# Finding the first zero element in a row and fill the entire row with the last non zero element

7 views (last 30 days)
MHN on 1 Apr 2016
Commented: MHN on 1 Apr 2016
I have an m*n matrix which has only zeros and natural numbers. If there is a zero in a row, all the remaining elements after the first zero are also zeros. I would like to replace all zero elements with the last non zero element in the row. I am looking for the fastest way to do it in Matlab. Is there anyway without a loop? if not, what is the best way? Thanks.
For example:
A = [1 3 1 0 0 0 0;
1 4 2 1 3 4 1;
2 3 1 2 3 3 0];
% and I would like to have
B = [1 3 1 1 1 1 1;
1 4 2 1 3 4 1;
2 3 1 2 3 3 3];

Stephen23 on 1 Apr 2016
Edited: Stephen23 on 1 Apr 2016
>> A = [1,3,1,0,0,0,0;1,4,2,1,3,4,1;2,3,1,2,3,3,0]
A =
1 3 1 0 0 0 0
1 4 2 1 3 4 1
2 3 1 2 3 3 0
>> A(:,2:end) = A(:,2:end) - cumsum(diff(A,1,2).*(A(:,2:end)==0),2)
A =
1 3 1 1 1 1 1
1 4 2 1 3 4 1
2 3 1 2 3 3 3
MHN on 1 Apr 2016
Wow, the way that you use cumsum and diff is very interesting! Could you please explain the logic behind it ?

MHN on 1 Apr 2016
Edited: MHN on 1 Apr 2016
It is my fastest way so far. Adding one column of zero at the end, finding the minimum of each row (which would be the first zero), and using a loop. Is there any way to use "find" in the similar manner? since we can set to find the first element (specially from the end) instead of minimum which will go through all the row and that must be faster.
temp = A;
temp(:,end+1) = 0;
[~, Ind] = min(temp, [],2 );
for i = 1:length(Ind)
temp(i,Ind(i):end) = temp(i,Ind(i)-1);
end
B = temp(:,1:end-1);