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

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andyc100
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?
Thanks in advance

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Stephen Cobeldick
Stephen Cobeldick on 28 Aug 2020
Vectorization is not always the fastest, nor always the clearest code:
A = double(rand(30000,1)>0.7);
%A = [0;0;0;1;0;1;1;1;0;0;0;0;0;1];
N = 1000;
tic
for ii=1:N
% simple loop
B0 = A;
cnt = 0;
for k = numel(A):-1:1
cnt = (cnt+1)*(1-A(k));
B0(k) = cnt;
end
end
toc
tic
for ii=1:N
% Rik's implementation
B=A;
pad=B(end)~=1;
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;
B1=cumsum(C)-1;
B1=flipud(B1);
if pad,B1(end)=[];end
end
toc
tic
for ii=1:N
% Bruno's implementation
i1=(size(A,1)+1)-flipud(find(A));
B2=ones(size(A));
B2(i1)=1-diff([0;i1]);
B2=flipud(cumsum(B2));
end
toc
isequal(B0,B1,B2)
Giving:
Elapsed time is 0.379344 seconds. % simple loop
Elapsed time is 1.111552 seconds. % Rik
Elapsed time is 0.554412 seconds. % Bruno Luong
ans =
1
Bruno Luong
Bruno Luong on 28 Aug 2020
Very convincing Stephen. In my laptop it's even more obvious
t Rik = 0.680010 [s]
t Bruno = 0.378046 [s]
t Stephen = 0.107799 [s]
In your for-loop code I would cast B initialization in double (in case A is logical)
B = double(A);
or
B = zeros(size(A));
Bruno Luong
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.

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Accepted Answer

Rik
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;
pad=B(end)~=1;
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);
if pad,out(end)=[];end
%only for display:
[A out]

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Bruno Luong
Bruno Luong on 27 Aug 2020
When I run this code with
A = [0;0;0]
I get
out=[3;2;1]
Does it meet the description "find the number of rows to the next row containing a 1".
To me no since there is no 1 in A it should return [0;0;0].
My code also have this flaw.
Rik
Rik on 27 Aug 2020
If that is indeed a problem that should be easy to fix. I chose to write a comment instead of changing that behavior, but maybe I should have made it more explicit. Thank you for drawing more attention to that point.
andyc100
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.

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More Answers (3)

Binbin Qi
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

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Binbin Qi
Binbin Qi on 27 Aug 2020
if data is larger, you can use the code following
A = [0 0 0 1 0 1 1 1 0 0 0 0 0 1]';
A = [A;1]; %
B = cell2mat(cellfun(@(x)(length(x)-1:-1:0)',...
mat2cell(A, diff([0;find(A==1)])),...
'UniformOutput',false));
B(end) = []
andyc100
andyc100 on 27 Aug 2020
Thanks so much for this. Always like one liner answers, but have chosen Rik's solution as it works a lot faster for the range of rows I'm working with.

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Bruno Luong
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

  1 Comment

andyc100
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.

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Bruno Luong
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');

  1 Comment

Rik
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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