MATLAB Answers

## elimination of consecutive regions

Asked by Michal Kvasnicka

### Michal Kvasnicka (view profile)

on 27 Sep 2018
Latest activity Commented on by Michal Kvasnicka

### Michal Kvasnicka (view profile)

on 29 Sep 2018
Accepted Answer by Matt J

### Matt J (view profile)

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).

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R2018a

## 3 Answers

Answer by Matt J

on 27 Sep 2018
Edited by Matt J

### Matt J (view profile)

on 27 Sep 2018
Accepted Answer

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;

Bruno Luong

### Bruno Luong (view profile)

on 27 Sep 2018
The above code is not working for N odd (e.g. N=1)
Matt J

### Matt J (view profile)

on 27 Sep 2018
I've modified it to handle odd N.
Michal Kvasnicka

### Michal Kvasnicka (view profile)

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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### Bruno Luong (view profile)

Answer by Bruno Luong

### Bruno Luong (view profile)

on 27 Sep 2018
Edited by Bruno Luong

### Bruno Luong (view profile)

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.

Michal Kvasnicka

### Michal Kvasnicka (view profile)

on 27 Sep 2018
Does it mean, that use of Huffman encoding brings some performance improvement?
Bruno Luong

### Bruno Luong (view profile)

on 27 Sep 2018
It depends how it's implemented. If it's a C-MEX file might be, pure MATLAB Huffman, no chance.
Otherwise my code is probably quite fast (but it creates few big intermediate arrays, you might add "clear...) while the code is running when an array is finished to be used).
Why not test yourself with different methods the link you have found?
Michal Kvasnicka

### Michal Kvasnicka (view profile)

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 (view profile)

Answer by Michal Kvasnicka

### Michal Kvasnicka (view profile)

on 27 Sep 2018
Edited by Michal Kvasnicka

### Michal Kvasnicka (view profile)

on 27 Sep 2018

good idea (only 1D) is here

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