I have a 3D matrix that consists of 3 matrices 500x500 elements. Now, I wanna take the third matrix and replace all its values that are, let's say, bigger than 100 with 0. If I have a matrix a, my code would simply be:
a(a>100)=0
However, in my case I need to take the third matrix of my 3D matrix, which would be a(:,:,3). If I now try to use the same code:
a(:,:,3)(a(:,:,3)>100)=0
I get the message "()-indexing must appear last in an index expression."
Any idea on how I can express that?
What about
a(:,:,3) = (a(:,:,3)<100).*a(:,:,3);
?
You can use linear indexes for that:
id = find(A(:,:,3)>100)+2*size(A, 1)*size(A, 2);
A(id)=0
Alternatively, you can reshape
array A
to 2D and:
AA = reshape(A, 500*500, 3);
AA(AA(:,3)>100,3) = 0;
A = reshape(AA, 500, 500,3);
which uses the original code of Acorbe, but works for 2D matrices in contrast to 3D :)
Just to add another alternative:
A(cat(3, false(size(A,1),size(A,2),2), A(:,:,3)>100)) = 0;
Alternatively, you can assign an index variable in 3D like so:
id(:,:,3) = A(:,:,3)>100;
A(id) = 0;
which has a much more cleaner syntax.
Now for some speed tests:
clc, clear all
b = 250*rand(500,500, 3);
% Me 1
tic
for ii = 1:1e2
A=b;
clear id
id = cat(3, false(size(A,1),size(A,2),2), A(:,:,3)>100);
A(id) = 0;
end
toc
% Acorbe
tic
for ii = 1:1e2
A=b;
A(:,:,3) = (A(:,:,3)<100).*A(:,:,3);
end
toc
% angainor 1
tic
for ii = 1:1e2
A=b;
clear id
id = find(A(:,:,3)>100) + 2*size(A, 1)*size(A, 2);
A(id)=0;
end
toc
% Me 2
tic
for ii = 1:1e2
A=b;
clear id
id(:,:,3) = A(:,:,3)>100;
A(id) = 0;
end
toc
% angainor 2
tic
for ii = 1:1e2
A=b;
clear id
AA = reshape(A, [], 3);
AA(AA(:,3)>100,3) = 0;
A = reshape(AA, size(A,1), size(A,2), 3);
end
toc
Results:
Elapsed time is 1.612787 seconds. % me #1
Elapsed time is 1.223496 seconds. % Acorbe
Elapsed time is 1.606858 seconds. % angainor #1
Elapsed time is 1.510153 seconds. % me #2
Elapsed time is 0.964423 seconds. % angainor #2
Seems the winner is angainor :)
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