How do I delete the zeros from the matrix?

Question

jakobjakob 2018-6-10

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评论： Monika Jaskolka 2018-6-11

采纳的回答： Monika Jaskolka

eerste_kijkmoment.mat

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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

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Rik 2018-6-10

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 2018-6-10

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@jakobjakob

You can remove 0s with regexprep .

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

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Answer 1

Monika Jaskolka 2018-6-10

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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

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Jan 2018-6-11

编辑：Jan 2018-6-11

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 2018-6-11

Thanks!

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Answer 2

Jan 2018-6-10

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编辑：Jan 2018-6-10

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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.

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Answer 3

Jan 2018-6-11

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编辑：Jan 2018-6-11

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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