Pythonic way to override variables in function with arguments

Question

New to python and have situation like this:

myfun.py:

def fun(x, ...):
   par1 = ...
   par2 = ...
   ... # many pars
   parn = ...
return #*long and complicated function of all the pars and x inputs*

Sometimes I'd like to modify one or more of the pars without modifying myfun.py itself. I know I could:

Define every par as an argument to fun and give it a default value
Make a dictionary par_dict = {'par1': ..., ...} then wherever I have par1 in my return code replace it with par_dict['par1']. Then use par_dict.update() to take a dictionary argument and update my pars.

Any clean and compact alternatives to these options? I'd like something like this:

fun(x_in, par10=5)
# output uses 5 where par10 occurs in the function, in other words the argument to the function overrides the values set inside the function.

Thank you.

Not sure I understand what is being asked but *args and/or **kwargs may be useful. — jodag
– jodag, Commented Mar 14, 2022 at 17:51
Illustrate your need with actual, simple example without omissions. — Parfait
– Parfait, Commented Mar 14, 2022 at 18:01
@jodag I'd like **kwargs to take priority over what is defined locally in the function. — user3547456
– user3547456, Commented Mar 14, 2022 at 18:15
I don't see the problem with defining parameters with default values. It does exactly what you want. — chepner
– chepner, Commented Mar 14, 2022 at 18:21
You're going to have to explicitly define all the default argument values anyway right? I also don't understand what is wrong with just defaulting all the optional arguments. — jodag
– jodag, Commented Mar 14, 2022 at 18:23

Aislan · Accepted Answer · 2022-03-14 19:39:02Z

1

The most pythonic way I find is something like this: First you define the function, such as

def fun(x, *, par1=1, par2=2, ..., parn=999):
    return # function using all params

The function has its default settings, with the predefined values for the parameters. The * as second argument is to prevent the use of positional arguments for further than the x variable.

Then you may use configurable dictionaries to alter the params:

params = {
    'par1': 10,
    'par2': 20,
    ...
    'parn': 0}

fun(X, **params)

The **params distributes the variables declared in the dictionary to the input parameters in the function.

EDIT It is also possible to use nested functions, like this:

def outer(par1=1, par2=2, ..., parn=999):
    def inner(x):
        return # function using x and pars...
    return inner

Notice that the params of the outer function don't need to have default values. Then you "instance" the function, using it with any set of new params.

params = {...}  # like the previous example
fun = outer(**params)
fun(X)

You can use outer to create different functions, that behaves as the params you input, for instance:

params1 = {...}
fun1 = outer(**params1)

params2 = {...}
fun2 = outer(**params2)

a = fun1(X)
b = fun2(X)

In this case a and b are different.

edited Mar 14, 2022 at 19:39

answered Mar 14, 2022 at 18:34

Aislan

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2 Comments

user3547456 Over a year ago

Would it necessarily be bad to do it the way I outline in my second point above?

Aislan Over a year ago

The method I presented expects that, everytime you want parameters different from the default, you should pass a dictionay. However you can use nested functions, then you define the parameters only once, when "instancing" it and you can use it as much as you want. Another method would involve using classes. But it would be similar, I will update the answer with an example of teh nested functions.

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