How to eliminate standalone 1 or 0 in binary matrix

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Simon Allosserie
Simon Allosserie 2020-11-23
评论： Setsuna Yuuki. 2020-11-24
Hello
I have large binary matrices (order 30000 x 5000 values) in which I have to eliminate stand-alone 1's or 0's.
E.g. when a row looks like this: 0 0 1 1 1 0 1 1 0 0 1 0 1 1 1
It should be adapted to: 0 0 1 1 1 1 1 1 0 0 0 0 1 1 1
Or thus, when there's only one 1 between some 0's or one 0 between some 1's, it should be changed to a 0 or 1 respectively.
I have no clue how to do this except from running through the entire matrix and keeping count of the length of the series of 1's or 0's - which seems utterly inefficiënt to me. Any ideas on functions or better ways to tackle this? Thanks!

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Rik
Rik 2020-11-23
Ran in:
I would suggest using a convolution with a flat kernel.
kernel=ones(1,3,1);kernel=kernel/sum(kernel(:));
A=[0 0 1 1 1 0 1 1 0 0 1 0 1 1 1];
B=convn(A,kernel,'same');
C=round(B);
clc,disp([A;B;C])%display the original, the convolution result and the final output
0 0 1.0000 1.0000 1.0000 0 1.0000 1.0000 0 0 1.0000 0 1.0000 1.0000 1.0000 0 0.3333 0.6667 1.0000 0.6667 0.6667 0.6667 0.6667 0.3333 0.3333 0.3333 0.6667 0.6667 1.0000 0.6667 0 0 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0 0 0 1.0000 1.0000 1.0000 1.0000
Note that the [0 0 1 0 1 1] is changed to [0 0 0 1 1 1]. If you don't want that, you can change the kernel to something like this:
kernel=[1 1 1 0 0];kernel=kernel/sum(kernel(:));
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Rik
Rik 2020-11-23
The more fundamental question is what the values should be. If you have a clear description of that, I might be able to help you find the right kernel.
Also note that this kernel only considers rows. It will work on a 2D array as well, but it will not check multiple rows at once. You can change the kernel to 2D if you want to change that.
Simon Allosserie
Simon Allosserie 2020-11-23
You're right, Rik.
The practical implementation is a laser that cannot switch on/off for just one pixel. So, one on/'1' in a series of off pixels or one zero/'0' pixel in a series of on pixels, should be eliminated
That means that we can keep looking in 1D rows, going from left to right.
The values should then be as follows:
1 0 0 1 1 0 1 0 1 0 0 %orig
0 0 0 1 1 1 1 1 1 0 0 %after convolution
or
0 0 1 0 0 1 0 1 0 1 1 %orig
0 0 0 0 0 0 0 0 0 1 1 %after convolution
in other words, the 0101 sequence adapts to the series of 1's or 0's before it.
Am I making it clear enough in this way? THanks for your help!

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Setsuna Yuuki.
Setsuna Yuuki. 2020-11-23
编辑：Setsuna Yuuki. 2020-11-23
You only need a loop and correct conditional statement, the conditional can be:
if(A(n) ~= A(n+1) && A(n) ~= A(n-1) && A(n) == 1)
A(n) = 0;
elseif(A(n) ~= A(n+1) && A(n) ~= A(n-1) && A(n) == 0)
A(n) = 1;
end
so you compare only with the previous bit and the next.
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Rik
Rik 2020-11-24
This code can be simplified (or at least be edited to remove duplicated code):
%option 1:
sz=size(A);
for m = 1:sz(1)
for n=1:sz(2)
if ...
( n==1 && ...
(A(m,n) ~= A(m,2 ) && A(m,n) ~= A(m,3 )) ) || ...
( n==sz(2) && ...
( A(m,n) ~= A(m,end-1)) ) || ...
( ( n~=1 && n~=sz(2) ) && ...
(A(m,n) ~= A(m,n+1) && A(m,n) ~= A(m,n-1 )) )
A(m,n) = abs(A(m,n)-1);
end
end
end
%option 2:
sz=size(A);
for m = 1:sz(1)
for n=1:sz(2)
if n==1
L = (A(m,n) ~= A(m,2 ) && A(m,n) ~= A(m,3 )) ;
elseif n==sz(2)
L = A(m,n) ~= A(m,end-1)) ;
else
L = (A(m,n) ~= A(m,n+1) && A(m,n) ~= A(m,n-1 )) ;
end
if L
A(m,n) = abs(A(m,n)-1);
end
end
end
Setsuna Yuuki.
Setsuna Yuuki. 2020-11-24
I did it this way:
A = [0 0 1 1 1 0 1 1 0 ;0 0 0 1 0 1 0 1 1]';
[r,c,~]=size(A);
A = reshape(A,[1, r*c]);
for n = 2:r*c-1
if(A(n) ~= A(n+1) && A(n) ~= A(n-1) && A(n) == 1)
A(n) = 0;
elseif(A(n) ~= A(n+1) && A(n) ~= A(n-1) && A(n) == 0)
A(n) = 1;
end
end
A = reshape(A,[r,c])';

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