这里有一个 更多 bsxfun 解决方案 -
S = size(shape,3);
shape = 1 - shape;
SHAPE = cumprod(shape(:,:,1:S-1),3);
vision(:,:,1:3,2:S) = bsxfun(@times,vision(:,:,1:3,2:S),permute(SHAPE,[1 2 4 3]));
output = sum(vision,4);
测试
由于代码具有vision(:,:,c,i) 并且迭代器c 来自c = 1:3,因此vision 的第三维很可能是3。为了验证提议的方法是否适用,我们将其保留为5。此外,为了进行适当的基准测试,让我们在其他维度上使用大数字,并在其中使用随机数。为了验证,最后我们会找到提议的方法和原始方法的输出之间的绝对最大差异。
基准测试和输出验证码-
% Inputs
shape = rand(150,160,170);
vision = rand(150,160,5,170);
shape = 1 - shape;
S = size(shape,3);
%// Proposed solution :
disp('----------------------- With Proposed solution')
tic
V = vision; %// Make a copy for using with proposed solution
SHAPE = cumprod(shape(:,:,1:S-1),3);
V(:,:,1:3,2:S) = bsxfun(@times,V(:,:,1:3,2:S),permute(SHAPE,[1 2 4 3]));
out = sum(V,4);
toc
%// Original solution :
disp('----------------------- With Original solution')
tic
S = size(shape,3);
for i = 2:S
SHAPE = prod(shape(:,:,1:i-1),3);
for c = 1:3
vision(:,:,c,i) = vision(:,:,c,i).*SHAPE;
end
end
output = sum(vision,4);
toc
error_value = max(abs(output(:) - out(:)))
命令输出 -
----------------------- With Proposed solution
Elapsed time is 0.802486 seconds.
----------------------- With Original solution
Elapsed time is 4.401897 seconds.
error_value =
0