Replace zero in a matrix with value in previous row

Hi,

Can you please help me on how can I replace all zeroes in a matrix with the value in previous row?

e.g. if value in row 3 column 4 is 0, it should pick value in row 2 column 4.

I can do it using a for loop but I dont want to use that.

Thanks.

Answer 1

[Azzi Abdelmalek]

A=[1 2 3 4;4 5 0 0;1 0 0 1 ;0 1 1 1]
while any(A(:)==0)
  ii1=A==0;
  ii2=circshift(ii1,[-1 0]);
  A(ii1)=A(ii2);
end

[Mohit]

Thanks

[Mido]

What should I do if I want to apply this process on a specific column. For example the third one?

[Jan]

@Mido: Please open a new thread for a new question.

match       = (A(:, 3)==0);
A(match, 3) = A(match, 2);

[Pardis]

Very helpful, thank you Azzi!

Answer 2

[Shivaputra Narke]

Now answer to your comment...

while(all(a(:))) a(find(a==0))=a(find(a==0)-1) end

[Mohit]

Thanks

[Captain Karnage]

That expression doesn't work as written in a single line. Without a ; after the assignment, it gets an error 'Error: Illegal use of reserved keyword "end".` If I add the ; however, it doesn't error but it still doesn't work. I'm not sure why, because logically it seems like it should, but I just get the original a out when I run it.

[DGM]

The loop is never entered at all. You could make some modifications.

a = [0 5 9 13; 2 6 0 0; 3 0 0 15; 0 8 12 16]

a = 4×4

0 5 9 13 2 6 0 0 3 0 0 15 0 8 12 16

na = numel(a);
while ~all(a(:)) % loop runs until there are no zeros
    idx = find(a==0);
    a(idx) = a(mod(idx-2,na)+1);
end
a

a = 4×4

16 5 9 13 2 6 9 13 3 6 9 15 3 8 12 16

The redundant find() can be removed. Since this is based on decrementing the linear indices, this will fill zeros at the top of a column with content from the bottom of the prior column. Using mod() allows the wrapping behavior to extend across the ends of the array. Note that a(1,1) is filled from a(16,16).

Whether this wrapping behavior is intended or acceptable is a matter for the reader to decide.

Answer 3

[Andrei Bobrov]

l = A == 0;
ii = bsxfun(@plus,size(A,1)*(0:size(A,2)-1),cumsum(~l));
out = A;
out(l) = A(ii(l));

Answer 4

[Shivaputra Narke]

May be this code can help...

% where a is your matrix a(find(a==0))=a(find(a==0)-1)

[Amit]

This will not work in many scenarios.

[Mohit]

Thanks a lot, it works!

What should I do in case I want to put another condition that if cell value above zero cell is also zero then go one cell up.

e.g. if value in row 6 column 4 is 0, it should pick value in row 5 column 4. If value in row 5 column 4 is also zero then it should pick value in row 4 column 4 and so on.

Answer 5

[Amit]

Lets say your matrix is A

[m,n] = size(A);
An = A';
valx = find(~An); % This will give you zeros elements linear index
valx = valx(valx-n > 0);
An(valx) = An(valx-n);
A = An';

[Mohit]

Thanks

Answer 6

[Paul]

    idx=find(A==0)
    A(idx)=A(idx-1)

[Amit]

This solution is similar to what Shivaputra posted. This solution will also fail. find(A==0) give linear indexing. Imaging that you have first row with some values 0. This approach will fill up those first row values with a value too.

Try these solution for a matrix like:

A = [1 0 0;2 1 0;2 3 1]

[Shivaputra Narke]

Thank you Amit. My solution wont work on such scenarios.

[Mohit]

Thanks