# Finding the number of rows to the next row containing a 1

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andyc100 on 27 Aug 2020
Edited: Bruno Luong on 28 Aug 2020
Hi
I have a column vector of 1s and 0s and I want to find find the number of rows to the next row containing a 1. For example:
A = [0 0 0 1 0 1 1 1 0 0 0 0 0 1]';
I would like the code to return
B = [3 2 1 0 1 0 0 0 5 4 3 2 1 0]';
Is there a vectorized way that this can be done?
Bruno Luong on 28 Aug 2020
Edited: Bruno Luong on 28 Aug 2020
As I'm slightly surprised by the performance of the for-loop (I would expext it's good but not THAT good), I then try to see how far it from a MEX implementation. And I'm stunned, it's almost as fast (even faster for smaller array).
t Rik = 0.651531 [s]
t Bruno = 0.379362 [s]
t Stephen = 0.104442 [s]
t MEX = 0.073168 [s]
The code (benchmark + Cmex) is in the attacheh file for those who wants to play with.
I must congratulat TMW for improving the for-loop performance over many years.

Rik on 27 Aug 2020
Edited: Rik on 27 Aug 2020
It took some time, but here is a solution that should also work for large matrices.
clc,clear
format compact
A = [0 0 0 1 0 1 1 1 0 0 0 0 0 1]';
% 3 2 1 0 1 0 0 0 5 4 3 2 1 0
B=A;
if pad,B(end+1)=1;end %this method requires the last position to be a 1
B=flipud(B);
C=zeros(size(B));
C(B==1)=[0;diff(find(B))];
C=ones(size(B))-C;
out=cumsum(C)-1;
out=flipud(out);
%only for display:
[A out]
andyc100 on 27 Aug 2020
This is not a concern for me as there will always be 1s in my implementation. Can always just do a check for not all zeros in the vector before running the code I guess.

Binbin Qi on 27 Aug 2020
A = [0 0 0 1 0 1 1 1 0 0 0 0 0 1]';
C = find(A);
D = (1:length(A)) - C;
D(D>0) = D(D>0) + inf';
min(abs(D))'
ans =
3
2
1
0
1
0
0
0
5
4
3
2
1
0
Andrei Bobrov on 27 Aug 2020
+1

Bruno Luong on 27 Aug 2020
Edited: Bruno Luong on 27 Aug 2020
As much as I love vectorization, this problem is a typical case where the for-loop method is easier, faster, more readable.
This code is ready for 2D array, it works along the first dimension independently.
A=rand(30000,1000)>0.7;
tic
Al=logical(A);
B=zeros(size(A));
b=B(1,:);
for k=size(B,1):-1:1
b = b+1;
b(Al(k,:))=0;
B(k,:)=b;
end
toc
andyc100 on 27 Aug 2020
Thank you Burno. This actually works very fast too and I like that it works for multiple columns. I might end up using it if I end up needing multiple columns.

Bruno Luong on 28 Aug 2020
Edited: Bruno Luong on 28 Aug 2020
Now I just discover CUMSUM has direction option, this is based on Rik's cumsum method, but avoid the double flipping.
B = ones(size(A));
i1 = find(A);
B(i1) = 1-diff([i1;size(A,1)+1]);
B = cumsum(B,'reverse');
Rik on 28 Aug 2020
Cool, I didn't remember that was an option. Turns out that is already an option as far back as R2015a. The release notes no longer allow you to look back further than R2015a, even if you try modifying the url. All I know is that it isn't possible in R2011a.

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