可以在MATLAB中对其进行优化以使其运行更快吗？_Matlab_Loops_Optimization_Vectorization

可以在MATLAB中对其进行优化以使其运行更快吗？

matlab loops optimization

可以在MATLAB中对其进行优化以使其运行更快吗？,matlab,loops,optimization,vectorization,Matlab,Loops,Optimization,Vectorization,这是密码 S = size(shape,3) shape = 1 - shape; 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); 也许有办法把它矢量化顺便说一句，shape和shape是由0和1组成的数组，因此它们可以以某种方式用作逻

这是密码

S = size(shape,3)
shape = 1 - shape;

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);

也许有办法把它矢量化

顺便说一句，

shape

和

shape

是由0和1组成的数组，因此它们可以以某种方式用作逻辑-

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

来自

c=1:3

，因此

vision

的第三维很可能是

。为了验证所提出的方法是否有效，让我们将其保持为

。另外，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。为了进行验证，最后我们将发现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代码-

% 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

这里还有一个解决方案-

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