Numbers to Strings in Python part 2

Question

After implementing most of the suggestions from the first question, This is the code I got:

from math import log

zero_to_nineteen = ("zero", "one", "two", "three", "four",
                    "five", "six", "seven", "eight", "nine",
                    "ten", "eleven", "twelve", "thirteen", "fourteen",
                    "fifteen", "sixteen", "seventeen", "eighteen", "nineteen")

tens = ("zero", "ten", "twenty", "thirty", "forty",
        "fifty", "sixty", "seventy", "eighty", "ninety")

# Powers of a thousand (US system of naming)
suffixes = ("zero", "thousand", "million", "billion", "trillion",
            "quadrillion", "quintillion", "sextillion", "septillion",
            "octillion", "nonillion", "decillion", "undecillion",
            "duodecillion", "tredecillion", "quattuordecillion",
            "quinquadecillion", "sedecillion", "septendecillion",
            "octodecillion", "novendecillion", "vigintillion",
            "unvigintillion", "duovigintillion", "tresvigintillion",
            "quattuorvigintillion", "quinquavigintillion", "sesvigintillion",
            "septemvigintillion", "octovigintillion", "novemvigintillion",
            "trigintillion", "untrigintillion", "duotrigintillion",
            "trestrigintilion", "quattuortrigintillion",
            "quinquatrigintillion", "sestrigintillion", "septentrigintillion",
            "octotrigintillion", "noventrigintillion", "quadragintillion")

def spell_out(number):
    """Returns a string representation of the number, in the US system"""
    try:
        number = int(number)
    except OverflowError:
        # This will be triggered with large float values such as 1e584
        return "infinity"

    if number < 0:
        return "negative " + spell_out(-1 * number)

    if number < 20:
        return zero_to_nineteen[number]

    if number < 100:
        tens_digit, ones_digit = divmod(number, 10)
        return _createNumber(tens[tens_digit], ones_digit, "-")

    if number < 1000:
        hundreds_digit, rest = divmod(number, 100)
        return _createNumber(spell_out(hundreds_digit) + " hundred", rest)

    suffix_index = int(log(number, 1000))

    if suffix_index < len(suffixes):
        suffix_value = 1000 ** suffix_index
        prefix, rest = divmod(number, suffix_value)
        prefix = spell_out(prefix) + " " + suffixes[suffix_index]
        return _createNumber(prefix, rest, ", ")

    return "infinity"

def _createNumber(prefix, suffix, seperator=" "):
    # Returns a number with given prefix and suffix (if needed)
    if not isinstance(prefix, str):
        prefix = spell_out(prefix)
    return prefix if suffix == 0 else prefix + seperator + spell_out(suffix)

I'd like to hear any comments about readability/performance/pythonicity and anything else you can think about that could use improving.

Answer 1

math.log works in floating point and thus has limited precision. On my system at least, your code goes to infinite recursion from spell_out(10**15-1) for example. To avoid this, I would break the number into groups using divmod in a loop.

Answer 2

It looks pretty nice! I can only nitpick.

Instead of:

return "negative " + spell_out(-1 * number)

You could simply:

return "negative " + spell_out(-number)

Some PEP8 recommendations:

• Use 2 blank lines before each function definition
• Use snake case in function names, _create_number instead of _createNumber