Generating a range of numbers

Question

For practice I wrote a function to generate a list of numbers from a start to an end by a step. I couldn't make it- like the built-in range- default start to 0, because all optional parameters must come after required ones, so def range(start=0, end, step=1) won't work.

Here's my script:

def Range(start, end, step=1):
 """Returns a list of numbers beginning from start and
 incremented by step to end (with end excluded).
 start: an integer specifies the start of the range.
 end: an integer specifies the end of the range.
 step: an integer specifies the number to increment by.
 """
 num = start
 result = []
 while num < end:
 result.append(num)
 num += step
 return result

How can it be refactored and optimized ?

Notes:

• I'm a hobbyist and beginner (in Python in particular, and programming in general).

Answer 1

I'm not sure how the actual python implementation for this works but you could query the length of *args to assign the parameters to the right variables:

def Range(*args):
 len_args = len(args)
 if len_args == 2:
 args = args + (1,)
 elif len_args == 1:
 args = (0, args[0], 1)
 start, end, step = args
 ...

In addition I would make this a generator (similar to how it is in python 3.x):

def Range(*args):
 len_args = len(args)
 if len_args == 2:
 args = args + (1,)
 elif len_args == 1:
 args = (0, args[0], 1)
 start, end, step = args
 while start < end:
 yield start
 start += step

• Mahmood Muhammad Nageeb: That is good! I've thought of using *args, but wasn't sure how to make that. Thank you!
• Mahmood Muhammad Nageeb: But when doing that how can we make the function usage easy? The user won't know in what order he should put arguments. I think the docstring need to be edited.
• 301_Moved_Permanently: You can try def Range(start, stop=None, step=1): if stop is None: start, stop = 0, start
• Mahmood Muhammad Nageeb: I've some questions. Is using len_args = len(args) better than using len(args) in each boolean expression? Does that avoid re-computation and make the script faster?
• Graipher: @Mahmud In theory, yes. in practice len is implemented very efficiently and calling it twice will hardly be worth the overhead of saving it in a variable.