Count upward in python with variable base

Issue

I would like to know how to do an equivalent of the range function in python, but with the ability to specify the base number. For example:

``````countUp(start=0, end=1010, base=2)
countUp(start=0, end=101, base=3)
countUp(start=0, end=22, base=4)
``````

Example output for base 2 counting:

``````[0, 1, 10, 11, 100, ...]
``````

Is there a function I’m missing that does this? Or what could I do instead?

Solution

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
``````

Answered By – Brian Cain

Answer Checked By – Robin (AngularFixing Admin)