You're right I had N and M defined outside of the function and rook and rsub
were picking up on that. The following is a bit better and more cleaned up
version with i-th order option:
rook = function(Y,i) {
N = nrow(Y);
M = ncol(Y);
rsub = function(Z,i) {
X = matrix(0,N,M);
X[1:(N-i),] = X[1:(N-i),] + Z[(1+i):N,];
X[(1+i):N,] = X[(1+i):N,] + Z[1:(N-i),];
X[,1:(M-i)] = X[,1:(M-i)] + Z[,(1+i):M];
X[,(1+i):M] = X[,(1+i):M] + Z[,1:(M-i)];
return(X);
}
return(rsub(Y,i)/rsub(matrix(1,N,M),i));
}
Notice that the variable "i" can be passed any value even one that causes an
error.
--Brett
-----Original Message-----
From: Mills, Jason [mailto:[EMAIL PROTECTED]
Sent: Friday, February 17, 2006 4:11 PM
To: Pontarelli, Brett
Subject: RE: [R] Matrix indexing in a loop
Hi Brett, thanks for the tip.
I tried this function on my sample matrix and got the error message "Error in
rsub(fun) : object "N" not found". Your code looks like it should work, so I
must be doing some wrong. I will continue to experiment.
As for the neighbor pattern, the convention follows the rules of chess.
I would consider a 2nd order rook case to include only 4 elements:
a[i-2,j]
a[i+2,j]
a[i,j-2]
a[i,j+2]
Even 3rd order Rook would still only include 4 elements. As another example, a
Queen neighborhood includes 8 elements, independent of the lag order.
I haven't started using 3D arrays yet. I am attempting spatio-temporal
analysis and I have thought about representing my landscape (two
dimensions) over time (the third dimension). For now, I'm trying to get a
handle on working in two dimensions.
Thanks.
Jason
-----Original Message-----
From: Pontarelli, Brett [mailto:[EMAIL PROTECTED]
Sent: Friday, February 17, 2006 4:04 PM
To: Mills, Jason; [email protected]
Subject: RE: [R] Matrix indexing in a loop
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)];
return(X);
}
return(rsub(Y)/rsub(matrix(1,nrow(Y),ncol(Y))));
}
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?
--Brett
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Mills, Jason
Sent: Friday, February 17, 2006 1:36 PM
To: [email protected]
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.
Example:
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:
a[i-1,j]
a[i+1,j]
a[i,j-1]
a[i,j+1]
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.
Thanks
J. Mills
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