通过Householder成功找到Q＆R后无法获得R中的线性回归系数

Question

我正在手动尝试计算回归系数，而不是对数据使用任何默认值 http://people.sc.fsu.edu/~jburkardt/datasets/regression/x31.txt

这是我的代码，它正确地生成了满足 A=QR 的 Q&R。但我无法找到系数作为 Q & R 产生问题的维度。任何专家都可以帮助我吗？当我有适当的问答时，查找系数怎么会出错？

library(xlsx)
data.df<-read.xlsx("regression.xlsx",2,header = F)
#Remove unneccesary Index position
data.df<-data.df[2:5]

#Decomposing
#coefficients [b]=inv(t(X)(Matrix Multiplication)(X))(Matrix Multiplication)t(X)(Matrix Multiplication)y
Y<-as.matrix(data.df[4])
#But note that if you need to express Y=Xb, the coefficient of b_0 must be X_0 
#which is 1
X_0<-as.data.frame(c(1,1,1,1,1,1,1,1,1,1))
X<-(cbind(X_0,data.df[1:3]))
names(X)<-c("X1","X2","X3","X4")
X<-as.matrix(X)
#Create copy for final evaluvations
A<-X  


x1<-as.matrix(X[,1])
deter_x<-sqrt(sum(x1^2))
n=dim(x1)[1]
deter_e1<-as.matrix(c(deter_x,rep(0,n-1)))
v1=x1+deter_e1
#c_i=2/(transpose(v_i)%*%(v_i))
c1<-as.numeric(2/(t(v1)%*%v1))
#H_i = I - c_i*v_i%*%transpose(v_i)
I<-diag(n)
H1<-I-c1*(v1%*%t(v1))
R1<-H1%*%X
#Check R1 and see if it is Upper Triangle Matrix
R1
#We will take rest of the interesting portion of matrix R1.
n=dim(R1)[1]
X<-as.matrix(as.data.frame(cbind(R1[2:n,2],R1[2:n,3],R1[2:n,4])))
x1<-as.matrix(X[,1])
deter_x<-sqrt(sum(x1^2))
n=dim(x1)[1]
deter_e1<-as.matrix(c(deter_x,rep(0,n-1)))
v1=x1-deter_e1
#c_i=2/(transpose(v_i)%*%(v_i))
c1<-as.numeric(2/(t(v1)%*%v1))
#H_i = I - c_i*v_i%*%transpose(v_i)
I<-diag(n)
H2<-I-c1*(v1%*%t(v1))
R2<-H2%*%X

#Check R2 and see if it is Upper Triangle Matrix, if no go for R3
n=dim(R2)[1]
X<-as.matrix(as.data.frame(cbind(R2[2:n,2],R2[2:n,3])))
x1<-as.matrix(X[,1])
deter_x<-sqrt(sum(x1^2))
n=dim(x1)[1]
deter_e1<-as.matrix(c(deter_x,rep(0,n-1)))
v1=x1+deter_e1
#c_i=2/(transpose(v_i)%*%(v_i))
c1<-as.numeric(2/(t(v1)%*%v1))
#H_i = I - c_i*v_i%*%transpose(v_i)
I<-diag(n)
H3<-I-c1*(v1%*%t(v1))
R3<-H3%*%X
R3

#Check R3 and see if it is Upper Triangle Matrix, if no go for R4
n=dim(R3)[1]
X<-as.matrix(as.data.frame(cbind(R3[2:n,2])))
x1<-as.matrix(X[,1])
deter_x<-sqrt(sum(x1^2))
n=dim(x1)[1]
deter_e1<-as.matrix(c(deter_x,rep(0,n-1)))
v1=x1-deter_e1
#c_i=2/(transpose(v_i)%*%(v_i))
c1<-as.numeric(2/(t(v1)%*%v1))
#H_i = I - c_i*v_i%*%transpose(v_i)
I<-diag(n)
H4<-I-c1*(v1%*%t(v1))
R4<-H4%*%X
R4
#As we can see R4 has all values except first element as zero
#Let us replace Matrices iteratively in R1 from R2 to R4 and round it of
R1[2:10,2:4]<-R2
R1[3:10,3:4]<-R3
R1[4:10,4]<-R4
R<-round(R1,5)
R
#Find Complete H1
#Q=H1%*%H2%*%H3%*%H4
H1_COM<-H1
# 
H_temp<-diag(10)
n=dim(H_temp)[1]
dim(H_temp[2:n,2:n])
dim(H2)
H_temp[2:n,2:n]<-H2
H2_COM<-H_temp
H2_COM
H_temp<-diag(10)
n=dim(H_temp)[1]
dim(H_temp[3:n,3:n])
dim(H3)
H_temp[3:n,3:n]<-H3
H3_COM<-H_temp
H3_COM

H_temp<-diag(10)
n=dim(H_temp)[1]
dim(H_temp[4:n,4:n])
dim(H4)
H_temp[4:n,4:n]<-H4
H4_COM<-H_temp
Q=H1_COM%*%H2_COM%*%H3_COM%*%H4_COM
# The following code properly reconstructs A Matrix proving proper Q & R
A=round(Q%*%R)
# When you try to find coefficients using Q&R you will end up in error.
solve(R)%*%t(Q)%*%Y
#Error in solve.default(R) : 'a' (10 x 4) must be square
#So trying to get matrix R without all 0 rows R[1:4,1:4]
solve(R[1:4,1:4])%*%t(Q)%*%Y
#Error in solve(R[1:4, 1:4]) %*% t(Q) : non-conformable arguments
dim(solve(R[1:4,1:4]))
dim(solve(R[1:4,1:4]))
#4 4
dim(t(Q))
#[1] 10 10
dim(Y)
#10  1

Answer 1

我想向您指出我已经回答（相当全面）的这个话题：。您的问题是否会被视为重复将由社区来判断。请注意，在那里的选项中，最后一个是使用我自己用纯 R 代码编写的 QR 分解。

鉴于您用于 QR 分解的玩具代码是正确的（正如您在代码中评论的那样），主要问题在于您的最后几行。

解决方法很简单：

solve(R) %*% (t(Q) %*% Y)[1:4,]

1:4选取单列矩阵的前4个元素t(Q) %*% Y.

如果您查看我的链接答案，您会发现我使用的不是 solve，而是 backsolve，因为这是一个三角方程组。

你还会发现我在可能的时候使用 crossprod 而不是 t 和 %*%。我最近的回答对这两种时尚有全面的讨论。