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First solution in Clear category for Friendly Number by yuwai
import math
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
i=len(powers)-1
while abs(number)< base**i and i>0:
i-=1
number/=base**i
if decimals==0:
number=int(number)
num=str(number)
else :
d='%.{}f'.format(decimals)
num=(d %number) #.format(decimals)
s=num+powers[i]+suffix
return s
#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'
June 17, 2018
