Default arguments with *args and **kwargs

In Python 2.x (I use 2.7), which is the proper way to use default arguments with *args and **kwargs?
I've found a question on SO related to this topic, but that is for Python 3:
Calling a Python function with *args,**kwargs and optional / default arguments

There, they say this method works:

def func(arg1, arg2, *args, opt_arg='def_val', **kwargs):
    #...

In 2.7, it results in a SyntaxError. Is there any recommended way to define such a function?
I got it working this way, but I'd guess there is a nicer solution.

def func(arg1, arg2, *args, **kwargs):
    opt_arg ='def_val'
    if kwargs.__contains__('opt_arg'):
        opt_arg = kwargs['opt_arg']
    #...

Solution 1:

Just put the default arguments before the *args:

def foo(a, b=3, *args, **kwargs):

Now, b will be explicitly set if you pass it as a keyword argument or the second positional argument.

Examples:

foo(x) # a=x, b=3, args=(), kwargs={}
foo(x, y) # a=x, b=y, args=(), kwargs={}
foo(x, b=y) # a=x, b=y, args=(), kwargs={}
foo(x, y, z, k) # a=x, b=y, args=(z, k), kwargs={}
foo(x, c=y, d=k) # a=x, b=3, args=(), kwargs={'c': y, 'd': k}
foo(x, c=y, b=z, d=k) # a=x, b=z, args=(), kwargs={'c': y, 'd': k}

Note that, in particular, foo(x, y, b=z) doesn't work because b is assigned by position in that case.

This code works in Python 3 too. Putting the default arg after *args in Python 3 makes it a "keyword-only" argument that can only be specified by name, not by position. If you want a keyword-only argument in Python 2, you can use @mgilson's solution.

Solution 2:

The syntax in the other question is python3.x only and specifies keyword only arguments. It doesn't work on python2.x.

For python2.x, I would pop it out of kwargs:

def func(arg1, arg2, *args, **kwargs):
    opt_arg = kwargs.pop('opt_arg', 'def_val')

Solution 3:

Similar approach to @yaccob, but clear and concise:

In Python 3.5 or greater:

def foo(a, b=3, *args, **kwargs):
  defaultKwargs = { 'c': 10, 'd': 12 }
  kwargs = { **defaultKwargs, **kwargs }
  print(a, b, args, kwargs)
  
  # Do something    

foo(1) # 1 3 () {'c': 10, 'd': 12}
foo(1, d=5) # 1 3 () {'c': 10, 'd': 5}
foo(1, 2, 4, d=5) # 1 2 (4,) {'c': 10, 'd': 5}

Note: you can use In Python 2

kwargs = merge_two_dicts(defaultKwargs, kwargs)

In Python 3.5

kwargs = { **defaultKwargs, **kwargs }

In Python 3.9

kwargs = defaultKwargs | kwargs  # NOTE: 3.9+ ONLY

Solution 4:

You could also use a decorator like this:

import functools
def default_kwargs(**defaultKwargs):
    def actual_decorator(fn):
        @functools.wraps(fn)
        def g(*args, **kwargs):
            defaultKwargs.update(kwargs)
            return fn(*args, **defaultKwargs)
        return g
    return actual_decorator

Then just do:

@default_kwargs(defaultVar1 = defaultValue 1, ...)
def foo(*args, **kwargs):
    # Anything in here

For instance:

@default_kwargs(a=1)
def f(*args, **kwargs):
    print(kwargs['a']+ 1)

f() # Returns 2
f(3) # Returns 4