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Simple solution in Clear category for Friendly Number by Pouf
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 + 1) and power < len(powers) - 1:
power += 1
number /= base**power
if not decimals:
number = int(number)
result = '{0:.{1}f}{2}{3}'.format(number, decimals, powers[power], suffix)
return result
March 20, 2017
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