Shifting a multidimensional matrix

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Alex Feinman
Alex Feinman el 12 de Oct. de 2012
Comentada: Pascal Loohuis el 17 de Sept. de 2019
I'm trying to offset a matrix by a certain distance, like dragging an image partially out of frame.
The 'new' area gets filled with zeroes or NaNs, and the 'extra' area gets clipped, so you end up with a new matrix the same size as the original.
In one dimension this is easy--just add 0s to the size of the offset:
offset = 3;
dest = [zeros(1, offset), original(1:end-offset)];
But I'm having trouble generalizing this to n dimensions. Is there an algorithmic way to handle this, or a built-in I've missed?
EDIT: To clarify, in the N dimensional case, offset is a vector of N elements, some of which can be negative.
For example:
A = ones([3 3]);
offset = [1 1];
_function_(A, offset) =
0 0 0
0 1 1
0 1 1
offset = [1 -1];
_function_(A, offset) =
0 0 0
1 1 0
1 1 0

Respuesta aceptada

Matt J
Matt J el 12 de Oct. de 2012
Editada: Matt J el 12 de Oct. de 2012
I think this might be the generalization you're looking for of Azzi's approach,
function B=noncircshift(A,offsets)
%Like circshift, but shifts are not circulant. Missing data are filled with
%zeros.
%
% B=noncircshift(A,offsets)
siz=size(A);
N=length(siz);
if length(offsets)<N
offsets(N)=0;
end
B=zeros(siz);
indices=cell(3,N);
for ii=1:N
for ss=[1,3]
idx=(1:siz(ii))+(ss-2)*offsets(ii);
idx(idx<1)=[];
idx(idx>siz(ii))=[];
indices{ss,ii}=idx;
end
end
src_indices=indices(1,:);
dest_indices=indices(3,:);
B(dest_indices{:})=A(src_indices{:});
  2 comentarios
Matt Fig
Matt Fig el 12 de Oct. de 2012
Nice!
Pascal Loohuis
Pascal Loohuis el 17 de Sept. de 2019
What if the shifts are different for each layer?

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Más respuestas (3)

Azzi Abdelmalek
Azzi Abdelmalek el 12 de Oct. de 2012
Editada: Azzi Abdelmalek el 12 de Oct. de 2012
offset=3
A=rand(10,12);
[n,m]=size(A)
out=zeros(n,m)
out(:,offset+1:m)=A(:,1:m-offset)
If your matrix is nxmxp
offset=3
A=rand(10,12,3);
[n,m,p]=size(A)
out=zeros(n,m,p)
out(:,offset+1:m,:)=A(:,1:m-offset,:)
  4 comentarios
Alex Feinman
Alex Feinman el 12 de Oct. de 2012
Editada: Alex Feinman el 12 de Oct. de 2012
This only appears to shift things along the first dimension of offset? (I apologize for being so unclear.)
Azzi Abdelmalek
Azzi Abdelmalek el 12 de Oct. de 2012
Ok, I did'nt read your full comment. I have a second answer

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Matt J
Matt J el 12 de Oct. de 2012
First, recognize that in 1D, this can be done by a sparse matrix multiplication
offset=3;
N=10;
x=(1:N).'
S=speye(N); %N is length of vector
S=circshift(S,[offset,0]);
S(1:offset,:)=0;
dest= S*x,
To generalize to 2D, multiply all the columns and rows by S
x=rand(N,N);
dest=S*x*S.';
Or, if you have different offsets in different dimensions, you'll need separate matrices Sx and Sy.
To generalize to 3D and higher, I recommend using my KronProd package
x=rand(N,N,N);
dest=KronProd({S},[1,1,1])*x;
where KronProd is available here
  2 comentarios
Alex Feinman
Alex Feinman el 12 de Oct. de 2012
What if offset contains negative numbers in some dimensions? I'm having trouble figuring that one out.
Matt J
Matt J el 12 de Oct. de 2012
Editada: Matt J el 12 de Oct. de 2012
Only change
S(end+1-(1:-offset),:)=0;
However, Azzi's method can be similarly generalized and is probably better, now that I think about it. That's assuming you're restricting yourself to integer shifts. If you need to do sub-pixel shifts, where you need to interpolate, then my approach is more easily generalized, I think.

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Azzi Abdelmalek
Azzi Abdelmalek el 12 de Oct. de 2012
A=randi(10,4,8,2,4,4,3);
offset=[2 2 1 2 1 2];
siz=size(A);
n=numel(siz);
out=zeros(siz);
idx1=sprintf('%d:%d,',[offset+1; siz]);
idx1(end)=[];
idx2=sprintf('%d:%d,',[ones(1,n); siz-offset]);
idx2(end)=[];
eval(['out(' idx1 ')=A(' idx2 ')'])
  3 comentarios
Alex Feinman
Alex Feinman el 15 de Oct. de 2012
It's possible that you could squeeze a lot of time out of your code but as they stand his is about 20% faster in my testing; just on that basis I'm going to accept his...but this is nice and compact!
Matt J
Matt J el 15 de Oct. de 2012
I think part of the compactness is due to the fact that this solution doesn't support negative offsets. It's interesting that you favor EVAL. Most TMW employees seem to discourage it

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