在椭圆协方差图上获取椭圆的顶点（由`car :: ellipse`创建）

Question

通过跟随this post，可以绘制具有给定形状矩阵（A）的椭圆：

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)

现在如何获得这个椭圆的主要/次要（主/短轴和椭圆的交叉点对）顶点？

Answer 1

我知道这个问题已经被解决了，但实际上有一个超级优雅的解决方案，只有几行如下。这种计算是精确的，没有任何数值优化。

## target covariance matrix
A <- matrix(c(20.43, -8.59,-8.59, 24.03), nrow = 2)

E <- eigen(A, symmetric = TRUE)  ## symmetric eigen decomposition
U <- E[[2]]  ## eigen vectors, i.e., rotation matrix
D <- sqrt(E[[1]])  ## root eigen values, i.e., scaling factor

r <- 1.44  ## radius of original circle
Z <- rbind(c(r, 0), c(0, r), c(-r, 0), c(0, -r))  ## original vertices on major / minor axes
Z <- tcrossprod(Z * rep(D, each = 4), U)  ## transformed vertices on major / minor axes

#          [,1]      [,2]
#[1,] -5.055136  6.224212
#[2,] -4.099908 -3.329834
#[3,]  5.055136 -6.224212
#[4,]  4.099908  3.329834

C0 <- c(-0.05, 0.09)  ## new centre
Z <- Z + rep(C0, each = 4)  ## shift to new centre

#          [,1]      [,2]
#[1,] -5.105136  6.314212
#[2,] -4.149908 -3.239834
#[3,]  5.005136 -6.134212
#[4,]  4.049908  3.419834

为了解释背后的数学，我将采取三个步骤：

1. 这个Ellipse来自哪里？
2. Cholesky分解方法及其缺点。
3. 特征分解方法及其自然解释。

这个椭圆来自哪里？



实际上，这个椭圆可以通过对单位圆x ^ 2 + y ^ 2 = 1进行一些线性变换来获得。

Cholesky分解方法及其缺点



## initial circle
r <- 1.44
theta <- seq(0, 2 * pi, by = 0.01 * pi)
X <- r * cbind(cos(theta), sin(theta))

## target covariance matrix
A <- matrix(c(20.43, -8.59,-8.59, 24.03), nrow = 2)

R <- chol(A)  ## Cholesky decomposition
X1 <- X %*% R  ## linear transformation

Z <- rbind(c(r, 0), c(0, r), c(-r, 0), c(0, -r))  ## original vertices on major / minor axes
Z1 <- Z %*% R  ## transformed coordinates

## different colour per quadrant
g <- floor(4 * (1:nrow(X) - 1) / nrow(X)) + 1

## draw ellipse
plot(X1, asp = 1, col = g)
points(Z1, cex = 1.5, pch = 21, bg = 5)

## draw circle
points(X, col = g, cex = 0.25)
points(Z, cex = 1.5, pch = 21, bg = 5)

## draw axes
abline(h = 0, lty = 3, col = "gray", lwd = 1.5)
abline(v = 0, lty = 3, col = "gray", lwd = 1.5)



我们看到线性变换矩阵R似乎没有自然解释。圆的原始顶点不会映射到椭圆的顶点。

特征分解方法及其自然解释



## initial circle
r <- 1.44
theta <- seq(0, 2 * pi, by = 0.01 * pi)
X <- r * cbind(cos(theta), sin(theta))

## target covariance matrix
A <- matrix(c(20.43, -8.59,-8.59, 24.03), nrow = 2)

E <- eigen(A, symmetric = TRUE)  ## symmetric eigen decomposition
U <- E[[2]]  ## eigen vectors, i.e., rotation matrix
D <- sqrt(E[[1]])  ## root eigen values, i.e., scaling factor

r <- 1.44  ## radius of original circle
Z <- rbind(c(r, 0), c(0, r), c(-r, 0), c(0, -r))  ## original vertices on major / minor axes

## step 1: re-scaling
X1 <- X * rep(D, each = nrow(X))  ## anisotropic expansion to get an axes-aligned ellipse
Z1 <- Z * rep(D, each = 4L)  ## vertices on axes

## step 2: rotation
Z2 <- tcrossprod(Z1, U)  ## rotated vertices on major / minor axes
X2 <- tcrossprod(X1, U)  ## rotated ellipse

## different colour per quadrant
g <- floor(4 * (1:nrow(X) - 1) / nrow(X)) + 1

## draw rotated ellipse and vertices
plot(X2, asp = 1, col = g)
points(Z2, cex = 1.5, pch = 21, bg = 5)

## draw axes-aligned ellipse and vertices
points(X1, col = g)
points(Z1, cex = 1.5, pch = 21, bg = 5)

## draw original circle
points(X, col = g, cex = 0.25)
points(Z, cex = 1.5, pch = 21, bg = 5)

## draw axes
abline(h = 0, lty = 3, col = "gray", lwd = 1.5)
abline(v = 0, lty = 3, col = "gray", lwd = 1.5)

## draw major / minor axes
segments(Z2[1,1], Z2[1,2], Z2[3,1], Z2[3,2], lty = 2, col = "gray", lwd = 1.5)
segments(Z2[2,1], Z2[2,2], Z2[4,1], Z2[4,2], lty = 2, col = "gray", lwd = 1.5)



在这里，我们看到在变换的两个阶段中，顶点仍然映射到顶点。它完全基于这样的属性，我们在一开始就给出了简洁的解决方案。

Answer 2

出于实际目的，@ Tensibai的答案可能已经足够好了。只需为segments参数使用足够大的值，以便这些点可以很好地逼近真实顶点。

如果你想要更严格的东西，你可以求解沿椭圆的位置，最大化/最小化距中心的距离，通过角度进行参数化。由于形状矩阵的存在，这比仅仅angle={0, pi/2, pi, 3pi/2}更复杂。但这并不太难：

# 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)

\n

Answer 3

仍然非常不确定这会真正回答这个问题，但这是我的尝试：

首先，将椭圆的中心定义为矢量供以后使用：

center<-c(x=-0.05, y=0.09)

绘制椭圆并获得带有enought值的“points”矩阵，以便与现实点接近：

tmp<-ellipse(c(-0.05, 0.09), shape=A, radius=1.44, segments=1e3, col="red", lty=2,add=FALSE)

用它创建一个data.table并计算每个点到中心的距离（point_x - center_x）²+（point_y - center_y）²：

dt <- data.table(tmp)
dt[,dist:={dx=x-center[1];dy=y-center[2];dx*dx+dy*dy}]

按距离排序顶点：

setorder(dt,dist)

获取最低和最高分数：

> tail(dt,2)
           x         y     dist
1:  4.990415 -6.138039 64.29517
2: -5.110415  6.318039 64.29517
> head(dt,2)
       x        y     dist
1:  4.045722  3.41267 27.89709
2: -4.165722 -3.23267 27.89709

不要添加太多的段，否则两个第一个值将是两个非常接近的点，而不是相反的。

带有视觉效果，听起来不那么精确：