Using default arguments before positional arguments

Question

I am learning to use positional arguments in python and also trying to see how they work when mixed up with default arguments:-

def withPositionalArgs(ae=9,*args         


        

Answer 1

In Python2, you are not allowed to put arguments which have a default value before positional arguments.

The positional arguments must come first, then the arguments with default values (or, when calling the function, the keyword arguments), then *args, and then **kwargs.

This order is required for both the function definition and for function calls.

In Python3, the order has been relaxed. (For example, *args can come before a keyword argument in the function definition.) See PEP3102.

Answer 2

Python3 has relaxed ordering.

Now you can do something like:

def withPositionalArgs(*args, ae=9):
    print('ae=', ae)
    print('args =', args)
a=1
b=2
c=[10, 20]
withPositionalArgs(a, b, c, ae=7)

Answer 3

I think we should make the distinction of default values vs. passing in arbitrary arguments/key-value pairs. The behaviour without default values is the same:

def f(ae,*args, **kwargs):
    print 'ae     = ', ae
    print 'args   = ', args
    print 'kwargs = ', kwargs

The way we have written this means that the first argument passed into f in the tuple args, that is f(1,2,3,a=1,b=2) (the sequence goes explicit arguments, *args, **kwargs.) Here: ae = 1, args = (2,3), kwargs = {'a': 1, 'b': 2}.

If we try to pass in f(1,2,3,a=1,ae=3) we are thrown a TypeError: f() got multiple values for keyword argument 'ae', since the value of ae is attempted to be changed twice.

.

One way around this is to only set ae when it is explicitly prescribed, we could (after the def line):

def g(*args, **kwargs):
    kwargs, kwargs_temp = {"ae": 9}, kwargs
    kwargs.update(kwargs_temp) #merge kwargs into default dictionary

and this time g(1,2,3,a=1,ae=3) sets args=(1,2,3), kwargs={a=1,ae=3}.

However, I suspect this is not best practice...