MATLAB: Finding a 2d block with highest average inside a 2d window

matrixmatrix manipulation

Consider a matrix A(52,34) How to find out which 2d block size(3,3) has the highest average in it

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[~, idx] = max(conv2(A, ones(3,3),'same'))
[r, c] = ind2sub(size(A), idx);

r and c is now the index of the center of the block that had the highest average.

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MATLAB: How to use output indexes from the MAX function

Hi Patrick, I think this is what you want to do:

arrSz = [5 6 7];                  % Size of matrix
cube1 = 10*rand(arrSz);           % Create the matrix
[vals1, idxs1] = max(cube1,[],3); % find first max across 3th dimension
% Make a 3D mask corresponding to indices in idxs1
idxsMask = bsxfun(@eq,reshape(1:arrSz(3), 1,1,[]), idxs1);
cube1(idxsMask) = -inf;           % Set first max vals to -inf
[vals2, idxs2] = max(cube1,[],3); % find second max across 3th dimension

The trick here is the bsxfun() call. I make a set of indices along the 3rd dimension, then bsxfun does its magic asking if the 2D matrix idxs1 matches that array. bsxfun() is a strange function if you haven't seen it before, but once you understand what it does it can be very useful. And very efficient - much faster than the loop.

Walter's question is a valid one... what if the max value isn't unique along the 3rd dimension? Should your second call to max() return the next index that had the same value as the first call? Or should the first call turn all elements that had the max value to -inf?

If you want to do this second way, simply replace the idxsMask line above with the following:

idxsMask = bsxfun(@eq,vals1,cube1);

MATLAB: Select first n nonzero elements in each row of matrix

Linear indexing does this simply and efficiently. The trick is to work down the columns, which requires transposing:

>> A = [1,3,0,2,5;0,2,0,1,0;2,0,0,0,0;3,1,1,0,0]
A =
   1   3   0   2   5
   0   2   0   1   0
   2   0   0   0   0
   3   1   1   0   0
>> N = 3;
>> Z = A.';
>> S = size(Z);
>> [~,R] = sort(Z==0,1);
>> [~,C] = ndgrid(1:N,1:S(2));
>> X = sub2ind(S,R(1:N,:),C);
>> B = Z(X).'
B =
   1   3   2
   2   1   0
   2   0   0
   3   1   1

Probably the most efficient approach would be to use a simple loop, e.g. (not particularly optimized):

R = size(A,1);
B = zeros(R,N);
for k = 1:R
    tmp = nonzeros(A(k,:));
    idx = 1:min(N,numel(tmp));
    B(k,idx) = tmp(idx);
end

Some timings (1e3 iterations):

Elapsed time is 5.358 seconds. % madhan ravi's with loop and CELLFUN
Elapsed time is 0.606 seconds. % my answer with SUB2IND
Elapsed time is 0.265 seconds. % my answer with loop and indexing

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Related Solutions

MATLAB: How to use output indexes from the MAX function

MATLAB: Select first n nonzero elements in each row of matrix

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