围绕MATLAB中的数据进行椭圆化 [英] Ellipse around the data in MATLAB

问题描述

我想在MATLAB中复制下图:

I would like to reproduce the following figure in MATLAB:

有两类具有X和Y坐标的点.我想用一个带有标准偏差参数的椭圆来包围每个类，该参数确定椭圆将沿着轴走多远.

There are two classes of points with X and Y coordinates. I'd like to surround each class with an ellipse with one parameter of standard deviation, which determine how far the ellipse will go along the axis.

该图是使用其他软件创建的，我不完全了解它是如何计算椭圆的.

The figure was created with another software and I don't exactly understand how it calculates the ellipse.

这是我用于此图的数据.第一列是类，第二列-X，第三列-Y.我可以使用gscatter来绘制点本身.

Here is the data I'm using for this figure. The 1st column is class, 2nd - X, 3rd - Y. I can use gscatter to draw the points itself.

A = [
    0   0.89287 1.54987
    0   0.69933 1.81970
    0   0.84022 1.28598
    0   0.79523 1.16012
    0   0.61266 1.12835
    0   0.39950 0.37942
    0   0.54807 1.66173
    0   0.50882 1.43175
    0   0.68840 1.58589
    0   0.59572 1.29311
    1   1.00787 1.09905
    1   1.23724 0.98834
    1   1.02175 0.67245
    1   0.88458 0.36003
    1   0.66582 1.22097
    1   1.24408 0.59735
    1   1.03421 0.88595
    1   1.66279 0.84183
];

gscatter(A(:,2),A(:,3),A(:,1))

仅供参考，此处是关于如何绘制椭圆的SO问题.因此，我们只需要知道所有参数即可绘制它.

FYI, here is the SO question on how to draw ellipse. So, we just need to know all the parameters to draw it.

更新:

我同意可以将中心计算为X和Y坐标的平均值.可能我必须对每个类使用主成分分析(PRINCOMP)来确定角度和形状.还在想...

I agree that the center can be calculated as the means of X and Y coordinates. Probably I have to use principal component analysis (PRINCOMP) for each class to determine the angle and shape. Still thinking...

推荐答案

考虑代码:

%# generate data
num = 50;
X = [ mvnrnd([0.5 1.5], [0.025 0.03 ; 0.03 0.16], num) ; ...
      mvnrnd([1 1], [0.09 -0.01 ; -0.01 0.08], num)   ];
G = [1*ones(num,1) ; 2*ones(num,1)];

gscatter(X(:,1), X(:,2), G)
axis equal, hold on

for k=1:2
    %# indices of points in this group
    idx = ( G == k );

    %# substract mean
    Mu = mean( X(idx,:) );
    X0 = bsxfun(@minus, X(idx,:), Mu);

    %# eigen decomposition [sorted by eigen values]
    [V D] = eig( X0'*X0 ./ (sum(idx)-1) );     %#' cov(X0)
    [D order] = sort(diag(D), 'descend');
    D = diag(D);
    V = V(:, order);

    t = linspace(0,2*pi,100);
    e = [cos(t) ; sin(t)];        %# unit circle
    VV = V*sqrt(D);               %# scale eigenvectors
    e = bsxfun(@plus, VV*e, Mu'); %#' project circle back to orig space

    %# plot cov and major/minor axes
    plot(e(1,:), e(2,:), 'Color','k');
    %#quiver(Mu(1),Mu(2), VV(1,1),VV(2,1), 'Color','k')
    %#quiver(Mu(1),Mu(2), VV(1,2),VV(2,2), 'Color','k')
end

如果希望椭圆表示特定水平的标准差，则正确的方法是缩放协方差矩阵:

If you want the ellipse to represent a specific level of standard deviation, the correct way of doing is by scaling the covariance matrix:

STD = 2;                     %# 2 standard deviations
conf = 2*normcdf(STD)-1;     %# covers around 95% of population
scale = chi2inv(conf,2);     %# inverse chi-squared with dof=#dimensions

Cov = cov(X0) * scale;
[V D] = eig(Cov);

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