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Simple solution in Clear category for Friendly Number by mindaugas.dadurkevicius
from math import *
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
i2 = 0
for i in range(len(powers)-1):
if abs(number) // base**(i+1) > 0:
i2 = i+1
form = '%.' + str(decimals) + 'f'
if decimals == 0 and number > 0:
return str(floor(number / base**i2)) + powers[i2] + suffix
elif decimals == 0:
return str(ceil(number / base**i2)) + powers[i2] + suffix
return form % round(number / base**i2, decimals) + powers[i2] + suffix
July 25, 2018
