Matlab计算3D相似度转换.适用于3D的fitgeotrans [英] Matlab calculate 3D similarity transformation. fitgeotrans for 3D

问题描述

如何在MatLab中计算3D中4个点之间的相似度转换? 我可以从中计算出变换矩阵

T*X = Xp，

但是由于点坐标中的小误差，它会给我仿射矩阵.如何使该矩阵适合相似性?我需要类似fitgeotrans的东西，但是要使用3D

谢谢

解决方案

假设您在3-dimensional空间中最多包含3个点，则@rayryeng的答案是正确的.如果需要变换n-dimensional空间(m>n)中的m点，则首先需要向这些m点添加m-n坐标，以使它们存在于m-dimensional空间中(即 @rayryeng中的矩阵变为方矩阵)...然后@rayryeng描述的过程将为您提供点的精确变换，然后只需要选择原始n-dimensional空间中已变换点的坐标即可.

例如，假设您要变换点:

(2 -2 2) -> (-3 5 -4)
(2 3 0) -> (3 4 4)
(-4 -2 5) -> (-4 -1 -2)
(-3 4 1) -> (4 0 5)
(5 -4 0) -> (-3 -2 -3)

请注意，您具有n=3维的m=5点.因此，您需要向这些点添加坐标，以使它们成为n=m=5维，然后应用@rayryeng描述的过程.

我已经实现了一个功能(请在下面找到).您只需要组织这些点，以使每个源点都是矩阵u中的一列，而每个目标点就是矩阵v中的一列.矩阵u和v将分别是3×5.

警告:

• 该函数中的矩阵A可能需要大量内存用于nP的许多点，因为它具有nP^4个元素.

• 要解决此问题，对于平方矩阵u和v，您可以简单地在MATLAB表示法中使用T=v*inv(u)或T=v/u.

• 代码运行速度可能很慢...

在MATLAB中:

u = [2 2 -4 -3 5;-2 3 -2 4 -4;2 0 5 1 0]; % setting the set of source points
v = [-3 3 -4 4 -3;5 4 -1 0 -2;-4 4 -2 5 -3]; % setting the set of target points
T = findLinearTransformation(u,v); % calculating the transformation

您可以通过以下方式验证T是正确的:

I = eye(5);
uu = [u;I((3+1):5,1:5)]; % filling-up the matrix of source points so that you have 5-d points
w = T*uu; % calculating target points
w = w(1:3,1:5); % recovering the 3-d points
w - v % w should match v ... notice that the error between w and v is really small

计算转换矩阵的函数:

function [T,A] = findLinearTransformation(u,v)
% finds a matrix T (nP X nP) such that T * u(:,i) = v(:,i)
% u(:,i) and v(:,i) are n-dim col vectors; the amount of col vectors in u and v must match (and are equal to nP)
%
    if any(size(u) ~= size(v))
        error('findLinearTransform:u','u and v must be the same shape and size n-dim vectors');
    end
    [n,nP] = size(u); % n -> dimensionality; nP -> number of points to be transformed
    if nP > n % if the number of points to be transform exceeds the dimensionality of points
        I = eye(nP);
        u = [u;I((n+1):nP,1:nP)]; % then fill up the points to be transformed with the identity matrix
        v = [v;I((n+1):nP,1:nP)]; % as well as the transformed points
        [n,nP] = size(u);
    end
    A = zeros(nP*n,n*n);
    for k = 1:nP
        for i = ((k-1)*n+1):(k*n)
            A(i,mod((((i-1)*n+1):(i*n))-1,n*n) + 1) = u(:,k)';
        end
    end
    v = v(:);
    T = reshape(A\v, n, n).';
end

How can I calculate in MatLab similarity transformation between 4 points in 3D? I can calculate transform matrix from

T*X = Xp,

but it will give me affine matrix due to small errors in points coordinates. How can I fit that matrix to similarity one? I need something like fitgeotrans, but in 3D

Thanks

解决方案

The answer by @rayryeng is correct, given that you have a set of up to 3 points in a 3-dimensional space. If you need to transform m points in n-dimensional space (m>n), then you first need to add m-n coordinates to these m points such that they exist in m-dimensional space (i.e. the a matrix in @rayryeng becomes a square matrix)... Then the procedure described by @rayryeng will give you the exact transformation of points, you then just need to select only the coordinates of the transformed points in the original n-dimensional space.

As an example, say you want to transform the points:

(2 -2 2) -> (-3 5 -4)
(2 3 0) -> (3 4 4)
(-4 -2 5) -> (-4 -1 -2)
(-3 4 1) -> (4 0 5)
(5 -4 0) -> (-3 -2 -3)

Notice that you have m=5 points which are n=3-dimensional. So you need to add coordinates to these points such that they are n=m=5-dimensional, and then apply the procedure described by @rayryeng.

I have implemented a function that does that (find it below). You just need to organize the points such that each of the source-points is a column in a matrix u, and each of the target points is a column in a matrix v. The matrices u and v are going to be, thus, 3 by 5 each.

WARNING:

• the matrix A in the function may require A LOT of memory for moderately many points nP, because it has nP^4 elements.

• To overcome this, for square matrices u and v, you can simply use T=v*inv(u) or T=v/u in MATLAB notation.

• The code may run very slowly...

In MATLAB:

u = [2 2 -4 -3 5;-2 3 -2 4 -4;2 0 5 1 0]; % setting the set of source points
v = [-3 3 -4 4 -3;5 4 -1 0 -2;-4 4 -2 5 -3]; % setting the set of target points
T = findLinearTransformation(u,v); % calculating the transformation

You can verify that T is correct by:

I = eye(5);
uu = [u;I((3+1):5,1:5)]; % filling-up the matrix of source points so that you have 5-d points
w = T*uu; % calculating target points
w = w(1:3,1:5); % recovering the 3-d points
w - v % w should match v ... notice that the error between w and v is really small

The function that calculates the transformation matrix:

function [T,A] = findLinearTransformation(u,v)
% finds a matrix T (nP X nP) such that T * u(:,i) = v(:,i)
% u(:,i) and v(:,i) are n-dim col vectors; the amount of col vectors in u and v must match (and are equal to nP)
%
    if any(size(u) ~= size(v))
        error('findLinearTransform:u','u and v must be the same shape and size n-dim vectors');
    end
    [n,nP] = size(u); % n -> dimensionality; nP -> number of points to be transformed
    if nP > n % if the number of points to be transform exceeds the dimensionality of points
        I = eye(nP);
        u = [u;I((n+1):nP,1:nP)]; % then fill up the points to be transformed with the identity matrix
        v = [v;I((n+1):nP,1:nP)]; % as well as the transformed points
        [n,nP] = size(u);
    end
    A = zeros(nP*n,n*n);
    for k = 1:nP
        for i = ((k-1)*n+1):(k*n)
            A(i,mod((((i-1)*n+1):(i*n))-1,n*n) + 1) = u(:,k)';
        end
    end
    v = v(:);
    T = reshape(A\v, n, n).';
end

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