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First solution in Clear category for Friendly Number by kurosawa4434
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
n = number
d = 0
while abs(n / base) >= 1 and d < len(powers) - 1:
if n % base:
n /= base
else:
n //= base
d += 1
if not decimals:
n = str(int(n))
else:
n = ('%0.' + str(decimals) + 'f') % round(n, decimals + 1)
return n + powers[d] + suffix
July 16, 2016
