Python 对数字的千分位处理方式_python_

>>> "{:,}".format(56381779049) '56,381,779,049' >>> "{:,}".format(56381779049.1) '56,381,779,049.1' >>>

>>> import re >>> subject = '1234567' >>> result = re.sub(r"(?<=\d)(?=(?:\d\d\d)+$)", ",", subject) >>> result '1,234,567'

>>> import re >>> subject = '1234567' >>> result = re.sub(r"(\d)(?=(\d\d\d)+(?!\d))", r"\1,", subject) >>> result '1,234,567'

Python 2.7 (r27:82500, Nov 23 2010, 18:07:12) [GCC 4.1.2 20070115 (prerelease) (SUSE Linux)] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> format(1234567890,',') '1,234,567,890' >>> 

import re def strConv(s):       s =  str(s)     while True:         (s,count) = re.subn(r"(\d)(\d{3})((:?,\d\d\d)*)$",r"\1,\2\3",s)         if count == 0 : break     return s print strConv(12345)