Loopless(cleaner) speedy solution for Friendly Number by JumpCutter - python coding challenges

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Loopless(cleaner) solution in Speedy category for Friendly Number by JumpCutter

from math import log

def friendly_number(number, base=1000, decimals=0, suffix='',
                    powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
   try:
    powered = min(int(log(abs(number), base)), len(powers) - 1)
    number /= (base**powered)
    if decimals == 0: number = int(number)
   except ValueError: powered = 0
   return "{:.{}f}".format(number, decimals) + powers[powered] + suffix

May 12, 2014

Comments:

colinmcnicholl on Aug. 30, 2021, 1:08 p.m.
<p>Nice use of try statement even though solutions for this mission do not require it.</p> <p>Your indentation is inconsistent - looks like 4 spaces between function definition and the try on line 5 then 1 space on line 6. You should fix that. 4 spaces would look much better than 1.</p>

Strings Theory

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