r - 在曲线中找到肘/膝

Question

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)

看起来像:

plot(x,y)

我想找到一种方法让肘/膝点在 x=6.5

左右

我认为拟合 loess 曲线然后取二阶导数可能可行，但是:

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

看起来它不会完成这项工作。

有什么想法吗？

Answer 1

我认为您想找到函数 y=f(x) 的导数在值上有巨大跳跃的点。您可以尝试以下操作，正如您所看到的，根据我们选择的阈值(对于巨大的跳跃)，可能会有一个或多个这样的点:

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')