当“类型"为“范数"时，如何计算ggplot stat_ellipse()的面积? [英] How to calculate the area of a ggplot stat_ellipse() when 'type = "norm"?

问题描述

当type ="norm"时，有什么方法可以计算此椭圆的面积吗?

Is there any way to calculate the area of this ellipse when type = "norm"?

默认值为type = "t". type = "norm"显示不同的椭圆，因为它假定多元正态分布而不是多元t分布

Default is type = "t". type = "norm" displays a different ellipse because it assumes a multivariate normal distribution instead of multivariate t-distribution

以下是代码和图解(使用与其他帖子类似的代码):

library(ggplot2)
set.seed(1234)
data <- data.frame(x = rnorm(1:1000), y = rnorm(1:1000))

ggplot (data, aes (x = x, y = y))+
  geom_point()+
  stat_ellipse(type = "norm")

上一个答案是:

#Plot object
p = ggplot (data, aes (x = x, y = y))+
  geom_point()+
  stat_ellipse(segments=201) # Default is 51. We use a finer grid for more accurate area.

#Get ellipse coordinates from plot

pb = ggplot_build(p)
el = pb$data[[2]][c("x","y")]

# Center of ellipse

ctr = MASS::cov.trob(el)$center 
# I tried changing this to 'stats::cov.wt' instead of 'MASS::cov.trob' 
#from what is saw from (https://github.com/tidyverse/ggplot2/blob/master/R/stat-ellipse.R#L98)

# Calculate distance to center from each point on the ellipse

dist2center <- sqrt(rowSums((t(t(el)-ctr))^2))

# Calculate area of ellipse from semi-major and semi-minor axes. 
These are, respectively, the largest and smallest values of dist2center. 

pi*min(dist2center)*max(dist2center)

更改为stats::cov.wt不足以获取标准"椭圆的面积(计算出的值相同).关于如何更改代码有任何想法吗?

Changing to stats::cov.wt wasn't enough to get the area of the "norm" ellipse (value calculated was the same). Any ideas on how to change the code?

谢谢！

推荐答案

很好的问题，我学到了一些东西.但是我无法重现您的问题，并且无法通过不同的方法获得(当然)不同的价值.

Nice question, I learned something. But I cannot reproduce your problem and get (of course) different values with the different approaches.

我认为链接答案中的方法不太正确，因为椭圆中心不是根据数据计算的，而是基于椭圆坐标的.我已更新，可以根据数据进行计算.

I think the approach in the linked answer is not quite correct because the ellipse center is not calculated with the data, but based on the ellipse coordinates. I have updated to calculate this based on the data.

library(ggplot2)

set.seed(1234)
data <- data.frame(x = rnorm(1:1000), y = rnorm(1:1000))

p_norm <- ggplot(data, aes(x = x, y = y)) +
  geom_point() +
  stat_ellipse(type = "norm")

pb <- ggplot_build(p_norm)
el <- pb$data[[2]][c("x", "y")]
ctr <- MASS::cov.trob(data)$center #updated previous answer here
dist2center <- sqrt(rowSums((t(t(el) - ctr))^2))
pi * min(dist2center) * max(dist2center)
#> [1] 18.40872

^由 reprex软件包(v0.3.0)创建于2020-02-27

更新，感谢Axeman的想法.

update thanks to Axeman for the thoughts.

可以通过首先计算特征值从协方差矩阵直接计算面积.您需要根据要获得的置信度来缩放方差/特征值. 此博客对我的理解有所帮助更好

The area can be directly calculated from the covariance matrix by calculating the eigenvalues first. You need to scale the variances / eigenvalues by the factor of confidence that you want to get. This blog helped me a lot to understand this a bit better

set.seed(1234)
dat <- data.frame(x = rnorm(1:1000), y = rnorm(1:1000))

cov_dat <- cov(dat) # covariance matrix

eig_dat <- eigen(cov(dat))$values #eigenvalues of covariance matrix

vec <- sqrt(5.991* eig_dat) # half the length of major and minor axis for the 95% confidence ellipse

pi * vec[1] * vec[2]  
#> [1] 18.38858

^{由 reprex软件包(v0.3.0)创建于2020-02-27}

在这种情况下，协方差为零，特征值或多或少是变量的方差.因此，您可以仅使用方差进行计算. -假设两者都是正态分布.

In this particular case, the covariances are zero, and the eigenvalues will be more or less the variance of the variables. So you can use just the variance for your calculation. - given that both are normally distributed.

set.seed(1234)
data <- data.frame(x = rnorm(1:1000), y = rnorm(1:1000))

pi * 5.991 * sd(data$x) * sd(data$y) # factor for 95% confidence = 5.991
#> [1] 18.41814

^{由 reprex软件包(v0.3.0)创建于2020-02-27}

因子5.991 代表数据95％置信度的卡方似然.有关更多信息，请参见此线程

The factor 5.991 represents the Chi-square likelihood for the 95% confidence of the data. For more information, see this thread

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