How to crop arrays using a vector? (reverse of padarray)

Asked by Kerwin on 11 Jul 2012
Latest activity Commented on by Kerwin on 20 Jul 2012

*Short version: I'm looking for the reverse of padarray, where you can do padarray(a,[2 2 2]), but instead have it crop [2 2 2] instead of pad [2 2 2]. (Or alternatively, imcrop for more then two dimensions); *

Long version: I have a cell arrays of unknown size K, each cell in the array contains an numerical array with dimensions (m,n,k,... )importantly:

          1. I don't know beforehand how many dimensions there will be for the 
             numerical arrays in the different cell arrays, but I
             know that within a cell array all the numerical arrays will have 
             the same number of dimensions. 
          2. While all the arrays have the same number of dimensions, they 
             differ in size. 

Basically I want to make a functions that crops the numerical arrays to the smallest dimensions, and then concatenates them, i.e. if

      a{1} = zeros(10,10)
      a{2} = zeros(16,6)
      a{3} = zeros(6,6)

I want the outcome to be an array c with size (6,6,3).

I guess the problem is that I don't know how to index the arrays for a variable amount of dimensions. That is, I can get the size to which I need to crop (or the number by which I need to crop) in vector form like [6 6] or [15 4 8], but I don't know how to use this vector to index my arrays.

Thanks in advance!

Edit: Just thought of something, but it's very ugly: if we know we want to crop to an array of size (30,30,30)

minSize = [30 30 30]
c = []
for ii = length(a);
  temp = a{ii}(:);
    mask = zeros(minSize)
    padding = (minSize - size(a{ii})./2;
    mask = padarray(mask,padding,1,'both');
    mask = logical(mask);
    temp(mask)= [];
    temp = reshape(temp,minSize);
    c = cat(ndim(a{ii})+1,c,temp);
  end

2 Comments

Jan Simon on 11 Jul 2012

Do you know an upper limit for the number of dimensions?

Kerwin on 13 Jul 2012

No, they can theoretically be of any nr.

Kerwin

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2 Answers

Answer by Sean de Wolski on 11 Jul 2012
Edited by Sean de Wolski on 11 Jul 2012
Accepted answer

So something along the lines of:

x = rand(10,6,8,9);
nd = ndims(x);
c = repmat({':'},nd-1,1);
for ii = 1:nd;
  x = shiftdim(x,1);
  x = x(3:end-2,c{:});
end

1 Comment

Kerwin on 12 Jul 2012

Definitely better than my solution, thanks!

On second inspection, for large arrays this operation is fairly slow, because shiftdim is quite intensive. I really wish there was a way to use vectors for dimensions.

Sean de Wolski
Answer by nanren888 on 13 Jul 2012
Edited by nanren888 on 13 Jul 2012

Sorry I do not know paddarray, so maybe I am answering the wrong question.

I can give you parts of a way to do it. There may be more elegant ways. Indexing any number of dimensions is supported by Matlab's cool mechanism of using cell arrays as parameters;

Short version:

(1) Find the size you want with size & min

(2) Create a cell array of the ranges you want indC = {1:4 1:5 1:6 1:2 ...}

(3) Use < for all cells k > c{k} = c{k}(indC{:});

Longer version:

It seems finding the size you want to crop to is easy, just go through all cells with some sort of min(), eg collect all sizes & use min(?,dim), or manually take minimum values.

Maybe for the indexing, this will help

gg = randn([2 3 4 2]);

szVec = size(gg);

nDim = length(szVec);

.... cropSize = [nDim,1] array of desired dimensions as above

ind = {};

for k = 1:nDim

 ind = [ind 1:cropSize(k)]; % I presume you want 1:cropSize

end

smallerGg = gg(ind{:});

Hope it helps

2 Comments

nanren888 on 14 Jul 2012

Yeah, Mine won't suffer to the same extent from the extremely slow moves, I guess.

Maybe you could profile them for us, on a reasonable number of trials?

I guess what you asked for "vectors for dimensions" I did with cells for dimensions.

%% timeMultiDimStuff.m

x = rand(10,6,8,9,5,6,4,5,5);

y = zeros(size(x));

nDim = ndims(x);

cropSize = [8,6,7,3,4,5,3,5,4];

ind = {};

for k = 1:nDim

   ind = [ind 1:cropSize(k)];

end

nTrial = 1000;

tic();

for k = 1:nTrial

    y = x(ind{:}); %#ok<NASGU>

end

toc();

Run with your solution & compare?

Elapsed time is 5.751869 seconds. (You have to run both on the same machine :))

Kerwin on 20 Jul 2012

sorry for the late reply!

Your solution works very well, I quickly tested it on some arrays, the largest being 4 arrays of each 5 dimensions, with the largest array having approx 40 million elements.

The earlier proposed solution with shiftdim took 9.87 seconds. Your solution took 2.01 seconds, a very nice speed-up! I expect that on my larger arrays this difference will become even more pronounced.

Thanks again!

nanren888

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