Keep only two digits after the first digit other than zero

Question

I know how to keep only two decimal places after the decimal point. For example (there are other methods):

>>> print(f'{10000.01908223295211791992:.2f}')
10000.02

But I want to keep the first two digits other than zero after the decimal point:

10000.01908223295211791992 will give: 10000.019

0.0000456576578765 will give: 0.000046

Is there an built-in method that I'm missing? Or is the only solution to code a test for (almost) every case?

Answer 1

The following should be able to match the string and then round the value by applying a magnitude transformation.

import math
import re

TWO_DIGITS_AFTER_ZERO = re.compile(r'(\d+\.(?:(?:[1-9]*)?|0+)0[1-9]{2})')

def truncate(n: float) -> float:
    if result := TWO_DIGITS_AFTER_ZERO.match(format(n, 'f')):
        value = result.group(0)
        magnitude = 10**len(value.split('.')[1])
        return round(float(value) * magnitude) / magnitude
    else:
        return n
 

if __name__ == '__main__':
    assert math.isclose(truncate(10000.01908223295211791992), 10000.019)
    assert math.isclose(truncate(0.0000456576578765), 0.000046)

Here is an example utilizing the ndigits keyword argument of the round() function:

from math import floor, isclose, log10, modf
from typing import Callable

f: Callable[[float],float] = lambda n: round(n, -int(floor(log10(abs(modf(n)[0]))))+1)

if __name__ == '__main__':
    assert isclose(f(10000.01908223295211791992), 10000.019)
    assert isclose(f(0.0000456576578765), 0.000046)