Simplifies "messy" large numbers

clean_exponent.py

from decimal import Decimal, DecimalTuple

NUMBER_EXPONENT_PLACES = 4


def clean_exponent(number, places=NUMBER_EXPONENT_PLACES, round_up=True):
    """
    Simplifies messy large numbers

    Example

    number = 345464354
    places = 2

    return = 340000000
    """

    input_type = type(number)
    dec_tuple = Decimal(number).as_tuple()
    clean_digits = [0 for _ in range(len(dec_tuple.digits))]
    for idx, digit in enumerate(dec_tuple.digits):
        if idx < places:
            clean_digits[idx] = digit
        elif round_up and digit > 5:
            clean_digits[idx-1] += 1
        else:
            break
    kwargs = {
        "sign": dec_tuple.sign,
        "digits": clean_digits,
        "exponent": dec_tuple.exponent,
    }
    clean_tuple = DecimalTuple(**kwargs)
    clean_decimal = Decimal(clean_tuple)
    return input_type(clean_decimal)

Hm, iterating all digits in python sets off my 'is it optimal?' spidey-sense...

googling got me here

https://gist.github.com/jackiekazil/6201722

Where examples use the decimal library itself -

Using option #2 you can do something like -

In [1]: from decimal import *
In [7]: d = Decimal(-16.0/7)
In [8]: d
Out[8]: Decimal('-2.28571428571428558740308289998210966587066650390625')

In [9]: Decimal(d).quantize(Decimal('.01'), rounding=ROUND_HALF_UP))
Out[9]: Decimal('-2.29')

@stuaxo the issue was more trying to create a fixed rounding point....