在3D空间中拟合椭圆的数据

Question

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我有一组数据，这些数据显然在3D空间中形成了一个椭圆（不是椭圆体，而是3D中的曲线）。 受到以下线程http://au.mathworks.com/matlabcentral/newsreader/view_thread/65773的启发并在某人的帮助下，我设法使优化代码运行并输出一组最佳参数x（向量）。 但是，当我尝试使用此x复制椭圆时，结果是空间中一条奇怪的直线。 我已经被这个问题堆积了好几天了。仍然不知道出了什么问题。。。。。。。我希望有人能对此有所启发。 椭圆的Mathematica公式与上述螺纹相同，其中

3D椭圆由下式给出：（x; y; z）=（z1; z2; z3）+ R（alpha，beta，gamma）。 （a cos（phi）; b * sin（phi）; 0）

其中：* z是转换向量。 * R是旋转矩阵（使用欧拉角，我们首先绕着x轴旋转alpha rad，然后绕着y轴旋转beta rad，最后再次绕着z轴旋转gamma rad）。 * a是椭圆的长轴* b是椭圆的短轴。

这是我要优化的目标函数（ellipsefit.m）

function [merit]= ellipsefit(x, vmatrix) % x is the initial parameters, vmatrix stores the datapoints
load vmatrix.txt % In vmatrix, the data are stored: N rows x 3 columns
a = x(1);
b = x(2);c
alpha = x(3);
beta = x(4);
gamma = x(5);
z = [x(6),x(7),x(8)];
t = z'
[dim1, dim2]=size(vmatrix);
% construction of rotation matrix R(alpha,beta,gamma)
R1 = [cos(alpha), sin(alpha), 0; -sin(alpha), cos(alpha), 0; 0, 0, 1];v
R2 = [1, 0, 0; 0, cos(beta), sin(beta); 0, -sin(beta), cos(beta)];
R3 = [cos(gamma), sin(gamma), 0; -sin(gamma), cos(gamma), 0; 0, 0, 1];
R = R3*R2*R1;
% first  compute  vector phi (in the length of the data) by minimizing for every
% point in the data set the distance of this point to the ellipse
% (with its initial parameters a,b,alpha,beta,gamma, z1,z2,z3 held fixed) with respect to phi.
for i=1:dim1
point=vmatrix(i,:)';
dist=@(phi)sum((R*[a*cos(phi); b*sin(phi); 0]+t-point)).^2; 
phi(i)=fminbnd(dist,0,2*pi);
end
v = [a*cos(phi); b*sin(phi); zeros(size(phi))];
P = R*v;
%The targetfunction is: g = (xi1,xi2,xi3)' -(z1,z2,z3)'-R(alpha,beta,gamma)(a cos(phi), b sin(phi),0)'
% Construction of distance function
merit = [vmatrix(:,1)-z(1)-P(1),vmatrix(:,2)-z(2)-P(2),vmatrix(:,3)-z(3)-P(3)]; 
merit = sqrt(sum(sum(merit.^2)))
end

这是参数初始化和调用opts（xfit.m）的主要功能

function [y] = xfit (x)
x= [1 1 1 1 1 1 1 1] % initial parameters
[x] = fminsearch(@ellipsefit,x) % set the distance minimum as the target function
y=x
end

用于在散点中重建椭圆的代码（ellipsescatter.txt）

x= [0.655,0.876,1.449,2.248,1.024,0.201,-0.11,0.002] % values obtained according to above routines
a = x(1);
b = x(2);
alpha = x(3);
beta = x(4);
gamma = x(5);
z = [x(6),x(7),x(8)];
R1 = [cos(alpha), sin(alpha), 0; -sin(alpha), cos(alpha), 0; 0, 0, 1];
R2 = [1, 0, 0; 0, cos(beta), sin(beta); 0, -sin(beta), cos(beta)];
R3 = [cos(gamma), sin(gamma), 0; -sin(gamma), cos(gamma), 0; 0, 0, 1];
R = R3*R2*R1;
phi=linspace(0,2*pi,100)
v = [a*cos(phi); b*sin(phi); zeros(size(phi))];
P = R*v;
u = P'

