围绕MATLAB中的数据椭圆

I would like to reproduce the following figure in MATLAB:

我想在MATLAB中再现如下图:

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.

有两类具有X和Y坐标的点。我想用一个标准差参数的椭圆来包围每一个类，这个椭圆决定了椭圆沿着轴走多远。

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

这个图形是用另一个软件创建的，我不太清楚它是如何计算椭圆的。

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.

这是我用于这个图的数据。第一列是第2 - X，第3 - y类，我可以使用gscatter来绘制点本身。

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))

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

这里有一个关于如何绘制椭圆的问题。我们只需要知道所有的参数就可以画出来。

---

Update:

更新:

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...

我同意中心可以作为X和Y坐标的均值来计算。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。还想……

3 个解决方案

#1

Consider the code:

考虑一下代码:

%# 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

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EDIT

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

如果希望椭圆表示特定水平的标准差，正确的方法是缩放协方差矩阵:

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);

#2

I'd try the following approach:

我将尝试以下方法:

1. Calculate the x-y centroid for the center of the ellipse (x,y in the linked question)
2. 计算椭圆中心的x-y质心(在相关问题中为x,y)
3. Calculate the linear regression fit line to get the orientation of the ellipse's major axis (angle)
4. 计算线性回归拟合线，得到椭圆主轴(角度)的方向
5. Calculate the standard deviation in the x and y axes
6. 计算x轴和y轴的标准差
7. Translate the x-y standard deviations so they're orthogonal to the fit line (a,b)
8. 把x-y标准差换算成正交于拟合直线(a,b)

#3

I'll assume there is only one set of points given in a single matrix, e.g.

我假设在一个矩阵中只有一组点。

B = A(1:10,2:3);

you can reproduce this procedure for each data set.

您可以为每个数据集复制此过程。

1. Compute the center of the ellipsoid, which is the mean of the points. Matlab function: mean
2. 计算椭圆体的中心，它是点的均值。Matlab功能:的意思是
3. Center your data. Matlab function bsxfun
4. 你的数据中心。Matlab函数bsxfun
5. Compute the principal axis of the ellipsoid and their respective magnitude. Matlab function: eig
6. 计算椭圆体的主轴和它们各自的大小。Matlab功能:eig

The successive steps are illustrated below:

后续步骤如下所示:

Center = mean(B,1);
Centered_data = bsxfun(@minus,B,Center);
[AX,MAG] = eig(Centered_data' * Centered_data);

The columns of AX contain the vectors describing the principal axis of the ellipsoid while the diagonal of MAG contains information on their magnitude. To plot the ellipsoid, scale each principal axis with the square root of its magnitude.

AX的列包含描述椭圆体主轴的向量，而MAG的对角线包含其大小的信息。要绘制椭圆体，要用其大小的平方根来绘制每个主轴。

Hope this helps.

希望这个有帮助。

A.

一个。