Can a variable number of arguments be passed to a function?

Yes. You can use *args as a non-keyword argument. You will then be able to pass any number of arguments.

def manyArgs(*arg):
  print "I was called with", len(arg), "arguments:", arg

>>> manyArgs(1)
I was called with 1 arguments: (1,)
>>> manyArgs(1, 2, 3)
I was called with 3 arguments: (1, 2, 3)

As you can see, Python will unpack the arguments as a single tuple with all the arguments.

For keyword arguments you need to accept those as a separate actual argument, as shown in Skurmedel's answer.

Adding to unwinds post:

You can send multiple key-value args too.

def myfunc(**kwargs):
    # kwargs is a dictionary.
    for k,v in kwargs.iteritems():
         print "%s = %s" % (k, v)

myfunc(abc=123, efh=456)
# abc = 123
# efh = 456

And you can mix the two:

def myfunc2(*args, **kwargs):
   for a in args:
       print a
   for k,v in kwargs.iteritems():
       print "%s = %s" % (k, v)

myfunc2(1, 2, 3, banan=123)
# 1
# 2
# 3
# banan = 123

They must be both declared and called in that order, that is the function signature needs to be *args, **kwargs, and called in that order.

Another way to go about it, besides the nice answers already mentioned, depends upon the fact that you can pass optional named arguments by position. For example,

def f(x,y=None):
    print(x)
    if y is not None:
        print(y)

Yields

In [11]: f(1,2)
1
2

In [12]: f(1)
1