How do I delete the zeros from the matrix?

2 views (last 30 days)
jakobjakob on 10 Jun 2018
Commented: Monika Jaskolka on 11 Jun 2018
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 on 10 Jun 2018
@jakobjakob
You can remove 0s with regexprep .
x = {10030050201};
x = regexprep(string(x{:}),'0','');
x = str2double(x);

Monika Jaskolka on 10 Jun 2018
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
Monika Jaskolka on 11 Jun 2018
Thanks!

Jan on 10 Jun 2018
Edited: Jan on 10 Jun 2018
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.

Jan on 11 Jun 2018
Edited: Jan on 11 Jun 2018
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

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