Compute the 'elbow' for a curve automatically and mathematically

Question

One example for curve is shown as below. The elbow point might be x=3 or 4. How to compute the elbow for a curve automatically and mathematically?

Answer 1

I created a Python package that attempts to implement the Kneedle algorithm.

To recreate the function above and detect the point of maximum curvature:

x = range(1,21) y = [0.065, 0.039, 0.030, 0.024, 0.023, 0.022, 0.019, 0.0185, 0.0187,      0.016, 0.015, 0.016, 0.0135, 0.0130, 0.0125, 0.0120, 0.0117, 0.0115, 0.0112, 0.013]  kn = KneeLocator(     x,     y,     curve='convex',     direction='decreasing',     interp_method='interp1d', )  print(kn.knee) 7 

import matplotlib.pyplot as plt plt.xlabel('x') plt.ylabel('f(x)') plt.xticks(range(1,21)) plt.plot(x, y, 'bx-') plt.vlines(kn.knee, plt.ylim()[0], plt.ylim()[1], linestyles='dashed') 

update
Kneed has an improved spline fitting method for handling local minima, use interp_method='polynomial'.

kn = KneeLocator(     x,     y,     curve='convex',     direction='decreasing',     interp_method='polynomial', )  print(kn.knee) 4 

And the new plot:

plt.xlabel('x') plt.ylabel('f(x)') plt.xticks(range(1,21)) plt.plot(x, y, 'bx-') plt.vlines(kn.knee, plt.ylim()[0], plt.ylim()[1], linestyles='dashed') 

Answer 2

You might want to look for the point with the maximum absolute second derivative which, for a set of discrete points x[i] as you have there, can be approximated with a central difference:

secondDerivative[i] = x[i+1] + x[i-1] - 2 * x[i]

As noted above, what you really want is the point with maximum curvature, but the second derivative will do, and this central difference is a good proxy for the second derivative.