Order of default and non-default arguments

It is not implemented by range. You can use *args or **args and treat the tuple or the dict as you want. For example:

def f(*args):
  if len(args) == 1:
     print "assuming the first is default"
  elif len(args) == 2:
     print "two arguments were passed"
  else:
     print "Complaining"

You can handle the Exceptions yourself if you really want that

def Range(start=0, end=None):
    if end is None:
        raise AttributeError("end value not specified")
     pass

There are a couple approaches. The first would be to switch the arguments in the function, if some of the arguments are "None". That would work like this.

def range1(value, end=None):
    if end == None:
        end = value
        value = 0
    return _generate_range_values(value, end)

The other primary method would be to have your function get a list of all arguments it receives. Then it can decide what to do, based on the number of arguments.

def range2(*args):
    if len(args) == 1:
        start = 0
        end = int(args[0])
    elif len(args) == 2:
        start = int(args[0])
        end = int(args[1])
    return _generate_range_values(start, end)

The third would be to encourage users to pass named arguments to your function, which makes the order less important.

def range3(end, start=0):
    return _generate_range_values(start, end)

Then users would call it with the named start argument when they wanted something besides 0. (Although the named argument would not be required, it keeps the code clear.

for i in range3(55, start=12)

Well, range is C code which can do this slightly better. Anyways, you can do this:

def range(start, stop=None):
    if stop is None: # only one arg, treat stop as start ...
        stop = start
        start = 0
    ...

and document the function accordingly.