How do I put a constraint on SciPy curve fit?

You can define your own residuals function, including a penalization parameter, like detailed in the code below, where it is known beforehand that the integral along the interval must be 2.. If you test without the penalization you will see that what your are getting is the conventional curve_fit:

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import matplotlib.pyplot as plt
import scipy
from scipy.optimize import curve_fit, minimize, leastsq
from scipy.integrate import quad
from scipy import pi, sin

x = scipy.linspace(0, pi, 100)
y = scipy.sin(x) + (0. + scipy.rand(len(x))*0.4)
def func1(x, a0, a1, a2, a3):
    return a0 + a1*x + a2*x**2 + a3*x**3

# here you include the penalization factor
def residuals(p, x, y):
    integral = quad(func1, 0, pi, args=(p[0], p[1], p[2], p[3]))[0]
    penalization = abs(2.-integral)*10000
    return y - func1(x, p[0], p[1], p[2], p[3]) - penalization

popt1, pcov1 = curve_fit(func1, x, y)
popt2, pcov2 = leastsq(func=residuals, x0=(1., 1., 1., 1.), args=(x, y))
y_fit1 = func1(x, *popt1)
y_fit2 = func1(x, *popt2)
plt.scatter(x, y, marker=".")
plt.plot(x, y_fit1, color="g", label="curve_fit")
plt.plot(x, y_fit2, color="y", label="constrained")
plt.legend()
plt.xlim(-0.1, 3.5)
plt.ylim(0, 1.4)
print('Exact integral:', quad(sin, 0, pi)[0])
print('Approx integral1:', quad(func1, 0, pi, args=(popt1[0], popt1[1], popt1[2], popt1[3]))[0])
print('Approx integral2:', quad(func1, 0, pi, args=(popt2[0], popt2[1], popt2[2], popt2[3]))[0])
plt.show()

#Exact   integral: 2.0
#Approx integral1: 2.60068579748
#Approx integral2: 2.00001911981

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