My apologies in advance for being a bit off-topic, but I could not quell my curiosity.
What might one do with a matrix of all order finite differences? It seems that such a matrix might be related to the Wronskian (its discrete analogue, perhaps). http://en.wikipedia.org/wiki/Wronskian Ravi. ------------------------------------------------------- Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu -----Original Message----- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Petr Savicky Sent: Wednesday, April 27, 2011 11:01 AM To: r-help@r-project.org Subject: Re: [R] matrix of higher order differences On Wed, Apr 27, 2011 at 11:25:42AM +0000, Hans W Borchers wrote: > 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) > } > # ---- Hi. The following avoids the inner loop and it was faster for x of length 100 and 1000. diffmatrix2 <- function(x){ n <- length(x) if (n == 1) return(x) A <- matrix(nrow=n+1, ncol=n) for(i in 1:n){ A[i, seq.int(along=x)] <- x x <- diff(x) } M <- matrix(A, nrow=n, ncol=n) M[upper.tri(M)] <- t(M)[upper.tri(M)] return(M) } Reorganizing an (n+1) x n matrix into an n x n matrix shifts i-th column by (i-1) downwards. In particular, the first row becomes the main diagonal. The initial part of each of the remaining rows becomes a diagonal starting at the first component of the original row. Petr Savicky. ______________________________________________ R-help@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. ______________________________________________ R-help@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.
