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Shifting a multidimensional matrix

Asked by Alex Feinman on 12 Oct 2012
Latest activity Commented on by 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

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4 Answers

Answer by Matt J
on 12 Oct 2012
Edited by Matt J
on 12 Oct 2012
 Accepted Answer

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 Comments

Matt Fig
on 12 Oct 2012
Nice!
What if the shifts are different for each layer?

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Answer by Azzi Abdelmalek
on 12 Oct 2012
Edited by 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)
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 Comments

Show 1 older comment
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,:)
This only appears to shift things along the first dimension of offset? (I apologize for being so unclear.)
Ok, I did'nt read your full comment. I have a second answer

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Answer by 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

  2 Comments

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.

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Answer by 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 ')'])

  3 Comments

I love the use of sprintf / eval...let me do some speed testing vs. Matt J.'s answer.
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

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