python - Default arguments with *args and **kwargs

Question

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']
    #...

Answer 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.