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First solution in Clear category for Friendly Number by Oleg_Domokeev
from math import trunc
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
temp = number
count = 0
while abs(temp) >= base and count < len(powers) - 1:
temp /= base
count += 1
number /= base ** count
if decimals == 0:
number = str(trunc(number))
else:
number = str(round(number, decimals))
if decimals > 0:
if '.' not in number:
number += '.' + '0' * decimals
else:
number += '0' * (number.find('.') + decimals + 1 - len(number))
return number + powers[count] + suffix
#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'
July 5, 2018
