scipy curve_fit error: divide by zero encountered

Question

I've been trying to fit a function to some data for a while using scipy.optimize.curve_fit:

from __future__ import (print_function,
                    division,
                    unicode_literals,
                    absolute_import)
import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as mpl
x = np.array([0, 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])
y = np.array([20.8, 20.9, 22.9, 25.2, 26.9, 28.3, 29.5, 30.7, 31.8, 32.9, 34.0, 35.3, 36.4, 37.5, 38.6, 39.6, 40.6, 41.6, 42.5, 43.2, 44.2, 45.0, 45.8, 46.5, 47.3, 48.0, 48.6, 49.2, 49.8, 50.4])
def f(x, a, b, c):
    return a/(1+b*x**c)
popt, pcov = curve_fit(f, x, y)
print(popt, np.sqrt(np.diag(pcov)), sep='\n')

But there always appears an error:

RuntimeWarning: divide by zero encountered in power
return a/(1+b*x**c)

Maybe someone can help me to avoid it? Any help would be much appreciated. Cheers!

Answer 1

Alright, two helpful tricks.

1st, replace 0 in your x with some really small number, such as 1e-8 (don't laugh, there is a core package in R actually does this, written by his name shall not be spoken and people use it all the time.) Actually I didn't get your RuntimeWarning at all. I am running scipy 0.12.0 and numpy 1.7.1. Maybe this is version dependent.

But we will get a very bad fit:

In [41]: popt, pcov
Out[41]: (array([  3.90107143e+01,  -3.08698757e+07,  -1.52971609e+02]), inf)

So, trick 2, instead of optimizing f function, we define a g function:

In [38]: def g(x, a, b, c):
   ....:     return b/a*x**c+1/a
   ....:

In [39]: curve_fit(g, x, 1/y) #Better fit
Out[39]:
(array([ 19.76748582,  -0.14499508,   0.44206688]),
 array([[ 0.29043958,  0.00899521,  0.01650935],
        [ 0.00899521,  0.00036082,  0.00070345],
        [ 0.01650935,  0.00070345,  0.00140253]]))

We can now use the resulting parameter vector as starting vector to optimize f(). As curve_fit is a nonlinear least square method, parameter optimizes g() is not necessary the parameter optimizes f(), but hopefully it will be close. The covariance matrices are very different of course.

In [78]: curve_fit(f, x, y, p0=curve_fit(g, x, 1/y)[0]) #Alternative Fit
Out[78]:
(array([ 18.0480446 ,  -0.22881647,   0.31200106]),
 array([[ 1.14928169,  0.03741604,  0.03897652],
        [ 0.03741604,  0.00128511,  0.00136315],
        [ 0.03897652,  0.00136315,  0.00145614]]))

The comparison of the results:

Now the result is pretty good.

Answer 2

Your x values start at 0. If for some reason the parameter c is negative during the calculation, then you will evaluate 0 raised to a negative exponent, which is a division by zero: for p>0 we have

0**(-p) = 1/(0**p)
        = 1/0