Replace zero in a matrix with value in previous row

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Mohit
Mohit 2014-2-1
编辑： DGM 2022-12-3
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.

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Azzi Abdelmalek
Azzi Abdelmalek 2014-2-1
编辑：Azzi Abdelmalek 2014-2-1
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
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Jan
Jan 2016-11-4
@Mido: Please open a new thread for a new question.
match = (A(:, 3)==0);
A(match, 3) = A(match, 2);

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Shivaputra Narke
Shivaputra Narke 2014-2-1
Now answer to your comment...
while(all(a(:))) a(find(a==0))=a(find(a==0)-1) end
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Captain Karnage
Captain Karnage 2022-12-2
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
DGM 2022-12-3
编辑：DGM 2022-12-3
Ran in:
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.

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Andrei Bobrov
Andrei Bobrov 2014-2-1
l = A == 0;
ii = bsxfun(@plus,size(A,1)*(0:size(A,2)-1),cumsum(~l));
out = A;
out(l) = A(ii(l));
Shivaputra Narke
Shivaputra Narke 2014-2-1
May be this code can help...
% where a is your matrix a(find(a==0))=a(find(a==0)-1)
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Mohit
Mohit 2014-2-1
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.

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Amit
Amit 2014-2-1
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';
Paul
Paul 2014-2-1
idx=find(A==0)
A(idx)=A(idx-1)
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