how to optimize the minimization of a vector function in python?

Question

I have an issue: I'm trying to find the minimum of a function which depends on several parameters that I'd like to change as well. let's take as a simplified example:

import numpy as np
import scipy.optimize as opt
def f(x, a, b, c):
    f = a * x**2 + b * x + c
return f

I'd like to find the x which minimizes the function for different set of values of a, b, c, let's say for

a = [-1, 0, 1]
b = [0, 1, 2]
c = [0, 1]

ATM I have three nested loops and a minimization:

for p1 in a:
    for p2 in b:
         for p3 in c:
            y = opt.minimize(f, x0=[0, ], args=(p1, p2, p3, ))
            print(y)

which is really slow for the calculation I'm doing, but I haven't found any better so far. So, does anyone know a way or a package that would allow me to improve the efficiency?

Answer 1

You could use a combination of different techniques to improve the efficiency of your script:

1. Use itertools.product to generate every possible combination in the list a, b, c
2. Use multiprocessing to execute the minimizations in parallel.

Other than this, i can't think of a way to optimize the efficiency of the code. As was pointed out in the comment, the constant value c has no influence on the minimization. But i'm sure the quadratic function is just an example.

I took the code of the multiprocessing part from here .

Here's the working code.

import numpy as np
import scipy.optimize as opt
import itertools
from multiprocessing import Pool

def f(x, a, b, c):
    f = a * x**2 + b * x + c
    return f

def mini(args):
    res = opt.minimize(f, x0=np.array([0]), args=args)
    return res.x

if __name__=="__main__":
    a = np.linspace(-1,2,100)
    b = np.linspace(0,2,100)
    c = [0, 1]
    args = list(itertools.product(a,b,c))
    print("Number of combos:" + str(len(args)))
    p = Pool(4)
    import time
    t0 = time.time()
    res = p.map(mini, args)
    print(time.time()-t0)

Even these 20000 combinations only need 5,28 seconds on my average laptop.

Answer 2

scipy.optimize.newton可以做到这一点。