在曲线中找到肘部/膝盖 [英] Finding the elbow/knee in a curve

查看:201
本文介绍了在曲线中找到肘部/膝盖的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有这些数据:

x <- c(6.626,6.6234,6.6206,6.6008,6.5568,6.4953,6.4441,6.2186,6.0942,5.8833,5.702,5.4361,5.0501,4.744,4.1598,3.9318,3.4479,3.3462,3.108,2.8468,2.3365,2.1574,1.899,1.5644,1.3072,1.1579,0.95783,0.82376,0.67734,0.34578,0.27116,0.058285)

y <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32)

如下所示:

plot(x,y)

我想找到一种方法来使肘/膝点在x=6.5

And I want to find a way to get the elbow/knee point at around x=6.5

我认为拟合loess曲线然后取二阶导数可能有用,但是:

I thought that fitting a loess curve and then taking the second derivative may work but:

plot(x,predict(loess(y ~ x)),type="l")

看起来不会完成工作.

有什么主意吗?

推荐答案

我想您想找到函数y=f(x)的导数具有巨大跳跃值的点.您可以尝试以下操作,因为您会看到根据选择的阈值(大幅跳动)可能存在一个或多个这样的点:

I think you want to find the points where the derivative of the function y=f(x) has a huge jump in value. you can try the following, as you can see there can be one or many such points depending on the threshold (for huge jump) we choose:

get.elbow.points.indices <- function(x, y, threshold) {
  d1 <- diff(y) / diff(x) # first derivative
  d2 <- diff(d1) / diff(x[-1]) # second derivative
  indices <- which(abs(d2) > threshold)  
  return(indices)
}

# first approximate the function, since we have only a few points
ap <- approx(x, y, n=1000, yleft=min(y), yright=max(y))
x <- ap$x
y <- ap$y

indices <- get.elbow.points.indices(x, y, 1e4) # threshold for huge jump = 1e4
x[indices]
#[1] 6.612851 # there is one such point
plot(x, y, pch=19)
points(x[indices], y[indices], pch=19, col='red')

 indices <- get.elbow.points.indices(x, y, 1e3) # threshold for huge jump = 1e3
 x[indices]
 #[1] 0.3409794 6.4353456 6.5931286 6.6128514 # there are 4 such points
 plot(x, y, pch=19)
 points(x[indices], y[indices], pch=19, col='red')

这篇关于在曲线中找到肘部/膝盖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!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