How to access args as a list?

Tim Chase python.list at
Sat Apr 3 19:20:34 EDT 2010

kj wrote:
> Suppose I have a function with the following signature:
> def spam(x, y, z):
>     # etc.
> Is there a way to refer, within the function, to all its arguments
> as a single list?  (I.e. I'm looking for Python's equivalent of
> Perl's @_ variable.)

It sounds like you want the "*" operator  for arguments:

   def foo(*args):
     print "Args:", repr(args)
     print " Type:", type(args)
     for i, arg in enumerate(args):
       print " Arg #%i = %s" % (i, arg)
   foo("abc", "def", 42)
   # pass a list as args
   lst = [1,2,3]

There's also a keyword form using "**":

   def foo(*args, **kwargs):
     print "Args:", repr(args)
     print " Type:", type(args)
     print "Keyword Args:", repr(kwargs)
     print " Type:", type(kwargs)
     for i, arg in enumerate(args):
       print " Arg #%i = %s" % (i+1, arg)
     for i, (k, v) in enumerate(kwargs.items()):
       print " kwarg #%i %r = %s" % (i+1, k, v)
   foo(1, "hello", something="Whatever", baz=42)
   # pass a list and a dict:
   lst = [1,2,3]
   dct = {100:10, 200:11}
   foo(*lst, **dct)

Both prevent introspection, so if you want the auto-generated 
help to populate with named arguments, you'd have to use the 
inspection method you're currently using.  But generally, if you 
want to be able to treat the args as lists/tuples/dicts, you 
don't care about the names given.


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