Count upward in python with variable base

You are apparently confusing numbers with the representation of numbers.

A number does not have a base... it's the number representation that has a base... for example the number represented as "101" in base 2 is the same as the number represented with "5" in base 10.

The range function will count successive numbers, and you can get their representation in any base you like with something like:

digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"

def int2str(x, base):
    if x < 0:
        return "-" + int2str(-x, base)
    return ("" if x < base else int2str(x//base, base)) + digits[x % base]

You can do it with a custom iterator:

I took the iterater code from here and the base conversion from here

import string
class BaseRange:
    def __init__(self, low, high, base):
        digs = string.digits + string.letters
        self.current = low
        self.high = high
        self.base = base
    def __iter__(self):
        return self
    def next(self):  # Python 3 requires this to be __next__
        if self.current > self.high:
            raise StopIteration
        else:
            self.current += 1
            return self.int2base(self.current - 1, self.base)
    def int2base(self, x, base):
        if x < 0: sign = -1
        elif x == 0: return digs[0]
        else: sign = 1
        x *= sign
        digits = []
        while x:
            digits.append(digs[x % base])
            x /= base
        if sign < 0:
            digits.append('-')
            digits.reverse()
        return ''.join(digits)

A Few Example runs produces:

>>> for c in BaseRange(0, 10, 2):
    print(c)


0
1
01
11
001
101
011
111
0001
1001
0101
>>> for c in BaseRange(0, 10, 3):
    print(c)


0
1
2
01
11
21
02
12
22
001
101