Multiple regression analysis in R using QR decomposition

Question

I am trying to write a function for solving multiple regression using QR decomposition. Input: y vector and X matrix; output: b, e, R^2. So far I`ve got this and am terribly stuck; I think I have made everything way too complicated:

QR.regression <- function(y, X) {
X <- as.matrix(X)
y <- as.vector(y)
p <- as.integer(ncol(X))
if (is.na(p)) stop("ncol(X) is invalid")
n <- as.integer(nrow(X))
if (is.na(n)) stop("nrow(X) is invalid")
nr <- length(y)
nc <- NCOL(X)

# Householder 
for (j in seq_len(nc)) {
id <- seq.int(j, nr)
sigma <- sum(X[id, j]^2)
s <- sqrt(sigma)
diag_ej <- X[j, j]
gamma <- 1.0 / (sigma + abs(s * diag_ej))
kappa <- if (diag_ej < 0) s else -s
X[j,j] <- X[j, j] - kappa
if (j < nc)
for (k in seq.int(j+1, nc)) {
yPrime <- sum(X[id,j] * X[id,k]) * gamma
X[id,k] <- X[id,k] - X[id,j] * yPrime
}
yPrime <- sum(X[id,j] * y[id]) * gamma
y[id] <- y[id] - X[id,j] * yPrime
X[j,j] <- kappa
} # end of Householder transformation

rss <- sum(y[seq.int(nc+1, nr)]^2)  # residuals sum of squares
e <- rss/nr
e <- mean(residuals(QR.regression)^2)
beta <- solve(t(X) %*% X, t(X) %*% y)
for (i in seq_len(ncol(X))) # set zeros in the lower triangular side of X
X[seq.int(i+1, nr),i] <- 0
Rsq <- (X[1:nc,1:nc])^2
return(list(Rsq=Rsq, y = y, beta = beta, e = e))
}


UPDATE:
my.QR <- function(y, X) {
X <- as.matrix(X)
y <- as.vector(y)
p <- as.integer(ncol(X))
if (is.na(p)) stop("ncol(X) is invalid")
n <- as.integer(nrow(X))
if (is.na(n)) stop("nrow(X) is invalid")
qr.X <- qr(X)
b <- solve(t(X) %*% X, t(X) %*% y)
e <- as.vector(y - X %*% beta) #e
R2 <- (X[1:p, 1:p])^2
return(list(b = b, e= e, R2 = R2 ))
}

X <- matrix(c(1,2,3,4,5,6), nrow = 2, ncol = 3)
y <- c(1,2,3,4)
my.QR(X, y)

Bittern

By "think I made this way too complicated" perhaps you mean: why didn't I just use lm(y~X)? – Grasp

I am allowed to use lm(y~X) only to prove that my function provided correct answers. Untill then I need to make mine work. – Bittern

@ZheyuanLi thank you, I was already reading that post. The task is rather vague, but I think I should better use some of the built-in functions. Also: my Householder is off since I keep on getting this error message: "Error in X[id, j] : subscript out of bounds". Any help would be very much appreciated, seriously. – Bittern

Answer 1

It all depends on how much of R's built-in facility you are allowed to use to solve this problem. I already know that lm is not allowed, so here is the rest of the story.

---

If you are allowed to use any other routines than lm

Then you can simply use lm.fit, .lm.fit or lsfit for QR-based ordinary least squares solving.

lm.fit(X, y)
.lm.fit(X, y)
lsfit(X, y, intercept = FALSE)

Among those, .lm.fit is the most light-weighed, while lm.fit and lsfit are pretty similar. The following is what we can do via .lm.fit:

f1 <- function (X, y) {
  z <- .lm.fit(X, y)
  RSS <- crossprod(z$residuals)[1]
  TSS <- crossprod(y - mean(y))[1]
  R2 <- 1 - RSS / TSS
  list(coefficients = z$coefficients, residuals = z$residuals, R2 = R2)
  }

In the question by your fellow classmate: Toy R function for solving ordinary least squares by singular value decomposition, I have already used this to check the correctness of SVD approach.

---

If you are not allowed to use R's built-in QR factorization routine qr.default

If .lm.fit is not allowed, but qr.default is, then it is also not that complicated.

f2 <- function (X, y) {
  ## QR factorization `X = QR`
  QR <- qr.default(X)
  ## After rotation of `X` and `y`, solve upper triangular system `Rb = Q'y` 
  b <- backsolve(QR$qr, qr.qty(QR, y))
  ## residuals
  e <- as.numeric(y - X %*% b)
  ## R-squared
  RSS <- crossprod(e)[1]
  TSS <- crossprod(y - mean(y))[1]
  R2 <- 1 - RSS / TSS
  ## multiple return
  list(coefficients = b, residuals = e, R2 = R2)
  }

If you further want variance-covariance of estimated coefficients, follow How to calculate variance of least squares estimator using QR decomposition in R?.

---

If you are not even allowed to use qr.default

Then we have to write QR decomposition ourselves. Writing a Householder QR factorization function in R code is giving this.

Using the function myqr there, we can write

f3 <- function (X, y) {
  ## our own QR factorization
  ## complete Q factor is not required
  QR <- myqr(X, complete = FALSE)
  Q <- QR$Q
  R <- QR$R
  ## rotation of `y`
  Qty <- as.numeric(crossprod(Q, y))
  ## solving upper triangular system
  b <- backsolve(R, Qty)
  ## residuals
  e <- as.numeric(y - X %*% b)
  ## R-squared
  RSS <- crossprod(e)[1]
  TSS <- crossprod(y - mean(y))[1]
  R2 <- 1 - RSS / TSS
  ## multiple return
  list(coefficients = b, residuals = e, R2 = R2)
  }

f3 is not extremely efficient, as we have formed Q explicitly, even though it is the thin-Q factor. In principle, we should rotate y along with the QR factorization of X, thus Q needs not be formed.

---

If you want to fix your existing code

This requires some debugging effort so would take some time. I will make another answer regarding this later.

Cinquain