Dynamically define a function

Question

I am trying to write a curve fitting function which returns the optimal parameters a, b and c, here is a simplified example:

import numpy
import scipy
from scipy.optimize import curve_fit

def f(x, a, b, c):
    return x * 2*a + 4*b - 5*c

xdata = numpy.array([1,3,6,8,10])
ydata = numpy.array([  0.91589774,   4.91589774,  10.91589774,  14.91589774,  18.91589774])
popt, pcov = scipy.optimize.curve_fit(f, xdata, ydata)

This works fine, but I want to give the user a chance to supply some (or none) of the parameters a, b or c, in which case they should be treated as constants and not estimated. How can I write f so that it fits only the parameters not supplied by the user?

Basically, I need to define f dynamically with the correct arguments. For instance if a was known by the user, f becomes:

def f(x, b, c):
    a = global_version_of_a
    return x * 2*a + 4*b - 5*c

unutbu · Accepted Answer · 2011-11-20 02:55:51Z

8

Taking a page from the collections.namedtuple playbook, you can use exec to "dynamically" define func:

import numpy as np
import scipy.optimize as optimize
import textwrap

funcstr=textwrap.dedent('''\
def func(x, {p}):
    return x * 2*a + 4*b - 5*c
''')
def make_model(**kwargs):
    params=set(('a','b','c')).difference(kwargs.keys())
    exec funcstr.format(p=','.join(params)) in kwargs
    return kwargs['func']

func=make_model(a=3, b=1)

xdata = np.array([1,3,6,8,10])
ydata = np.array([  0.91589774,   4.91589774,  10.91589774,  14.91589774,  18.91589774])
popt, pcov = optimize.curve_fit(func, xdata, ydata)
print(popt)
# [ 5.49682045]

Note the line

func=make_model(a=3, b=1)

You can pass whatever parameters you like to make_model. The parameters you pass to make_model become fixed constants in func. Whatever parameters remain become free parameters that optimize.curve_fit will try to fit.

For example, above, a=3 and b=1 become fixed constants in func. Actually, the exec statement places them in func's global namespace. func is thus defined as a function of x and the single parameter c. Note the return value for popt is an array of length 1 corresponding to the remaining free parameter c.

Regarding textwrap.dedent: In the above example, the call to textwrap.dedent is unnecessary. But in a "real-life" script, where funcstr is defined inside a function or at a deeper indentation level, textwrap.dedent allows you to write

def foo():
    funcstr=textwrap.dedent('''\
        def func(x, {p}):
            return x * 2*a + 4*b - 5*c
        ''')

instead of the visually unappealing

def foo():
    funcstr='''\
def func(x, {p}):
    return x * 2*a + 4*b - 5*c
'''

Some people prefer

def foo():
    funcstr=(
        'def func(x, {p}):\n'
        '    return x * 2*a + 4*b - 5*c'
        )

but I find quoting each line separately and adding explicit EOL characters a bit onerous. It does save you a function call however.

edited Nov 20, 2011 at 2:55

answered Nov 19, 2011 at 20:47

unutbu

886k197 gold badges1.9k silver badges1.7k bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Benjamin Over a year ago

Basic question, but I'm having trouble defining funcstr in deeper indentation levels (inside a class). It's throwing indentation errors...

Benjamin Over a year ago

One more thing. What about passing other variables, for example if "c" was an array I needed to pass in as a global, but do not want to have it in the parameters of f? Thanks.

unutbu Over a year ago

If I understand you correctly, just pass it in to make_model: e.g. func=make_model(c=array). (It will not show up in the parameters of f. It will be accessible to f, however.)

Benjamin Over a year ago

btw, that last example does not work, it is lacking \n newlines between the strings.

unutbu Over a year ago

Oh yes, another reason I don't like this method... :) (Thanks for catching the error.)

robince · Accepted Answer · 2011-11-21 12:08:34Z

1

I usually use a lambda for this purpose.

user_b, user_c = get_user_vals()
opt_fun = lambda x, a: f(x, a, user_b, user_c)
popt, pcov = scipy.optimize.curve_fit(opt_fun, xdata, ydata)

answered Nov 21, 2011 at 12:08

robince

11k3 gold badges38 silver badges49 bronze badges

1 Comment

Benjamin Over a year ago

Does that work if the lambda uses numpy functions? How could I also make the opt_fun call dynamic (I want the user to set paramaters at the top of the script, setting to None where they want the parameter to be fitted).

Jan · Accepted Answer · 2011-11-19 21:39:34Z

0

If you want a simple solution based on curve_fit, I'd suggest that you wrap your function in a class. Minimal example:

import numpy
from scipy.optimize import curve_fit

class FitModel(object):
    def f(self, x, a, b, c):
        return x * 2*a + 4*b - 5*c

    def f_a(self, x, b, c):
        return self.f(x, self.a, b, c)

# user supplies a = 1.0
fitModel = FitModel()
fitModel.a = 1.0

xdata = numpy.array([1,3,6,8,10])
ydata = numpy.array([  0.91589774,   4.91589774,  10.91589774,  14.91589774,  18.91589774])
initial = (1.0,2.0)
popt, pconv = curve_fit(fitModel.f_a, xdata, ydata, initial)

answered Nov 19, 2011 at 21:39

Jan

7311 gold badge13 silver badges25 bronze badges

1 Comment

Benjamin Over a year ago

This would start to get messy with 5 parameters, the number of functions I need to write goes up fast...

Honza Alby Michalčák · Accepted Answer · 2016-11-27 14:42:38Z

0

There is already a package that does this:

https://lmfit.github.io/lmfit-py/index.html

From the README:

"LMfit-py provides a Least-Squares Minimization routine and class with a simple, flexible approach to parameterizing a model for fitting to data. Named Parameters can be held fixed or freely adjusted in the fit, or held between lower and upper bounds. In addition, parameters can be constrained as a simple mathematical expression of other Parameters."

edited Nov 27, 2016 at 14:42

Honza Alby Michalčák

637 bronze badges

answered Nov 21, 2011 at 17:14

deprecated

2,31519 silver badges12 bronze badges

Comments

urschrei · Accepted Answer · 2011-11-19 20:09:22Z

-2

def f(x, a = 10, b = 15, c = 25):
    return x * 2*a + 4*b - 5*c

If the user doesn't supply an argument for the parameter in question, whatever you specified on the right-hand side of the = sign will be used:

e.g.:
f(5, b = 20) will evaluate to return 5 * 2*10 + 4*20 - 5*25 and
f(7) will evaluate to return 7 * 2*10 + 4*15 - 5*25

answered Nov 19, 2011 at 20:09

urschrei

27.3k13 gold badges53 silver badges91 bronze badges

Collectives™ on Stack Overflow

Dynamically define a function

5 Answers 5

5 Comments

1 Comment

1 Comment

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

5 Comments

1 Comment

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related