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First solution in Clear category for Friendly Number by alexandrov.net
import math
def toFixed(numObj, digits):
return f"{numObj:.{digits}f}"
def friendly_number(numbers, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
if numbers==0:
return toFixed(0,decimals)+powers[0]+suffix
i=int(math.log(math.fabs(numbers),base))
if i>len(powers)-1:
i=len(powers)-1
x=numbers/(base**i)
if decimals==0:
return str(int(x))+powers[i]+suffix
else:
return toFixed(x,decimals)+powers[i]+suffix
Sept. 6, 2018
