How do I delete the zeros from the matrix?

I would like to delete all the zeros. I want that the non-zero number shift to the left, like this:

00100200300 --> 123--
10030050201 --> 13521

[Paolo]

Why is 123 followed by " -- " ? Do you always want five alphanumeric characters?

[jakobjakob]

Not all rows have the same number of non-zeros, so there will be cells which have no number, that's what i meant with -. I just want all the zeros out.

[Image Analyst]

Even a cell array has to be rectangular. You can't have ragged edges. However you can have cells in the array that are empty. OR you can have row vector cell arrays and have as many of those as you have rows in your original matrix.

I don't see how any of those options are as convenient as just sticking with your original matrix. Why do you want to do this?

[Rik]

Arrays in Matlab must be square. What data type do you want? A double array? A cell array? You can use find to find non-zero elements. You can also use eerste_kijkmoment(eerste_kijkmoment~=0)=[]; to remove all non-zero elements and convert the matrix to a vector.

[Paolo]

@jakobjakob

You can remove 0s with regexprep .

 x = {10030050201};
 x = regexprep(string(x{:}),'0','');
 x = str2double(x);

Answer 1

[Monika Jaskolka]

Instead of "-" the following function uses NaN.

function B = removeMatZeros(A)
    B = [];
    for i = 1: size(A, 1)
        r = A(i,:);
        r(r==0) = []; % remove zeros

          % handle expansion
          ncolR = size(r, 2);
          ncolB = size(B, 2);
          diffcol = ncolR - ncolB;
          if (diffcol > 0) % previous rows need more cols
              for j = ncolB+1:ncolR
                  B(:,j) = NaN;
              end
          elseif (diffcol < 0) % this row needs more cols
              r = [r, NaN(1, abs(diffcol))];
          end
          B(i,:) = r;
      end
  end

Example:

A =

       0     0     1     0     0     2     0     0     3     0     0
       1     0     0     3     0     0     5     0     2     0     1

>> removeMatZeros(A)

ans =

       1     2     3   NaN   NaN
       1     3     5     2     1

[Jan]

See my second answer for a simplified version of your code. With a pre-allocation of B with NaN values, the iterative filling can be omitted.

[Monika Jaskolka]

Thanks!

Answer 2

[Jan]

Your data is a numerical matrix - the upper left 5x5 submatrix is:

    6.5600         0         0         0         0
         0    9.1300    9.9200   10.2000   11.2400
         0         0         0         0         0
         0         0         0         0         0
         0         0         0         0         0

Then the explanation is not clear:

00100200300 --> 123--
10030050201 --> 13521

Maybe you want:

C = num2cell(eerste_kijkmoment, 2);
C = cellfun(@(a) a(a~=0), C, 'UniformOutput', 0);

Or less nice, but with double speed:

C = cell(size(eerste_kijkmoment, 1), 1);
for iC = 1:numel(C)
  a     = eerste_kijkmoment(iC, :);
  C{iC} = a(a ~= 0);
end

Now the cell array C contains the row vectors with different lengths.

Answer 3

[Jan]

You can pre-allocate the output the avoid the time-consuming iterative growing. This simplifies the code:

s1 = size(A, 1);
s2 = max(sum(A ~= 0, 2));  % Maximum row width
B  = nan(s1, s2);          % Pre-allocation
for k = 1:s1
  r = A(k, :);             % Get non-zero values
  r = r(r ~= 0);
  B(k, 1:length(r)) = r;   % Insert it in NaN matrix
end