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First solution in Clear category for Friendly Number by igor.v.dudenko
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
power = 0
while (abs(number) / base ** power >= base and
len(powers) - 1 > power):
power += 1
number = number / base ** power
number = '{:.{dec}f}'.format(number, dec=decimals) if decimals else int(number)
return f'{number}{powers[power]}{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'
Feb. 18, 2019
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