>>> format(1234567890,',')'1,234,567,890'

import redef formatNum(num):    num=str(num)    pattern=r'(/d+)(/d{3})((,/d{3})*)'    while True:        num,count=re.subn(pattern,r'/1,/2/3',num)        if count==0:            break    return numif __name__=='__main__':    PRint formatNum(1234455)

def formatNum(num):    num=str(num)    result=''    count=0    for i in num[::-1]:        count+=1        result+=i        if count%3==0:            result+=','      return result[::-1].str