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Friendly number solution in Clear category for Friendly Number by dominieq
import math
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
result = ""
dec = ""
counter = 0
isBelow = False
if number < 0:
isBelow = True
number = abs(number)
tem = number
while number >= base:
number = number/base
counter = counter + 1
if counter >= len(powers) - 1:
break
if tem != number*(base**counter):
number = tem/(base**counter)
if decimals == 0:
number = math.floor(number)
else:
number = round(number, decimals)
new = str(number)
index = new.find(".")
if index != -1:
dec = new[index+1:]
if len(dec)
Oct. 29, 2016