在椭圆协方差图上获得椭圆的顶点(由`car :: ellipse`创建) [英] Obtain vertices of the ellipse on an ellipse covariance plot (created by `car::ellipse`)

问题描述

通过遵循这篇文章，可以绘制具有给定形状矩阵的椭圆(A) :

By following this post one can draw an ellipse with a given shape matrix (A):

library(car)
A <- matrix(c(20.43, -8.59,-8.59, 24.03), nrow = 2)
ellipse(c(-0.05, 0.09), shape=A, radius=1.44, col="red", lty=2, asp = 1)

现在如何获取该椭圆的主/副(长/副轴与椭圆的相交点对)顶点?

Now how to get the major/minor (pair of intersect points of the major/minor axis and the ellipse) vertices of this ellipse?

推荐答案

出于实际目的，@ Tensibai的答案可能足够好.只需对segments参数使用足够大的值，以使这些点能很好地逼近真实顶点.

For practical purposes, @Tensibai's answer is probably good enough. Just use a large enough value for the segments argument so that the points give a good approximation to the true vertices.

如果您想要更严格一些的东西，可以解决沿椭圆的位置，该位置最大化/最小化到中心的距离(以角度为参数).由于存在形状矩阵，这比仅获取angle={0, pi/2, pi, 3pi/2}要复杂.但这并不困难:

If you want something a bit more rigorous, you can solve for the location along the ellipse that maximises/minimises the distance from the center, parametrised by the angle. This is more complex than just taking angle={0, pi/2, pi, 3pi/2} because of the presence of the shape matrix. But it's not too difficult:

# location along the ellipse
# linear algebra lifted from the code for ellipse()
ellipse.loc <- function(theta, center, shape, radius)
{
    vert <- cbind(cos(theta), sin(theta))
    Q <- chol(shape, pivot=TRUE)
    ord <- order(attr(Q, "pivot"))
    t(center + radius*t(vert %*% Q[, ord]))
}

# distance from this location on the ellipse to the center 
ellipse.rad <- function(theta, center, shape, radius)
{
    loc <- ellipse.loc(theta, center, shape, radius)
    (loc[,1] - center[1])^2 + (loc[,2] - center[2])^2
}

# ellipse parameters
center <- c(-0.05, 0.09)
A <- matrix(c(20.43, -8.59, -8.59, 24.03), nrow=2)
radius <- 1.44

# solve for the maximum distance in one hemisphere (hemi-ellipse?)
t1 <- optimize(ellipse.rad, c(0, pi - 1e-5), center=center, shape=A, radius=radius, maximum=TRUE)$m
l1 <- ellipse.loc(t1, center, A, radius)

# solve for the minimum distance
t2 <- optimize(ellipse.rad, c(0, pi - 1e-5), center=center, shape=A, radius=radius)$m
l2 <- ellipse.loc(t2, center, A, radius)

# other points obtained by symmetry
t3 <- pi + t1
l3 <- ellipse.loc(t3, center, A, radius)

t4 <- pi + t2
l4 <- ellipse.loc(t4, center, A, radius)

# plot everything
MASS::eqscplot(center[1], center[2], xlim=c(-7, 7), ylim=c(-7, 7), xlab="", ylab="")
ellipse(center, A, radius, col="red", lty=2)
points(rbind(l1, l2, l3, l4), cex=2, col="blue", lwd=2)

这篇关于在椭圆协方差图上获得椭圆的顶点(由`car :: ellipse`创建)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！