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elimination of conscutive regions (generalization: ones with zeros between)

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I need effectively eliminate (by zeroing) the consecutive "1's" between "-1's" and start/end of column at each column of matrix A, which now can be separated by any number of zeroes. The number of consecutive "1's" between "-1's" and start/end of column is > N. This is a non-trivial generalization of my previous Question.

Again, typical size(A) = [100000,1000].

See example:

A =
     1    -1     0
     0     1     1
     0     1     1
     1     1     0
     0     0     1
     1    -1     0
    -1     1     1
    -1     0    -1
     1     1     1
     0     1    -1

For N = 2 the expected result is

Aclean =
     0    -1     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0    -1     0
    -1     0     0
    -1     0    -1
     1     0     1
     0     0    -1

For N = 3 the expected result is

Aclean =
    1    -1     0
    0     1     0
    0     1     0
    1     1     0
    0     0     0
    1    -1     0
   -1     1     0
   -1     0    -1
    1     1     1
    0     1    -1
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Accepted Answer

Bruno Luong
Bruno Luong on 1 Oct 2018
A = [1 -1 0;
0 1 1;
0 1 1;
1 1 0;
0 0 1;
1 -1 0;
-1 1 1;
-1 0 -1;
1 1 1;
0 1 -1];
N = 3;
[m,n] = size(A);
Aclean = A;
for j=1:n
Aj = [-1; A(:,j); -1];
i = find(Aj == -1);
c = histc(find(Aj==1),i);
b = c <= N;
im = i(b);
ip = i([false; b(1:end-1)]);
a = accumarray(im,1,[m+2,1])-accumarray(ip,1,[m+2,1]);
mask = cumsum(a);
mask(i) = 1;
Aclean(:,j) = Aclean(:,j).*mask(2:end-1);
end
Aclean
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More Answers (2)

Michal Kvasnicka
Michal Kvasnicka on 1 Oct 2018
Edited: Michal Kvasnicka on 1 Oct 2018
I am still looking for better (faster) solution. Any idea how to improve so far best solution:
A = [1 -1 0;
0 1 1;
0 1 1;
1 1 0;
0 0 1;
1 -1 0;
-1 1 1;
-1 0 -1;
1 1 1;
0 1 -1];
N = 3;
sep = A==-1;
sep(1,:) = true;
idx = cumsum(sep(:));
sep(1,:) = A(1,:)==-1;
num = accumarray(idx, A(:)==1);
iff = num <= N;
Aclean = reshape(sep(:)|iff(idx), size(A)) .* A;
Aclean
Big test matrix:
A = double(rand(100000,1000)>.4) - double(rand(100000,1000)>.65);
Bruno's code (N = 5): Elapsed time is 4.848747 seconds.
My code (N = 5): Elapsed time is 3.257089 seconds.
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Bruno Luong
Bruno Luong on 2 Oct 2018
That's true, somehow it's a 1D scanning problem.
I think we are close to the limit of MATLAB can do, if faster speed is still needed, then one should go to MEX programming route instead of torturing MATLAB to squeeze out the last once of speed.

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