Question

由于for循环次数过多，我经常会遇到缓慢的Matlab处理问题。也许你可以帮助我这个例子，这在其他情况下也会帮助我：是否可以将它完全转移到矩阵中？

R = [1 2 3; 1 2 3; 1 2 3];
u1 = 200; u2 = 300; v1 = 100; v2 = 222; w1 = 123; w2 = 312;
px = 20; py = 30; pz = 25;
a = 20; b = 30; c = 25;
Ellipse = zeros(500,600,500)

for i=u1:u2
    for j=v1:v2
        for k=w1:w2
            x=[i-px;j-py;k-pz];
            x=R*x;
            if (x(1)/a)^2+(x(2)/b)^2+(x(3)/c)^2<1                      
                Ellipse(i,j,k)=1;
            end
        end
    end
end

Answer 1

这是一种运行速度快2个数量级的方法

R = [1 2 3; 1 2 3; 1 2 3];
u1 = 200; u2 = 300; v1 = 100; v2 = 222; w1 = 123; w2 = 312;
px = 20; py = 30; pz = 25;
a = 20; b = 30; c = 25;
Ellipse = zeros(500,600,500);

% Create mesh of points to process
vx = (u1:u2) - px;
vy = (v1:v2) - py;
vz = (w1:w2) - pz;
[X,Y,Z] = meshgrid(vx, vy, vz);

% Compute R*x
V = [X(:)'; Y(:)'; Z(:)'];
V = R * V;

% Divide by ellipse axes
M = diag([1/a, 1/b, 1/c]);
V = M * V;

% Determine if norm criteria is met
index = norm(V') < 1;
Ellipse(X(index), Y(index), Z(index)) = 1;

同意@Dan，一个较小的工作示例会使测试变得更容易。

Answer 2

如果你做的不仅仅是简单的数学运算，那么矩阵可能会很棘手。在这里你可以（应该）减小Ellipse的大小，因为你没有使用它的大部分条目。你所需要的只是（101,123,190）并且事情会快得多，特别是你访问矩阵条目的方式，记住MATLAB最终会将矩阵存储为一个数组。

Answer 3

@dpmcmlxxvi非常感谢您的回答。首先，抱歉这个大例子。我使用了你的答案进行了一些修改，因为norm(XY)对我来说不起作用。（较小的例子和值适合于正常运行）

R = [1 2 3; 1 2 3; 1 2 3];
u1 = 50; u2 = 90; v1 = 20; v2 = 50; w1 = 30; w2 = 60;
px = 60; py = 25; pz = 50;
a = 10; b = 20; c = 15;
Ellipse = zeros(100,200,100);

% Create mesh of points to process
vx = (u1:u2) - px;
vy = (v1:v2) - py;
vz = (w1:w2) - pz;
[X,Y,Z] = meshgrid(vx, vy, vz);

% Compute R*x
V = [X(:)'; Y(:)'; Z(:)'];
V = R * V;

% Divide by ellipse axes
M = diag([1/a, 1/b, 1/c]);
V = M * V;

% Apply criteria
Vsq = V.*V;
crit = Vsq(1,:)+Vsq(2,:)+Vsq(3,:);

% Look for indices in the meshgrid fullfilling criteria
xe = X(find(crit<1));
ye = Y(find(crit<1));
ze = Z(find(crit<1));

xi = ceil(abs(xe+px));
yi = ceil(abs(ye+py));
zi = ceil(abs(ze+pz));

% Set values at correpsonding indices to 1
linind = sub2ind(size(Ellipse),xi,yi,zi);

Ellipse(linind) = 1;

它工作正常，所有for循环都已删除。

3D / 4D矩阵操作

3 个答案: