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First solution in Clear category for Friendly Number by frantisek.jahoda
from math import trunc
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
"""
Format a number as friendly text, using common suffixes.
"""
i = 1
while abs(number) >= base**i and i < len(powers):
i += 1
number /= base**(i-1)
number = round(number, decimals) if decimals != 0 else trunc(number)
return ('%%.%df' % decimals) % number + powers[i-1] + suffix
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert friendly_number(10**32) == '100000000Y'
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'
Oct. 2, 2016
