finding the coordinates of a value in a cell array

for x = 1:100
    for y =  1:61
             b = abs(bsxfun(@minus,q{x,y},kevin))
             averagesAcrossColumns = mean(b, 2);
             b = abs(bsxfun(@minus,b,averagesAcrossColumns));
             b = sum(b, 2);
             s{x,y} = b;
      end
  end

q is a cell array that has has many matrices. kevin is a variable that is a 1x60 matrix.

I want to add an additional line to the code so that I find the smallest three values ie smallest three sums in variable s. The output should give me the position in the cell array, row and column number in the matrix. Also shortening the code to make it more cpu and memory effective would be appreciated.

thanks

Answer 1

[Image Analyst]

There is no reason why S should be a cell array. The last two lines should simply be

s(x,y) = sum(b, 2);

Make it just a normal numerical array (double). Then try this:

sorted_s = sort(s(:), 'ascend');
% Find rows and columns where it's the min
% There may be more than one location, particularly if S is integer.
[rows1, columns1] = find(s == sorted_s(1))
% Now find the second and third smallest locations.
[rows2, columns2] = find(s == sorted_s(2))
[rows3, columns3] = find(s == sorted_s(3))

I really don't know what you're doing with bsxfun, q, kevin, and all that. I urge you to avoid cell arrays unless they're really necessary such as when you have collections of different length strings, or collections of variables where the type of the variable (char, integer, structure, etc.) varies on a element by element basis.

[AA]

I get this error message Conversion to cell from double is not possible.

[Image Analyst]

Yes, that's why I said "I urge you to avoid cell arrays" - they're always more complicated.

[AA]

s(x,y) = sum(b, 2); this command gives me an error. can you give me a solution with a cell array? thanks

Answer 2

[AA]

mn = cellfun(@(x) min(x(x>0)),q,'Un',0);
[mn,idx] = min(fliplr([mn{:}]));  
mn  % Show the minimum positve value.
idx = length(q) - idx + 1  % Which cell has the min.
L = cellfun(@(x) find(x==mn),q,'Un',0);
L = L{idx} % And the positions.