尋找肘/膝曲線

我有這些數據：尋找肘/膝曲線

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)

看起來像：

plot(x,y)

而且我想找到一種方式來獲得肘部/拐點在x=6.5

我以爲裝修一個loess曲線，然後取二階導數可以工作，但：

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

看起來並不像它會做的工作。

有什麼想法？

來源

2017-01-07 dan

可能[R：使用ggplot2在LOESS曲線中標記斜率變化]的副本（http://stackoverflow.com/questions/8245792/r-marking-slope-changes-in-loess-curve-using-ggplot2） – Uwe

我想你想找到函數y=f(x)的派生值具有巨大跳躍的點。你可以嘗試以下方法，你可以看到有可根據閾值一個或多個這樣的點（巨大的跳躍），我們選擇：

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')

來源

2017-01-07 08:01:17

尋找肘/膝曲線

回答

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