From: Hans W Borchers <hwborchers_at_googlemail.com>

Date: Wed, 27 Apr 2011 11:25:42 +0000

}

# ----

R-help_at_r-project.org mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Wed 27 Apr 2011 - 11:28:51 GMT

Date: Wed, 27 Apr 2011 11:25:42 +0000

Jeroen Ooms <jeroenooms <at> gmail.com> writes:

*>
*

> Is there an easy way to turn a vector of length n into an n by n matrix, in

*> which the diagonal equals the vector, the first off diagonal equals the
**> first order differences, the second... etc. I.e. to do this more
**> efficiently:
**>
**> diffmatrix <- function(x){
**> n <- length(x);
**> M <- diag(x);
**> for(i in 1:(n-1)){
**> differences <- diff(x, dif=i);
**> for(j in 1:length(differences)){
**> M[j,i+j] <- differences[j]
**> }
**> }
**> M[lower.tri(M)] <- t(M)[lower.tri(M)];
**> return(M);
**> }
**>
**> x <- c(1,2,3,5,7,11,13,17,19);
**> diffmatrix(x);
**>
*

I do not know whether you will call the appended version more elegant, but at least it is much faster -- up to ten times for length(x) = 1000, i.e. less than 2 secs for generating and filling a 1000-by-1000 matrix. I also considered col(), row() indexing:

M[col(M) == row(M) + k] <- x

Surprisingly (for me), this makes it even slower than your version with a double 'for' loop.

- Hans Werner

# ----

diffmatrix <- function(x){

n <- length(x) if (n == 1) return(x) M <- diag(x) for(i in 1:(n-1)){ x <- diff(x) # use 'diff' in a loop for(j in 1:(n-i)){ # length is known M[j, i+j] <- x[j] # and reuse x } } M[lower.tri(M)] <- t(M)[lower.tri(M)] return(M)

}

# ----

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