[英]Replacing elements of a 3D matrix in MATLAB
我有一個3D矩陣,由3個矩陣500x500元素組成。 現在,我想取第三個矩陣並用0代替它所有的值,比如大於100。如果我有一個矩陣a,我的代碼就是:
a(a>100)=0
但是,在我的情況下,我需要采用我的3D矩陣的第三個矩陣,它將是(:,:,3)。 如果我現在嘗試使用相同的代碼:
a(:,:,3)(a(:,:,3)>100)=0
我收到消息“() - 索引必須出現在索引表達式的最后。”
關於我如何表達的任何想法?
關於什么
a(:,:,3) = (a(:,:,3)<100).*a(:,:,3);
?
您可以使用線性索引:
id = find(A(:,:,3)>100)+2*size(A, 1)*size(A, 2);
A(id)=0
或者,您可以將陣列A
reshape
為2D並且:
AA = reshape(A, 500*500, 3);
AA(AA(:,3)>100,3) = 0;
A = reshape(AA, 500, 500,3);
它使用Acorbe的原始代碼,但與3D相比,適用於2D矩陣:)
只是添加另一種選擇:
A(cat(3, false(size(A,1),size(A,2),2), A(:,:,3)>100)) = 0;
或者,您可以像這樣在3D中分配索引變量:
id(:,:,3) = A(:,:,3)>100;
A(id) = 0;
它的語法更清晰。
現在進行一些速度測試:
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
結果:
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
似乎勝利者是angainor :)
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