数据拟合3D空间中的椭圆-Java 学习之路

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我有一组数据显然在3D空间中形成一个椭圆（不是椭球，而是3D中的曲线） . 受到跟随线程http://au.mathworks.com/matlabcentral/newsreader/view_thread/65773的启发并在某人的帮助下，我设法运行优化代码并输出一组最佳参数x（向量） . 但是，当我尝试使用此x来复制椭圆时，结果是空间中的奇怪直线 . 我已经被困在这里好几天了，仍然不知道出了什么问题......非常沮丧......我希望有人可以对此有所了解 . 椭圆的Mathematica公式与上面的线程相同，其中

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

其中：* z是翻译向量 . * R是旋转矩阵（使用欧拉角，我们首先围绕x轴旋转αrad，然后绕y轴旋转βrad，最后再绕z轴旋转γrad） . * a是椭圆的长轴* b是椭圆的短轴 .

这是我的优化目标函数（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的主要功能（xfit.m）

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'

并持续数据点（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

1 回答

这个答案并不直接适用于3D，而是首先涉及数据的旋转，以使点的平面与xy平面重合，然后拟合2D中的数据 .

% 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

这是函数 fellipse ，基于提供的功能：

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

此外，再次注意，这可以直接在3D中变成一个拟合，但是如果您可以假设数据点大约是这样，那么这个答案也同样好 . 在2D平面中 . 当前的解决方案可能比使用附加参数的3D解决方案更有效 .

希望代码是不言自明的 . 我建议查看OP中包含的链接，它解释了循环的目的 phi .

这就是你如何检查拟合的结果：

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 . 它似乎不是很稳健，因为它对起始参数的选择非常敏感 . 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

函数fellipse3d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-')

回复于 2024-04-29T01:55:17+08:00

数据拟合3D空间中的椭圆

1 回答

编辑

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