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First solution in Clear category for Friendly Number by mehWincenty
import math
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
p = 0
rstr = ""
while abs(number) >= base and p < len(powers)-1 :
number = number/base
p += 1
if decimals == 0:
if number < 0:
number += 1
number += 0.01
rstr = str(math.floor(number))
else :
rstr += ('{:.'+str(decimals)+'f}').format(round(number, decimals))
rstr += powers[p]+suffix
return rstr
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert friendly_number(102) == '102', '102'
assert friendly_number(10240) == '10k', '10k'
assert friendly_number(12341234, decimals=1) == '12.3M', '12.3M'
assert friendly_number(12461, decimals=1) == '12.5k', '12.5k'
assert friendly_number(1024000000, base=1024, suffix='iB') == '976MiB', '976MiB'
Nov. 11, 2016
