# [R] matrix of higher order differences

peter dalgaard pdalgd at gmail.com
Wed Apr 27 22:58:43 CEST 2011

```On Apr 27, 2011, at 21:34 , Ravi Varadhan wrote:

> 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

Not quite, I think. This is one function at different values of x, the Wronskian is about n different functions.

Tables of higher-order differences were used fundamentally for interpolation and error detection in tables of function values (remember those?), but rarely computed to the full extent - usually only until the effects of truncation set in and the differences start alternating in sign.

>
> Ravi.
> -------------------------------------------------------
> Assistant Professor,
> Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University
>
> Ph. (410) 502-2619
>
>
> -----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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