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How to eliminate standalone 1 or 0 in binary matrix

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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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Accepted Answer

Rik
Rik on 23 Nov 2020
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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Simon Allosserie
Simon Allosserie on 23 Nov 2020
We do run into issues when we have consecutive 0 1 0 1 patterns, then the convolution just shifts it to 1 0 1 0:
1 0 0 0 0 1 1 0 1 0 1 0 1 0 1
0.333333333333333 0.333333333333333 0 0 0.333333333333333 0.666666666666667 0.666666666666667 0.666666666666667 0.333333333333333 0.666666666666667 0.333333333333333 0.666666666666667 0.333333333333333 0.666666666666667 0.333333333333333
0 0 0 0 0 1 1 1 0 1 0 1 0 1 0
Still figuring out how to clear up this one.
Rik
Rik on 23 Nov 2020
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 on 23 Nov 2020
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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More Answers (1)

Bastian Arturo Romero Mardones
Edited: Bastian Arturo Romero Mardones on 23 Nov 2020
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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Simon Allosserie
Simon Allosserie on 24 Nov 2020
@Bastian
I updated your code to be complete for an entire matrix, FYI:
for m = 1:rows
for n = 1
if (A(m,n) ~= A(m,n+1) && A(m,n) ~= A(m,n+2))
A(m,n) = abs(A(m,n)-1);
end
end
for n = 2:cols-1
if (A(m,n) ~= A(m,n+1) && A(m,n) ~= A(m,n-1))
A(m,n) = abs(A(m,n)-1);
end
end
for n = cols
if A(m,n) ~= A(m,n-1)
A(m,n) = abs(A(m,n)-1);
end
end
end
Rik
Rik on 24 Nov 2020
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
Bastian Arturo Romero Mardones
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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