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 Kommentare

Jan
Jan am 25 Apr. 2022
What is the wanted output for this example?
WEN SHIN LU
WEN SHIN LU am 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.

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 Akzeptierte Antwort

Matt J
Matt J am 25 Apr. 2022

0 Stimmen

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

Weitere Antworten (1)

Matt J
Matt J am 25 Apr. 2022
Bearbeitet: Matt J am 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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Gefragt:

am 25 Apr. 2022

Beantwortet:

am 25 Apr. 2022

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