[R] matrix of higher order differences

[Ravi Varadhan]

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: rvaradhan at jhmi.edu


-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Petr Savicky
Sent: Wednesday, April 27, 2011 11:01 AM
To: r-help at 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.

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