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First solution in Clear category for Friendly Number by stanwys
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
prefix, zn = '', ''
if number < 0:
zn = "-"
number, answer = abs(number), number
for i, power in enumerate(powers):
if number // (base**i) > 0:
answer = number // (base**i)
prefix = powers[i]
if decimals:
i = powers.index(prefix)
frac = round(float(number) / (base**i), decimals)
a_lst = str(frac).split(".")
answer = a_lst[0] + "." + a_lst[1].zfill(decimals)
return zn+str(answer)+prefix+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'
Oct. 29, 2016