包含 R 中给定点百分比的椭圆

发布于 2024-11-19 09:39:36 字数 1501 浏览 7 评论 0原文

我正在绘制 F1/F2 元音图(一个例子是在这里)。每个元音都有几个点/值,我想在这些点周围画一个椭圆,这样:

  • 椭圆覆盖至少 80% 的点(即上图中“i”有几个值,但它们包含在椭圆内)。
  • 位于最小/最大值方向。

我可能把事情复杂化了,但三角学和数学对我来说是希腊语。以下是我尝试过的。

Ellipsoidhull()

Ellipsoidhull () 位于包“cluster”中。如果我将具有 F1 和 F2 的矩阵传递给函数,它似乎计算了椭圆的中心,但方向值很大。例如:

> olm
      ol.f1 ol.f2 # f1/f2 data
 [1,] 501.3 850.5
 [2,] 488.5 906.5
 [3,] 456.3 857.0
 [4,] 505.8 895.3
 [5,] 499.5 898.0
 [6,] 431.8 891.5
 [7,] 416.3 870.5
 [8,] 506.0 887.8
 [9,] 500.3 985.8
[10,] 513.5 955.3
[11,] 531.5 958.0
[12,] 483.0 847.3
[13,] 533.3 982.8
[14,] 480.8 881.8
[15,] 484.3 884.5

如果传递给 ellipsoidhull :

> ellipsoidhull(olm)
'ellipsoid' in 2 dimensions:
 center = ( 480.69 904.33 ); squared ave.radius d^2 =  2 
 and shape matrix =
       ol.f1  ol.f2
ol.f1 2115.5 1449.5
ol.f2 1449.5 3558.2
  hence, area  =  14636 

我想弄清楚如何绘制椭圆并不难,但是“形状矩阵”(最大/最小半径值?)太高了。顺便说一句,感谢 Freednode 上的 #R 提供的提示。

来自 EMU-R 的源代码

然后,我研究了 EMU-R 的代码,它是与 EMU,除其他外,可以用椭球体绘制 F1/F2。似乎可以做到这一点的代码是 这里 但我不明白椭圆是如何绘制的。

任何帮助表示赞赏。

I'm drawing F1/F2 vowel graph (an example is here). Each vowel has several points/values, and I'd like to draw an ellipse around the points, so that:

  • ellipse covers at least 80% of points (ie. in the picture above "i" has several values, but they are contained within the ellipse).
  • is positioned in the direction on min/max values.

I may be complicating the stuff, but trigonometry and maths are Greek to me. Below is what I've tried.

Ellipsoidhull()

Ellipsoidhull() is in the package package "cluster". If I pass to a function a matrix with F1 and F2, it seems to calculate the center of the ellipse, but the directional values are huge. For example:

> olm
      ol.f1 ol.f2 # f1/f2 data
 [1,] 501.3 850.5
 [2,] 488.5 906.5
 [3,] 456.3 857.0
 [4,] 505.8 895.3
 [5,] 499.5 898.0
 [6,] 431.8 891.5
 [7,] 416.3 870.5
 [8,] 506.0 887.8
 [9,] 500.3 985.8
[10,] 513.5 955.3
[11,] 531.5 958.0
[12,] 483.0 847.3
[13,] 533.3 982.8
[14,] 480.8 881.8
[15,] 484.3 884.5

If passed to ellipsoidhull:

> ellipsoidhull(olm)
'ellipsoid' in 2 dimensions:
 center = ( 480.69 904.33 ); squared ave.radius d^2 =  2 
 and shape matrix =
       ol.f1  ol.f2
ol.f1 2115.5 1449.5
ol.f2 1449.5 3558.2
  hence, area  =  14636 

I guess it wouldn't be hard to figure out how to draw an ellipse, but the "shape matrix" (max/min radius values?) is too high. Btw, thanks to #R on Freednode for the tips.

Source code from EMU-R

Then, I've taken a look into the code of EMU-R, R package that works with EMU that can, amongst other things, draw F1/F2 with ellipsoids. The code that seems to do that is here but I don't understand how the ellipse is drawn.

Any help appreciated.

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黑白记忆 2024-11-26 09:39:36
require(car)
 x=rnorm(100)
 y=1+.3*x+.3*rnorm(100)
 dataEllipse(x,y, levels=0.80)

因此,对于您的数据:

with(olm ,dataEllipse(ol.f1, ol.f2, levels=0.8) )

另一个包 mixtools 具有类似的功能,但使用 alpha 级别而不是 1-alpha:

 mu <- with(olm, c(mean(ol.f1), mean(ol.f2)) )
 sigma <- var(olm)  # returns a variance-covariance matrix.
 sigma
#          ol.f1     ol.f2
#ol.f1 1077.2098  865.9306
#ol.f2  865.9306 2090.2021

require(mixtools)
#Loading required package: mixtools
#Loading required package: boot
# And you get a warning that ellipse from car is masked.

ellipse(mu, sigma, alpha=0.2, npoints = 200, newplot = FALSE)

这将用新的估计覆盖早期的图(在本例中稍微窄一些。这是两种方法的比较

require(car)
 x=rnorm(100)
 y=1+.3*x+.3*rnorm(100)
 dataEllipse(x,y, levels=0.80)

So with your data:

with(olm ,dataEllipse(ol.f1, ol.f2, levels=0.8) )

Another package, mixtools, has similar capabilities but uses the alpha level rather than the 1-alpha:

 mu <- with(olm, c(mean(ol.f1), mean(ol.f2)) )
 sigma <- var(olm)  # returns a variance-covariance matrix.
 sigma
#          ol.f1     ol.f2
#ol.f1 1077.2098  865.9306
#ol.f2  865.9306 2090.2021

require(mixtools)
#Loading required package: mixtools
#Loading required package: boot
# And you get a warning that ellipse from car is masked.

ellipse(mu, sigma, alpha=0.2, npoints = 200, newplot = FALSE)

Which would overlay the earlier plot with the new estimate (which is slightly narrower in this case.This is the comparison of the two methods

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