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First solution in Clear category for Friendly Number by rodka81
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
exps = [(i, power) for i, power in enumerate(powers)]
for i, prefix in reversed(exps):
num = number / (base ** i)
if abs(num) >= 1:
if decimals > 0:
num = round((10 ** decimals) * num) / (10 ** decimals)
else:
num = int(num)
return "{0:.{1}f}{2}{3}".format(num, decimals, prefix, suffix)
return "{0:.{1}f}{2}{3}".format(number, decimals, prefix, suffix)
Nov. 25, 2016