最后的数据点（vmatrix）

0.002037965 0.004225765 0.002020202
0.005766671 0.007269193 0.004040404
0.010004802 0.00995638  0.006060606
0.014444336 0.012502725 0.008080808
0.019083408 0.014909533 0.01010101
0.023967745 0.017144827 0.012121212
0.03019849  0.01969697  0.014591289
0.038857283 0.022727273 0.017839321
0.045443501 0.024730475 0.02020202
0.051213405 0.026346492 0.022222222
0.061038174 0.028787879 0.02555121
0.069408829 0.030575164 0.028282828
0.075785861 0.031818182 0.030321465
0.088818543 0.033954681 0.034343434
0.095538223 0.03490652  0.036363636
0.109421234 0.036499949 0.04040404
0.123800737 0.037746182 0.044444444
0.131206601 0.038218171 0.046464646
0.146438211 0.038868525 0.050505051
0.162243245 0.039117883 0.054545455
0.178662839 0.03893748  0.058585859
0.195740664 0.038296774 0.062626263
0.204545539 0.037790433 0.064646465
0.222781268 0.036340005 0.068686869
0.23715887  0.034848485 0.071748051
0.251787024 0.033009003 0.074747475
0.26196429  0.031542949 0.076767677
0.278510276 0.028787879 0.079919236
0.294365342 0.025757576 0.082799669
0.306221108 0.023197784 0.084848485
0.31843759  0.020305704 0.086868687
0.331291367 0.016967964 0.088888889
0.342989936 0.013636364 0.090622484
0.352806191 0.010606061 0.091993214
0.36201461  0.007575758 0.093211986
0.376385537 0.002386324 0.094949495
0.386214665 -0.001515152    0.096012
0.396173756 -0.005800677    0.096969697
0.406365393 -0.010606061    0.097799682
0.417897899 -0.016666667    0.098516141
0.428059375 -0.022727273    0.098889844
0.436894505 -0.028787879    0.09893196
0.444444123 -0.034848485    0.098652697
0.45074522  -0.040909091    0.098061305
0.455830971 -0.046969697    0.097166076
0.457867157 -0.05   0.096591789
0.46096663  -0.056060606    0.095199991
0.461974832 -0.059090909    0.094368708
0.462821268 -0.063709158    0.092929293
0.46279206  -0.068181818    0.091323015
0.462224312 -0.071212121    0.090097745
0.461247257 -0.074242424    0.088770148
0.459194871 -0.07812596 0.086868687
0.456406121 -0.0818267  0.084848485
0.45309565  -0.085162601    0.082828283
0.449335762 -0.088184223    0.080808081
0.445185841 -0.090933095    0.078787879
0.440695103 -0.093443633    0.076767677
0.435904796 -0.095744683    0.074747475
0.429042582 -0.098484848    0.072052312
0.419877272 -0.101489369    0.068686869
0.41402731  -0.103049401    0.066666667
0.407719192 -0.104545455    0.064554798
0.395265308 -0.106881864    0.060606061
0.388611992 -0.107880111    0.058585859
0.374697979 -0.10945186 0.054545455
0.360058411 -0.11051623 0.050505051
0.352443612 -0.11084211 0.048484848
0.336646801 -0.111097219    0.044444444
0.320085063 -0.110817414    0.04040404
0.31150078  -0.110465333    0.038383838
0.293673303 -0.109300395    0.034343434
0.275417637 -0.107575758    0.030396076
0.265228963 -0.106361993    0.028282828
0.251914589 -0.104545455    0.025603647
0.234385536 -0.101745907    0.022222222
0.223443994 -0.099745394    0.02020202
0.212154519 -0.097501571    0.018181818
0.20046153  -0.09497557 0.016161616
0.188298809 -0.092121085    0.014141414
0.17558878  -0.088883868    0.012121212
0.162241674 -0.085201142    0.01010101
0.148154337 -0.081000773    0.008080808
0.136529019 -0.077272727    0.006507255
0.127611912 -0.074242424    0.005361311
0.116762238 -0.070350086    0.004040404
0.103195122 -0.065151515    0.002507114
0.095734279 -0.062121212    0.001725236
0.081719652 -0.056060606    0.000388246
0   0   0

Answer 1

这个答案不是3D的直接拟合，而是首先旋转数据，使点的平面与xy平面重合，然后再拟合2D的数据。

% input: data, a N x 3 array with one set of Cartesian coords per row

% remove the center of mass
CM = mean(data);
datap = data - ones(size(data,1),1)*CM;


% now rotate all points into the xy plane ...
% start by finding the plane:

[u s v]=svd(datap);

% rotate the data into the principal axes frame:

datap = datap*v;


