elimination of conscutive regions (generalization: ones with zeros between)
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Michal
el 1 de Oct. de 2018
Comentada: Bruno Luong
el 2 de Oct. de 2018
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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Respuesta aceptada
Bruno Luong
el 1 de Oct. de 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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Bruno Luong
el 2 de Oct. de 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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