How to exclude zeros in a matrix?

Hello,

I am trying to get rid of the zero values in a matrix I have in V(:,:,i) in my for loop. However, I am getting an error, "Unable to perform assignment because the size of the left side is 297-by-448 and the size of the right side is 10611-by-1. Error in DIC_Data_Extraction (line 17) V(:,:,i) = nonzeros(data_dic_save.displacements(i).plot_v_ref_formatted);". Therfore, if you are able to help me out in resolving this error, it will be most appreciated, thanks. Below is a copy of my code. Also, unfortunatley, I can not attach my work space, so I hope this is enough information for you to help me out, thanks.

-Robert

% The displacements in the Y-direction 
V = zeros(297,448);
% Vyy = zeros{20863,1};

for i = 1:8 
    V(:,:,i) = nonzeros(data_dic_save.displacements(i).plot_v_ref_formatted);
%     Vyy{i} = {nonzeros(V(:,:,i))}
    avg(i) = abs(mean(mean(V(:,:,i))));
    med(i) = median(median(V(:,:,i)));
   
end

[Rik]

How are you getting rid of the zeros? Arrays in Matlab must be rectangular. You can use a cell array with empty cells in them, or signaling values like NaN.

[Adam]

The NaN approach is definitely the best choice in most cases as working with numeric arrays is vastly simpler and quicker than working with cell arrays.

[Adam Danz]

From your description, it seems like zeros aren't the problem. Your nonzeros() function is producing a columnar vector with 10611 elements and you're trying to store that in a 297x448 matrix.

[Robert Flores]

Adam Danz, that is exactly my issue. I am sorry for my late response, but that is the issue I am dealing with.

Answer 1

[Jan]

The cell is the right approach:

for i = 1:8 
    Vyy{i} = nonzeros(data_dic_save.displacements(i).plot_v_ref_formatted);
    avg(i) = abs(mean(Vyy{i}));
    med(i) = median(Vyy{i});
end

Other have explained already, that arrays must be rectangular.

Another option would be to set the zeros to NaN:

V = zeros(297, 448, 8);  % Pre-allocate all 3 dimensions!
for i = 1:8 
    aV          = data_dic_save.displacements(i).plot_v_ref_formatted;
    aV(aV == 0) = NaN;
    V(:, :, i)  = aV;
    
    avg(i) = abs(mean(aV, 'all', 'omitnan'));
    med(i) = median(aV, 'all', 'omitnan');  % See comments
end

Attention: I guessed, that you want the median of all values. Then median(X, 'all') is equivalent to median(X(:)), which is not necessarily the same as median(median(X, 1), 2). With nonzeros the output is a vector and there is no difference. Please check this explicitly.

[madhan ravi]

+1, cell is the best and uncertainty friend of a programmer.