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elimination of consecutive regions

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I need to effectively eliminate consecutive regions in vector "a" or better in rows/columns of matrix "A" with length of separate ones regions greater than positive integer N <= length(A):
See following example:
N = 2 % separate consecutive regions with length > 2 are zeroed
a = [0 1 1 0 0 1 1 1 0 0 1 1 1 1 0 1]
a_elim = [0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1]
or 2D case:
N = 2
A = [1 0 1
1 1 0
1 1 0
0 0 1
1 1 1]
% elimination over columns
A_elim= 0 0 1
0 1 0
0 1 0
0 0 1
1 1 1
% elimination over rows
A_elim= 1 0 1
1 1 0
1 1 0
0 0 1
0 0 0
I am looking for effective vectorized function performing this task for size(A) ~ [100000, 1000] (over columns case).

Accepted Answer

Matt J
Matt J on 27 Sep 2018
Edited: Matt J on 27 Sep 2018
e=ones(N+1,1);
if mod(N,2) %even mask
mask=~conv2(conv2(A,e,'valid')>=N+1 ,[zeros(N,1);e])>0;
mask=mask(N+1:end,:);
else %odd mask
mask=~(conv2( conv2(A,e,'same')>=N+1, e,'same')>0);
end
A_elim=A.*mask;
  3 Comments
Michal Kvasnicka
Michal Kvasnicka on 29 Sep 2018
Hi Matt … now your solution works very well. Is faster and significantly less memory consuming than Bruno's code. Moreover, the "mask" is very useful to know.
Thanks!!!

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More Answers (2)

Bruno Luong
Bruno Luong on 27 Sep 2018
Edited: Bruno Luong on 27 Sep 2018
You could use Huffman encoding (there might be some on the FEX), but the idea is similar to this direct code:
N = 2
A = [1 0 1;
1 1 0;
1 1 0;
0 0 1;
1 1 1];
% Engine for working along the column (1st dimension)
[m,n] = size(A);
z = zeros(1,n);
Apad = [z; A; z];
d = diff(Apad,1,1);
[i1,j1] = find(d==1);
[i0,j0] = find(d==-1);
lgt = i0-i1;
keep1 = lgt <= N;
keep0 = keep1 & i0 <= m;
i1 = [i1,j1];
i0 = [i0,j0];
C1 = accumarray(i1(keep1,:),1,[m n]);
C0 = accumarray(i0(keep0,:),-1,[m n]);
Aclean = cumsum(C1+C0,1);
Aclean
If you want to filter along the 2nd dimension, transpose A, apply the above, then transpose the result Aclean.
  3 Comments
Michal Kvasnicka
Michal Kvasnicka on 27 Sep 2018
Yes, you are right, the pure MATLAB Huffman encoding is not quite fast. I will test all options. Anyway, your code looks as very good method.
Thanks!

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Michal Kvasnicka
Michal Kvasnicka on 27 Sep 2018
Edited: Michal Kvasnicka on 27 Sep 2018
good idea (only 1D) is here

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