Shifting a multidimensional matrix

12 次查看（过去 30 天）
Alex Feinman
Alex Feinman 2012-10-12
评论： Pascal Loohuis 2019-9-17
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

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2012-10-12
编辑：Matt J 2012-10-12
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{:});

更多回答（3 个）

Azzi Abdelmalek
Azzi Abdelmalek 2012-10-12
编辑：Azzi Abdelmalek 2012-10-12
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 个评论
显示 2更早的评论
Alex Feinman
Alex Feinman 2012-10-12
编辑：Alex Feinman 2012-10-12
This only appears to shift things along the first dimension of offset? (I apologize for being so unclear.)
Azzi Abdelmalek
Azzi Abdelmalek 2012-10-12
Ok, I did'nt read your full comment. I have a second answer

请先登录，再进行评论。

Matt J
Matt J 2012-10-12
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
http://www.mathworks.com/matlabcentral/fileexchange/25969-efficient-object-oriented-kronecker-product-manipulation
  2 个评论
Alex Feinman
Alex Feinman 2012-10-12
What if offset contains negative numbers in some dimensions? I'm having trouble figuring that one out.
Matt J
Matt J 2012-10-12
编辑：Matt J 2012-10-12
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.

请先登录，再进行评论。

Azzi Abdelmalek
Azzi Abdelmalek 2012-10-12
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 个评论
显示 1更早的评论
Alex Feinman
Alex Feinman 2012-10-15
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 2012-10-15
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

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Resizing and Reshaping Matrices 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by