How to flip non-zero elements of an array and keep zero elements at initial position?

I have an array (A) with non-zero elements and zero elements.

I want to flip all non-zero elements of A but keep all zero elements at their initial position to get B.

I tried this and it works but I am sure there is a one-liner-ish solution to my problem:

A       = [1 2 3 4 5 0];
nz      = nnz(A);           %nz=5
N       = numel(A);         %N=6
numZend = N-nz;             %numZend = 1
Zend    = zeros(numZend);   %Zend = 0
A1      = flip(nonzeros(A));%A1 = [5 4 3 2 1]'
B       = [A1',Zend];       %B = [5 4 3 2 1 0]

Note that my zero elements are always at the end of my array. I can have 0 to 5 zero elements at the end of my array. Normally, length(A) ranges between 17 and 22.

Thank you :)

[Image Analyst]

Give a general example. I have no idea how you can flip non-zero numbers if they are randomly located in a matrix. If you flipped the value right-to-left so that col now goes to (totalColumns-col+1) then it might happen that a non-zero number would like right exactly on top of a zero (which is not allowed to move). Like

m = randi([0, 9], 5, 10)

m = 5×10

0 1 4 9 1 3 8 7 3 1 4 2 8 0 6 2 4 0 6 8 5 4 5 1 3 5 6 1 3 6 1 6 5 0 2 8 7 7 5 2 6 7 6 5 3 7 6 7 5 5

What would be the desired output for that matrix?

[A LL]

In my case, my zeros are always located at the end of my matrix with something like this:

A=[randi([1,9],5,3);zeros(2,3)]

I tried this, it works but it seems a bit too complicated for what I am trying to do:

A       =[randi([1,9],5,3);zeros(2,3)];
nz      = nnz(A(:,1));
N       = numel(A(:,1));
numZend = N-nz;
Zend    = zeros(numZend,numel(A(1,:)));
A1      = flip(A,1);
A2      = A1(numZend+1:end,:);
B       = [A2;Zend];

Would you have an idea to make this simpler?

Answer 1

[Les Beckham]

Another possibility (still requires a loop):

A = [ 134   159   122   175   126     0     0
      151   170   189   196     0     0     0
      155   114   115   126   184   125     0 ]; % test data

B = zeros(size(A));
for i = 1:size(A, 1)
    inz  = A(i,:)~=0; % logical indices of non-zero elements
    B(i,inz) = flip(A(i,inz)) % flip the non-zero elements in this row
end

[Les Beckham]

If all rows of the array have the same number of columns of zeros it is possible to do this without the loop.

A = [ 134   159   122   175   126     0     0
      151   170   189   196   125     0     0
      155   114   115   126   184     0     0 ]; % test data -- modified 
B = zeros(size(A));
[r, c] = find(A~=0);
B(A~=0) = fliplr(A(unique(r), unique(c)))

B = 3×7

126 175 122 159 134 0 0 125 196 189 170 151 0 0 184 126 115 114 155 0 0

Not exactly pretty, but it works. I think I like the loop better since it works for different numbers of zeros in each row and is a little easier to follow.

Answer 2

[David Hill]

A = [1 2 3 4 5 0];
a=flip(A(A~=0));
a=[a,zeros(1,length(A)-length(a))];

[A LL]

Thank you for your answer. :)

Do you have a way to generalize this?

My A is in fact a 20x3 matrix with 2x3 zeros at the end.

Using the nonzeros function (A~=0) returns a full column vector so this will be problematic with A that has dimension greater than 1.

Answer 3

[DGM]

I'm sure there's something simpler, but this is what my sleeplessness created:

A = [1:5 0; 11:14 0 0; 21:23 0 0 0; 31 0 32 0 33 0];
A = A.';
mask = A~=0;
B = zeros(size(A));
B(flipud(mask)) = A(mask);
B = flipud(B).'

B = 4×6

5 4 3 2 1 0 14 13 12 11 0 0 23 22 21 0 0 0 33 0 32 0 31 0

Answer 4

[Image Analyst]

Does this do what you want:

% Generate sample data.
A = zeros(6, 10);
for row = 1 : size(A, 1)
    numNonZeros = randi([3,7]); % Between 3 and 7 non-zeros to start each row.
    A(row, 1:numNonZeros) = randi(9, 1, numNonZeros);
end
A

A = 6×10

5 9 8 3 2 0 0 0 0 0 6 4 8 0 0 0 0 0 0 0 9 2 8 9 6 0 0 0 0 0 9 6 9 2 0 0 0 0 0 0 1 4 8 4 7 3 0 0 0 0 7 3 2 6 0 0 0 0 0 0

% Now we have A, let's flip the non-zeros row-by-row
AFlipped = zeros(size(A));
for row = 1 : size(A, 1)
    lastCol = find(A(row,:) ~= 0, 1, 'last');
    AFlipped(row, 1:lastCol) = fliplr(A(row, 1:lastCol)); % Do the flip of non-zeros ONLY
end
AFlipped % Show in command window.

AFlipped = 6×10

2 3 8 9 5 0 0 0 0 0 8 4 6 0 0 0 0 0 0 0 6 9 8 2 9 0 0 0 0 0 2 9 6 9 0 0 0 0 0 0 3 7 4 8 4 1 0 0 0 0 6 2 3 7 0 0 0 0 0 0

[A LL]

Yes, it does! Thank you very much!

It is even more general than what I need since, in my case, I always have a fix number of zero for each row.