问题
I have these data:
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)
which look like:
plot(x,y)
And I want to find a way to get the elbow/knee point at around x=6.5
I thought that fitting a loess
curve and then taking the second derivative may work but:
plot(x,predict(loess(y ~ x)),type="l")
doesn't look like it'll do the job.
Any idea?
回答1:
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')
回答2:
You can now find the knees/elbows with different methods by using the maxcurv function in the soilphysics package.
来源:https://stackoverflow.com/questions/41518870/finding-the-elbow-knee-in-a-curve