How to eliminate standalone 1 or 0 in binary matrix

Question

Simon Allosserie 2020-11-23

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评论： Setsuna Yuuki. 2020-11-24

采纳的回答： Rik

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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Answer 1

Rik 2020-11-23

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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 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 2020-11-23

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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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Answer 2

Setsuna Yuuki. 2020-11-23

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编辑：Setsuna Yuuki. 2020-11-23

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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 2020-11-24

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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. 2020-11-24

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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])';