Pass tuple as input argument for scipy.optimize.curve_fit

Question

I have the following code:

import numpy as np
from scipy.optimize import curve_fit


def func(x, p): return p[0] + p[1] + x


popt, pcov = curve_fit(func, np.ara         


        

Answer 1

scipy.optimize.curve_fit

scipy.optimize.curve_fit(f, xdata, ydata, p0=None, sigma=None, **kw)

Use non-linear least squares to fit a function, f, to data.

Assumes ydata = f(xdata, *params) + eps

Explaining the idea

The function to be fitted should take only scalars (not: *p0). Remember that the result of the fit depends on the initialization parameters.

Working example

import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt

def func(x, a0, a1):
    return a0 + a1 * x

x, y = np.arange(10), np.arange(10) + np.random.randn(10)/10
popt, pcov = curve_fit(func, x, y, p0=(1, 1))

# Plot the results
plt.title('Fit parameters:\n a0=%.2e a1=%.2e' % (popt[0], popt[1]))
# Data
plt.plot(x, y, 'rx')
# Fitted function
x_fine = np.linspace(x[0], x[-1], 100)
plt.plot(x_fine, func(x_fine, popt[0], popt[1]), 'b-')
plt.savefig('Linear_fit.png')
plt.show()