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First solution in Clear category for Friendly Number by Crachton
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
powers_ = powers.copy()
result, power = number, powers_.pop(0)
while abs(result) >= base and powers_:
if decimals:
result /= base
elif result < 0:
result = int(result / base)
else:
result //= base
power = powers_.pop(0)
return '{:.{dec}f}{pw}{sf}'.format(result, pw=power, sf=suffix, dec=decimals)
#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'
Oct. 12, 2017