is there a way to calculate the curvature in a specific point having a smooth.spline curve (or similar)in R? The curve is calculate from a set of x,y points.
Thank you in advance.
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问题:
回答1:
This one is actually very easy if you know that there is a predict() method for objects created by smooth.spline() and that this method has an argument deriv which allows you to predict a given derivative (in your case the second derivative is required) instead of points on the spline.
cars.spl <- with(cars, smooth.spline(speed, dist, df = 3)) with(cars, predict(cars.spl, x = speed, deriv = 2)) Which gives:
$x [1] 4 4 7 7 8 9 10 10 10 11 11 12 12 12 12 13 13 13 13 14 14 14 14 15 15 [26] 15 16 16 17 17 17 18 18 18 18 19 19 19 20 20 20 20 20 22 23 24 24 24 24 25 $y [1] -6.492030e-05 -6.492030e-05 3.889944e-02 3.889944e-02 5.460044e-02 [6] 7.142609e-02 6.944645e-02 6.944645e-02 6.944645e-02 9.273343e-02 [11] 9.273343e-02 1.034153e-01 1.034153e-01 1.034153e-01 1.034153e-01 [16] 5.057841e-02 5.057841e-02 5.057841e-02 5.057841e-02 1.920888e-02 [21] 1.920888e-02 1.920888e-02 1.920888e-02 1.111307e-01 1.111307e-01 [26] 1.111307e-01 1.616749e-01 1.616749e-01 1.801385e-01 1.801385e-01 [31] 1.801385e-01 1.550027e-01 1.550027e-01 1.550027e-01 1.550027e-01 [36] 2.409237e-01 2.409237e-01 2.409237e-01 2.897166e-01 2.897166e-01 [41] 2.897166e-01 2.897166e-01 2.897166e-01 1.752232e-01 1.095682e-01 [46] -1.855994e-03 -1.855994e-03 -1.855994e-03 -1.855994e-03 4.478382e-05 where $y is the second derivative of the fitted spline, evaluated at the observed speed values in the data used to fit the spline. Of course you can plug in any values you want here, such as 100 values equally spaced over the range of speed. For example:
newspeed <- with(cars, seq(min(speed), max(speed), length = 100)) curvature <- predict(cars.spl, x = newspeed, deriv = 2) plot(curvature, type = "l")