I am trying to find the root y of a function called f using Python.

Here is my code:

def f(y):w,p1,p2,p3,p4,p5,p6 = y[:7] t1 = w - 0.99006633*(p1**0.5) - (-1.010067)*((1-p1))t2 = w - 22.7235687*(p2**0.5) - (-1.010067)*((1-p2))t3 = w - 9.71323491*(p3**0.5) - (-1.010067)*((1-p3))t4 = w - 2.43852877*(p4**0.5) - (-1.010067)*((1-p4))t5 = w - 3.93640207*(p5**0.5) - (-1.010067)*((1-p5))t6 = w - 9.22688144*(p6**0.5) - (-1.010067)*((1-p6))t7 = p1 + p2 + p3 + p4 + p5 + p6 - 1return [t1,t2,t3,t4,t5,t6,t7]x0 = np.array([-0.01,0.4,0.1,0.2,0.1,0.1,0.1])
sol = scipy.optimize.root(f, x0)
print sol 

Python does not find the root. However there is one, I found it with the function fsolve in Matlab.

It is:

[ 0.3901, 0.6166, 0.0038, 0.0202, 0.2295, 0.1076, 0.0223]

I really want to use Python. Can anyone explain why scipy.optimize.root in Python does not converge while fsolve in Matlab does?

For info, scipy.optimize.solve does not converge either.

Try a different method. For me, method="lm" (I'm guessing Levenberg-Marquardt, but I'm not entirely sure) works very well:

import numpy as np
import scipy.optimizedef f(y):w,p1,p2,p3,p4,p5,p6 = y[:7]t1 = w - 0.99006633*(p1**0.5) - (-1.010067)*((1-p1))t2 = w - 22.7235687*(p2**0.5) - (-1.010067)*((1-p2))t3 = w - 9.71323491*(p3**0.5) - (-1.010067)*((1-p3))t4 = w - 2.43852877*(p4**0.5) - (-1.010067)*((1-p4))t5 = w - 3.93640207*(p5**0.5) - (-1.010067)*((1-p5))t6 = w - 9.22688144*(p6**0.5) - (-1.010067)*((1-p6))t7 = p1 + p2 + p3 + p4 + p5 + p6 - 1return [t1,t2,t3,t4,t5,t6,t7]x0 = np.array([-0.01,0.4,0.1,0.2,0.1,0.1,0.1])
sol = scipy.optimize.root(f, x0, method='lm')assert sol['success']
print 'Solution: ', sol.x
print 'Misfit: ', f(sol.x)

This yields:

Solution: [ 0.39012036  0.61656436  0.00377616  0.02017937  0.22954825 0.10763827  0.02229359]
Misfit: [0.0, 0.0, 1.1102230246251565e-16, -1.1102230246251565e-16,   1.1102230246251565e-16, 0.0, -2.2204460492503131e-16]

I'm actually a bit surprised Levenberg-Marquardt isn't the default. It's usually one of the first "gradient-descent" style solvers one would try.