Is there a simple way to have optional output arguments in a python function?

Question

I want to do something like the following.

I want to define a function func(), as:

def func(**kwargs):
    if 'a' in kwargs:
        a = funcA()
    if 'b' in kwargs:
        b = funcB()
    ...

or

def func(**kwargs):
    if 'a' in kwargs:
        kwargs['a'] = funcA()
    if 'b' in kwargs:
        kwargs['b'] = funcB()
    ...

where funcA() and funcB() are defined elsewhere. I want to then call the function func() with a variable number of arguments:

>>> func(a = x)
>>> func(b = y)
>>> func(a = x, b = y)
...

These calls should assign values to x, y, etc. What is the best way to accomplish this?

I'm not sure what a = funcA(a) would allow you to do, but func(funcA) allows you to pass in function references as parameters. — OneCricketeer
– OneCricketeer, Commented May 2, 2016 at 23:47
"These calls should assign values to a, b, etc." - calling a function will not cause assignments to variables in the caller's scope. Also, those example calls are passing a and b positionally, so func wouldn't be able to tell which one is intended as a and which one is intended as b. Python's execution model doesn't work the way you're expecting. You'll need to get used to return values and the parameter passing mechanics. — user2357112
– user2357112, Commented May 2, 2016 at 23:49
Assigning to a parameter variable doesn't modify the variable that was used to call the function. — Barmar
– Barmar, Commented May 2, 2016 at 23:49
python3 or python2? 3 let's you do keyword-only arguments more easily and expressively. — Ryan Haining
– Ryan Haining, Commented May 2, 2016 at 23:54
@RyanHaining I want to do this in Python 2.7. If it can only be done well in Python 3.x, I'd like to hear about that, as well. — Coleman Alleman
– Coleman Alleman, Commented May 3, 2016 at 0:31

Barmar · Accepted Answer · 2016-05-03 00:10:55Z

3

You can't modify the caller's variable by assigning to the parameter like that, you need to return a value with the return statement, and the caller can assign back to the variable.

def func(**kwargs):
    if 'a' in kwargs and 'b' in kwargs:
        return (funcA(kwargs['a']), funcB(kwargs['b']))
    elseif 'a' in kwargs:
        return funcA(kwargs['a'])
    elseif 'b' in kwargs:
        return funcB(kwargs['b'])

a = func(a = a)
b = func(b = b)
a, b = func(a = a, b = b)

A more general, scalable method would be to return a dictionary.

def func(**kwargs):
    retval = {}
    if 'a' in kwargs:
        retval['a'] = funcA(kwargs['a'])
    if 'b' in kwargs:
        retval['b'] = funcB(kwargs['b'])
    return retval

a = func(a = a)['a']
b = func(b = b)['b']
vals = func(a = a, b = b)
a = vals['a']
b = vals['b']

edited May 3, 2016 at 0:10

answered May 2, 2016 at 23:55

Barmar

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2 Comments

Coleman Alleman Over a year ago

This is something like what I had in mind, but I want to avoid enumerating all the possible return combinations. Is there a way around that? I edited my question to hopefully be a little clearer about what I'm trying to do.

Barmar Over a year ago

A more scalable method would be to return a dictionary.

Kasravnd · Accepted Answer · 2016-05-03 00:02:43Z

As a general approach, for refusing of writing multiple if condition you better to create a dictionary of your arguments with their relative functions then class the relative function if the intended argument has been passed to the function caller:

def func(**kwargs):
    arg_dict = {'a': funcA, 'b': funcB}
    for arg, value in kwargs.items():
        try:
            yield arg_dict.get(arg)(value)
        except TypeError:
            # do stuff or just pass

This function will return a generator contain the values of relative functions for arguments.

Note that if you don't want to pass the value of your arguments instead of the aforementioned arg_dict you should use arg_dict = {'a': funcA(), 'b': funcB()} and just yield the arg_dict.get(arg).

Jivan · Accepted Answer · 2016-05-02 23:48:42Z

0

You're using positional arguments when calling func, which do not pass in **kwargs.

Instead, you have to specify keyword arguments to func too.

Like this:

def func(**kwargs):
    if 'a' in kwargs:
        funcA(kwargs['a'])
    if 'b' in kwargs:
        funB(kwargs['b'])

And then:

func(a='hello')
func(b='thing')
func(a='hello', b='thing')

See the docs for positional and keyword arguments.

answered May 2, 2016 at 23:48

Jivan

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