通过 2d 平面绕一个轴旋转生成柱面对称 3D 数据

Question

根据之前提出的问题，我想创建一个 3d 体积，即 ，f(x,y,z)从 2D 矩阵开始I(x,y)，而不仅仅是一条曲线。

例如，假设I=peaks(10)，如何绕其中一个轴（例如 y 轴）旋转它以获得 3D 矩阵？如果我有的话会更容易吗I(r,theta)？我可以在 3D 中旋转平面，但这不会是 3D 矩阵的一部分，只是新的 x、y、z 坐标。

Answer 1

我认为检查预期 3d 数组的索引，在几何上找到I(x,y)它们对应的值（是的，本质上是笛卡尔到圆柱变换），然后根据需要填充每个值是非常简单的。

这是我的意思的最小版本。假设大小为 的输入数组[N,N]对应于中的坐标x, 。假设输入数组沿平面定向，并绕轴旋转以生成输出数组。因此的维度对应于沿和以及沿。因此 的维度是。y0:N-1xzzVV-(N-1):N-1xy0:N-1zV[2*N-1, 2*N-1, N]

演示两种方法（分别使用griddata和interp2），并完成可重复性绘图：

% size of the problem
N = 10;

% input data
I = peaks(N);

% two sets of output V for two methods
V1 = zeros(2*N-1,2*N-1,N);
V2 = zeros(2*N-1,2*N-1,N);    
[i1,i2,i3] = ndgrid(1:2*N-1,1:2*N-1,1:N);
% [i1(:), i2(:), i3(:)] are the contiguous indices of V
% z dimension is the same as of I: rotate around z axis

% it will be assumed that input 1:N span elements from 0 to N-1
% output V spans -(N-1):N-1 along x and y
x = i1-N; % -(N-1):N-1
y = i2-N; % -(N-1):N-1
z = i3-1; %      0:N-1
% input array I is in xz plane, rotated along z axis, geometrically speaking

% identify the cylindrical coordinates of each voxel [i1,i2,i3]
[~,r_out,z_out] = cart2pol(x,y,z); % theta is redundant; z_out===z

% identify the coordinates of each input pixel with the above
[j1,j2] = meshgrid(1:N,1:N);
r_in = j1-1; % Cartesian input x <-> cylindrical output r
z_in = j2-1; % Cartesian input y <-> cylindrical output z
% note that j1 and j2 are swapped with respect to x and y
% but this is what interp2 will expect later

% interpolate each voxel based on r and z
method = 'nearest'; %probably the least biased
%method = 'cubic'; %probably the prettiest

V1(:) = griddata(r_in,z_in,I,...
        r_out(:),z_out(:),method);

V2(:) = interp2(r_in,z_in,I,...
        r_out(:),z_out(:),method,...
        0); % extrapolation value, otherwise NaNs appear outside


% plot two slices: xz plane and general rotated one
figure;
% generate rotated versions of the xz plane by rotating with phi around z
for phi=[0, -25, -90]/180*pi
    [xp0,zp] = meshgrid(-(N-1):0.1:N-1,0:0.1:N-1);
    xp = xp0*cos(phi);
    yp = xp0*sin(phi);

    subplot(121);
    slice(y,x,z,V1,xp,yp,zp);
    title('griddata nearest');
    shading flat;
    axis equal vis3d;
    hold on;

    subplot(122);
    slice(y,x,z,V2,xp,yp,zp);
    title('interp2 nearest, extrap 0');
    shading flat;
    axis equal vis3d;
    hold on;
end

正如您所看到的，slice直接在 中绘制数据V，因此这是生成的 3d 数组的准确表示。

作为参考，这里是输入的单个样本I = peaks(10)：