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 [[alternative HTML version deleted]]
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: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at
stat.math.ethz.ch] On Behalf Of Mills, Jason
Sent: Friday, February 17, 2006 1:36 PM
To: r-help at stat.math.ethz.ch
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
[[alternative HTML version deleted]]
______________________________________________
R-help at stat.math.ethz.ch mailing list
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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:Jason.Mills at afhe.ualberta.ca]
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:bpontar at amazon.com]
Sent: Friday, February 17, 2006 4:04 PM
To: Mills, Jason; r-help at stat.math.ethz.ch
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: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Mills, Jason
Sent: Friday, February 17, 2006 1:36 PM
To: r-help at stat.math.ethz.ch
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
[[alternative HTML version deleted]]
______________________________________________
R-help at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
For 2d here is a solution based on zoo. It turns the matrix into
a time series and lags it forwards and backwards and does the
same for its transpose in order to avoid index machinations.
The function is called rook2 and it first defines three local
functions, one that converts NAs to zero, one that does
a lag using na.pad = TRUE and one to invoke Lag and
add up the lags:
library(zoo)
rook2 <- function(x, i = 1) {
na2zero <- function(x) ifelse(is.na(x), 0, x)
Lag <- function(x, i) na2zero(lag(zoo(x), i, na.pad = TRUE))
Rook <- function(x, i) Lag(x, i) + Lag(x, -i) + t(Lag(t(x), i) +
Lag(t(x), -i))
Rook(x, i) / Rook(1+0*x, i)
}
# test
m <- matrix(1:24, 6)
rook2(m)
On 2/17/06, Mills, Jason <Jason.Mills at afhe.ualberta.ca>
wrote:>
> 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
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
>