% fit the equation for an ellipse to the rotated points

x= [0.25 0.07 0.037 0 0]'; % initial parameters    
options=1;
xopt = fmins(@fellipse,x,options,[],datap) % set the distance minimum as the target function

这是基于提供的功能的fellipse函数：

function [merit]= fellipse(x,data) % x is the initial parameters, data stores the datapoints

a = x(1);
b = x(2);
alpha = x(3);    
z = x(4:5);

R = [cos(alpha), sin(alpha), 0; -sin(alpha), cos(alpha), 0; 0, 0, 1];
data = data*R; 

merit = 0;

[dim1, dim2]=size(data);
for i=1:dim1
    dist=@(phi)sum( ( [a*cos(phi);b*sin(phi)] + z - data(i,1:2)').^2 ); 
    phi=fminbnd(dist,0,2*pi);
    merit = merit+dist(phi);
end

end

另外，请再次注意，可以直接将其转换为3D拟合，但是如果您可以假设数据点位于大约3D范围内，则此答案也一样好。 在2D平面中。 当前的解决方案可能比带有附加参数的3D解决方案效率更高。

希望代码是不言自明的。 我建议查看OP中包含的链接，它说明了phi循环的目的。

这就是您可以检查拟合结果的方式：

a = xopt(1);
b = xopt(2);
alpha = xopt(3);
z = [xopt(4:5) ; 0]';

phi = linspace(0,2*pi,100)';
simdat = [a*cos(phi) b*sin(phi) zeros(size(phi))];
R = [cos(alpha), -sin(alpha), 0; sin(alpha), cos(alpha), 0; 0, 0, 1];
simdat = simdat*R  + ones(size(simdat,1), 1)*z ; 


figure, hold on
plot3(datap(:,1),datap(:,2),datap(:,3),'o')
plot3(simdat(:,1),simdat(:,2),zeros(size(simdat,1),1),'r-')

---

编辑

以下是3D方法。 它似乎对启动参数的选择非常敏感，因此并不十分健壮。 可能需要一些改进。

CM = mean(data);
datap = data - ones(size(data,1),1)*CM;
xopt = [  0.07 0.25 1 -0.408976 0.610120 0 0  0]';
options=1;
xopt = fmins(@fellipse3d,xopt,options,[],datap) % set the distance minimum as the target function

功能fallipse3d是

function [merit]= fellipse3d(x,data) % x is the initial parameters, data stores the datapoints

a = abs(x(1));
b = abs(x(2));
alpha = x(3);
beta = x(4);
gamma = x(5);
z = x(6:8)';

[dim1, dim2]=size(data);

R1 = [cos(alpha), sin(alpha), 0; -sin(alpha), cos(alpha), 0; 0, 0, 1];
R2 = [1, 0, 0; 0, cos(beta), sin(beta); 0, -sin(beta), cos(beta)];
R3 = [cos(gamma), sin(gamma), 0; -sin(gamma), cos(gamma), 0; 0, 0, 1];
R = R3*R2*R1;

data = (data - z(ones(dim1,1),:))*R; 

merit = 0;
for i=1:dim1
    dist=@(phi)sum( ([a*cos(phi);b*sin(phi);0]  - data(i,:)').^2 ); 
    phi=fminbnd(dist,0,2*pi);
    merit = merit+dist(phi);
end
end

您可以使用以下方式可视化结果

a = xopt(1);
b = xopt(2);
alpha = -xopt(3);
beta = -xopt(4);
gamma = -xopt(5);
z = xopt(6:8)' + CM;

dim1 = 100;
phi = linspace(0,2*pi,dim1)';

simdat = [a*cos(phi) b*sin(phi) zeros(size(phi))];

R1 = [cos(alpha), sin(alpha), 0; ...
     -sin(alpha), cos(alpha), 0; ... 
        0, 0, 1];

R2 = [1, 0, 0;  ...
      0, cos(beta), sin(beta);  ...
      0, -sin(beta), cos(beta)];

R3 = [cos(gamma), sin(gamma), 0;  ...
      -sin(gamma), cos(gamma), 0;  ...
           0, 0, 1];

R = R1*R2*R3;

simdat = simdat*R + z(ones(dim1,1),:); 

figure, hold on
plot3(data(:,1),data(:,2),data(:,3),'o')
plot3(simdat(:,1),simdat(:,2),simdat(:,3),'r-')