT
DECLARE @sxx FLOAT,@syy FLOAT, @sxy FLOAT,@sxsy FLOAT, @sxsx FLOAT, @sysy FLOAT
DECLARE @n FLOAT, @A FLOAT, @B FLOAT, @Rsq FLOAT
SELECT @sx =SUM(D.X) ,@sy =SUM(D.Y), @sxx=SUM(D.X*D.X),@syy=SUM(D.Y*D.Y),
@sxy =SUM(D.X*D.Y),@n =COUNT(*)
From @XYData D
SET @sxsx =@sx*@sx
SET @sxsy =@sx*@sy
SET @sysy = @sy*@sy
SET @A = (@n*@sxy -@sxsy)/(@n*@sxx -@sxsx)
SET @B = @sy/@n - @A*@sx/@n
SET @Rsq = POWER((@n*@sxy -@sxsy),2)/((@n*@sxx-@sxsx)*(@n*@syy -@sysy))
INSERT INTO @ABData (A,B,Rsquare) VALUES(@A,@B,@Rsq)
RETURN
END
我希望以下答案有助于理解某些解决方案的来源。我将用一个简单的例子来说明它，但只要你知道如何使用索引表示法或矩阵，对许多变量的泛化在理论上是直截了当的。为了实现超过3个变量的解决方案，你可以使用Gram-Schmidt（参见上面的Colin Campbell的答案）或其他矩阵求逆算法。
由于我们需要的所有函数都是方差，协方差，平均值，求和等都是SQL中的聚合函数，因此可以轻松实现解决方案。我在HIVE中已经完成了对Logistic模型得分的线性校准 - 在许多优点中，一个是你可以完全在HIVE中运行，而无需从某些脚本语言中退出。
您的数据点由i索引的数据模型（x_1，x_2，y）是
y（x_1，x_2）= m_1 * x_1 + m_2 * x_2 + c
该模型看起来是“线性的”，但不一定是，例如，x_2可以是x_1的任何非线性函数，只要它在其中没有自由参数，例如， x_2 = Sinh（3 *（x_1）^ 2 + 42）。即使x_2是“仅”x_2并且模型是线性的，回归问题也不是。只有当您确定问题是找到参数m_1，m_2，c以便它们最小化L2错误时，才会出现线性回归问题。
L2误差是sum_i（（y [i] -f（x_1 [i]，x_2 [i]））^ 2）。最小化这个w.r.t. 3个参数（设置偏导数w.r.t.每个参数= 0）产生3个未知数的3个线性方程。这些方程在参数中是LINEAR（这就是使其成为线性回归的方法）并且可以通过分析求解。对于简单模型（1个变量，线性模型，因此有两个参数）这样做是直截了当且具有指导性的。对误差向量空间的非欧几里德度量范数的推广是直截了当的，对角线特殊情况相当于使用“权重”。
回到我们的模型中有两个变量：
y = m_1 * x_1 + m_2 * x_2 + c
取期望值=>
= m_1 * + m_2 * + c（0）
现在采取协方差w.r.t. x_1和x_2，并使用cov（x，x）= var（x）：
（y，x_1）= m_1 * var（x_1）+ m_2 * covar（x_2，x_1）（1）
（y，x_2）= m_1 * covar（x_1，x_2）+ m_2 * var（x_2）（2）
这是两个未知数的方程，您可以通过反转2X2矩阵来求解。
矩阵形式：...可以倒置以产生......其中
it = var（x_1）* var（x_2） - covar（x_1，x_2）^ 2
（哦，barf，到底是什么“声望点？如果你想看方程，请给我一些。”
在任何情况下，既然你有m1和m2的封闭形式，你可以求解（0）c。
我检查了上面的分析解决方案到Excel的求解器，得到了具有高斯噪声的二次方，剩余误差与6位有效数字一致。
如果你想在SQL中做大约20行的离散傅里叶变换，请联系我。