# Shifting a multidimensional matrix

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Alex Feinman on 12 Oct 2012
Commented: Pascal Loohuis on 17 Sep 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

Matt J on 12 Oct 2012
Edited: Matt J on 12 Oct 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{:});

Matt Fig on 12 Oct 2012
Nice!
Pascal Loohuis on 17 Sep 2019
What if the shifts are different for each layer?

Azzi Abdelmalek on 12 Oct 2012
Edited: Azzi Abdelmalek on 12 Oct 2012
offset=3
A=rand(10,12);
[n,m]=size(A)
out=zeros(n,m)
out(:,offset+1:m)=A(:,1:m-offset)
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,:)

Show 1 older comment
Azzi Abdelmalek on 12 Oct 2012
for multidimension >3
A=randi(10,4,8,2,4,2) % example
siz=size(A)
out=zeros(siz)
i1=offset+1:siz(2)
i2=1:siz(2)-offset
out(:,i1,:)=A(:,i2,:)
Alex Feinman on 12 Oct 2012
This only appears to shift things along the first dimension of offset? (I apologize for being so unclear.)
Azzi Abdelmalek on 12 Oct 2012

Matt J on 12 Oct 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

Alex Feinman on 12 Oct 2012
What if offset contains negative numbers in some dimensions? I'm having trouble figuring that one out.
Matt J on 12 Oct 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.

Azzi Abdelmalek on 12 Oct 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 ')'])

Alex Feinman on 15 Oct 2012
I love the use of sprintf / eval...let me do some speed testing vs. Matt J.'s answer.
Alex Feinman on 15 Oct 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 on 15 Oct 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