If you want to be tricky about it, you can do it in one line in Python. Let d be your original integer. Then [(d%10**(n+1) - d%10**n) / 10**n for n in r

If you want to be tricky about it, you can do it in one line in Python. Let d be your original integer. Then [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))] will be a list containing the base 10 digits of d as integers. Technically, you need to do from math import log10 to make this work, but you can fit that on the same line with a little hacking around using __import__, such as this quick and very dirty attempt: [ [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))] for log10 in [__import__("math", fromlist=["log10"]).log10]][0] Edit: Just to carry this a little further, if we interpret "a new integer that's the result of each of the original integer's digits squared" as doing something like this: 129 -> 1481 (i.e. 9 gets replaced with 81), then we can do it in one line like this: int(''.join([str(x**2) for x in [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))]]))