计算R中二维置信椭圆的面积 [英] calculate area of 2-dimensional confidence ellipse in R

问题描述

所以我在一个简单的表格中排列了对数转换的测量数据:

So I have log-transformed measurement data arranged in a simple table:

      x         y
1.158362492 1.322219295
1.1430148   1.267171728
1.11058971  1.252853031
1.120573931 1.260071388
1.149219113 1.278753601
1.123851641 1.276461804
1.096910013 1.222716471

我知道有一些函数可以为这些数据绘制置信椭圆，但是如何计算生成形状的面积?

I know there are functions for plotting a confidence ellipse for these data, but how to I calculate the area of the generated shape?

谢谢

推荐答案

先计算椭圆，然后确定长短轴的长度，然后 计算面积.

First calculate the ellipse, then determine the lengths of the major and minor axes, and then calculate the area.

这是一个无脑的近似.

首先，您的数据.

dat <- structure(list(x = c(1.158362492, 1.1430148, 1.11058971, 1.120573931, 
          1.149219113, 1.123851641, 1.096910013), y = c(1.322219295, 1.267171728, 
          1.252853031, 1.260071388, 1.278753601, 1.276461804, 1.222716471
          )), .Names = c("x", "y"), class = "data.frame", row.names = c(NA, 
          -7L))

然后加载包car；dataEllipse 可用于使用数据的二元正态近似计算椭圆.

Then load the package car; dataEllipse can be used to calculate an ellipse using a bivariate normal approximation to the data.

require(car)
dataEllipse(dat$x, dat$y, levels=0.5)

调用 ellipse 可以沿着 dataEllipse 绘制的椭圆给出点.

A call to ellipse can give points along the ellipse that dataEllipse plots.

me <- apply(dat, 2, mean)
v <- var(dat)
rad <- sqrt(2*qf(0.5, 2, nrow(dat)-1))
z <- ellipse(me, v, rad, segments=1001)

然后我们可以计算椭圆上每个点到中心的距离.

We can then calculate the distance from each point on the ellipse to the center.

dist2center <- sqrt(rowSums((t(t(z)-me))^2))

这些距离的最小值和最大值是短轴和长轴的半长.所以我们可以得到如下面积.

The minimum and maximum of these distances are the half-lengths of the minor and major axes. So we can get the area as follows.

pi*min(dist2center)*max(dist2center)

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