Minimize function with parameters

Question

Currently I have the following code that defines the function f.

a = #something
b = #something
c = #something
def f(x):
    """Evaluates some function that depends on parameters a, b, and c"""
    someNumber = #some calculation
    return someNumber

Ideally I would do def f(x, a, b, c), BUT I am minimizing f with respect to x and SciPy's optimization toolbox doesn't allow for one to minimize functions with parameters in the arguments. That said I would like to run my minimization code for multiple values of a, b and c. Is there a way I can do this?

"Freeze" the function with particular a, b, c values using functools.partial ? — PM 2Ring
– PM 2Ring, Commented Mar 25, 2017 at 14:54
@PM2Ring this would work, and I'm glad I know about this tool now, but as @Liteye pointed out minimize actually allows me to pass it parameters. Thanks! — Nate Stemen
– Nate Stemen, Commented Mar 25, 2017 at 16:17
No worries. partial is great when you need it, but I don't know SciPy, so it was only a suggestion. Obviously, Liteye's answer is the proper way to do this. :) — PM 2Ring
– PM 2Ring, Commented Mar 25, 2017 at 16:34

Evgeny · Accepted Answer · 2018-12-11 12:43:40Z

37

You can specify additional arguments in args

from scipy.optimize import minimize 
minimize(f, x0, args=(a, b, c))

edited Dec 11, 2018 at 12:43

Evgeny

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answered Mar 25, 2017 at 15:00

Liteye

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1 Comment

Robb1 Over a year ago

Do parameters in args need to be called in the same way they are called in the body of the function f? Does also f remain f(x) or does it become f(x,a,b,c)?

hpaulj · Accepted Answer · 2017-04-14 17:58:14Z

17

This is a straightforward question and answer about using minimize. In case other users need something more concrete, here's a simple example.

A generalized quadratic equation:

In [282]: def fun(x, a,b,c):
     ...:     return a*x**2 + b*x + c

In [283]: optimize.minimize(fun, 10, args=(1,0,0))
Out[283]: 
      fun: 1.7161984122524196e-15
 hess_inv: array([[ 0.50000001]])
      jac: array([ -6.79528891e-08])
  message: 'Optimization terminated successfully.'
     nfev: 15
      nit: 4
     njev: 5
   status: 0
  success: True
        x: array([ -4.14270251e-08])

In [284]: optimize.minimize(fun, 10, args=(1,1,1))
Out[284]: 
      fun: 0.7500000000000221
 hess_inv: array([[ 0.49999999]])
      jac: array([  3.12924385e-07])
  message: 'Optimization terminated successfully.'
     nfev: 12
      nit: 2
     njev: 4
   status: 0
  success: True
        x: array([-0.49999985])

The function could take arrays as input as well, but still needs to return a single (scalar) value:

In [289]: optimize.minimize(fun, [10,10,10], args=(np.array([1,2,3]), 1, 1))
Out[289]: 
      fun: 2.541666666667115
 hess_inv: array([[ 0.50021475, -0.00126004,  0.00061239],
       [-0.00126004,  0.25822101, -0.00259327],
       [ 0.00061239, -0.00259327,  0.16946887]])
      jac: array([ -8.94069672e-08,   4.47034836e-07,  -2.20537186e-06])
  message: 'Optimization terminated successfully.'
     nfev: 55
      nit: 9
     njev: 11
   status: 0
  success: True
        x: array([-0.50000006, -0.2499999 , -0.16666704])

In [286]: def fun(x, a,b,c):
 ...:     return (a*x**2 + b*x + c).sum()

It's a good idea to make sure the function runs with the proposed x0 and args, e.g.

In [291]: fun(np.array([10,10,10]), np.array([1,2,3]), 1, 1)
Out[291]: 633

If you can't call the objective function, or are confused as to how its arguments work, minimize isn't a magic bullet. This minimization is only as good as your understanding of the objective function.

answered Apr 14, 2017 at 17:58

hpaulj

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3 Comments

user1936752 Over a year ago

Sorry to dig up an old answer but I think another question is probably a duplicate. How would you add a constraint that depends on both the minimization variable, x, and one of the arguments, say a. I have a similar problem with the additional constraint that x + a < 1.

hpaulj Over a year ago

@user1936752, I think the constraints are called with the same arguments as the objective function, i.e. (x,) + args (tuple join). So they have to have the same signature, or at least a compatible one.

user1936752 Over a year ago

thanks for the comment. Maybe I was unable to express my question properly so I've asked a separate question here: stackoverflow.com/questions/54611746/…. If you have time to look at it, I'd be most grateful!

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