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First solution in Speedy category for Friendly Number by michael.kej
def friendly_number(number,base=1000,decimals=0,suffix='',powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
power, number = 0, float(number)
while abs(number) >= base and power+1 < len(powers):
number /= base
power += 1
if decimals == 0:
number=int(number)
return '%.*f'%(decimals,number) + powers[power] + suffix
Sept. 21, 2015
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