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First solution in Clear category for Friendly Number by atikin88
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
p=0
while abs(number) >= base:
number = number/base
p += 1
if p+2>len(powers):
break
## round(number,1) - fix shit with floats
number = number if decimals else int(round(number,1))
return f"%.{decimals}f" % (number) + powers[p]+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, 2020
