Re: [R] Matrix indexing in a loop

From: Pontarelli, Brett <>
Date: Sat 18 Feb 2006 - 10:04:18 EST

Do you have to use a loop? The following function should do what you want for the 1st order:

rook = function(Y) {

	rsub = function(Z) {
		X = matrix(0,nrow(Z),ncol(Z));
		X[1:(N-1),1:M] = X[1:(N-1),1:M] + Z[2:N,1:M];
		X[2:N,1:M] = X[2:N,1:M] + Z[1:(N-1),1:M];
		X[1:N,1:(M-1)] = X[1:N,1:(M-1)] + Z[1:N,2:M];
		X[1:N,2:M] = X[1:N,2:M] + Z[1:N,1:(M-1)];

I'm not sure I understand how the higher orders work. For example, an interior element for the 1st order is always divided by 4. Is an interior element for a 3rd order divided by 4 or 8 or something else? Also, how are you implementing your 3D matrices?


-----Original Message-----
From: [] On Behalf Of Mills, Jason Sent: Friday, February 17, 2006 1:36 PM
Subject: [R] Matrix indexing in a loop

How do you specify matrix location a[i,j] (or a[i-1,j], etc.) in a "for" loop?

I am looking for a flexible method of indexing neighbors over a series of lags (1,2,3...) and I may wish to extend this method to 3D arrays.


Data matrix
> fun

     [,1] [,2] [,3]

[1,]    1    5    9
[2,]    2    6   10
[3,]    3    7   11
[4,]    4    8   12

For each element a[i,j] in "fun", sum the 1st order (Rook's) neighbors:





Then divide by the number of elements included as neighbors-- this number depends on the location of a[i,j] in the matrix.

Insert the product of the neighbor calculation for each a[i,j] into the corresponding position b[i,j] in an empty matrix with the same dimensions as "fun".

For example, element [2,2] in "fun" should yield element [2,2] in a new matrix equal to 24/4=6. Of course, element [1,1] in the new matrix should be the product of only two numbers.


J. Mills

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