Set same elements in a matrix to zero

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WEN SHIN LU
WEN SHIN LU il 25 Apr 2022
Risposto: Matt J il 25 Apr 2022
Hi,
I have a large matrix which has duplicate elements in neighborhood.
For example,
A = [0 0 0 01 1 0 0 0 2 2 2 0 0 0 5 5 5 5 5;
0 0 0 1 1 0 0 0 0 0 2 2 0 0 0 5 0 5 0 5;
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 5 0 5 0]
I have to keep just a unique element in an arbitrary position. Could I use loop to solve this problem?
  2 Commenti
Jan
Jan il 25 Apr 2022
What is the wanted output for this example?
WEN SHIN LU
WEN SHIN LU il 25 Apr 2022
sorry didn't explain clearly. The output can be:
A = [0 0 0 01 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
I want to keep just one repeated element in its neighborhood.

Accedi per commentare.

Risposta accettata

Matt J
Matt J il 25 Apr 2022
Without the Image Processing Toolbox,
A = [0 0 0 0 1 1 0 0 0 2 2 2 0 0 0 5 5 5 5 5;
0 0 0 1 1 0 0 0 0 0 2 2 0 0 0 5 0 5 0 5;
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 5 0 5 0]
A = 3×20
0 0 0 0 1 1 0 0 0 2 2 2 0 0 0 5 5 5 5 5 0 0 0 1 1 0 0 0 0 0 2 2 0 0 0 5 0 5 0 5 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 5 0 5 0
[I,J,S]=find(A);
IJS=num2cell(splitapply(@(x) x(1,:), [I,J,S],findgroups(S)),1);
[I,J,S]=deal(IJS{:});
B=accumarray([I,J],S,size(A))
B = 3×20
0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 5 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Più risposte (1)

Matt J
Matt J il 25 Apr 2022
Modificato: Matt J il 25 Apr 2022
Something like this, perhaps?
A = [0 0 0 0 1 1 0 0 0 2 2 2 0 0 0 5 5 5 5 5;
0 0 0 1 1 0 0 0 0 0 2 2 0 0 0 5 0 5 0 5;
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 5 0 5 0]
A = 3×20
0 0 0 0 1 1 0 0 0 2 2 2 0 0 0 5 5 5 5 5 0 0 0 1 1 0 0 0 0 0 2 2 0 0 0 5 0 5 0 5 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 5 0 5 0
reg=regionprops(A~=0,'PixelIdxList');
B=zeros(size(A));
for i=1:numel(reg),
j=reg(i).PixelIdxList(1);
B(j)=A(j);
end
B
B = 3×20
0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 5 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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