使用不同的sep将字符串@拆分一次

一个想法是这样的(未经测试):

years, months, days = the_string.split('-')
days, time = days.split(' ')
time = time.split(':')

或者这个更适合你的数据。

date, time = the_string.split(' ')
years, months, days = date.split('-')
hours, minute, seconds = time.split(":")

我猜你还想在中间的空间上分裂:

import re
values = re.split(r'[- :]', "1122-33-44 55:66:77")
print values
# Prints ['1122', '33', '44', '55', '66', '77']

也许你真正想要的是分析日期

>>> from time import strptime
>>> strptime( '2000-01-01 00:00:00', '%Y-%m-%d %H:%M:%S') 
time.struct_time(tm_year=2000, tm_mon=1, tm_mday=1, tm_hour=0, tm_min=0, tm_sec=0, tm_wday=5, tm_yday=1, tm_isdst=-1)
>>>

使用regexps,并使用模式匹配所需的格式,

>>> pat = r"(\d{4})-(\d\d)-(\d\d) (\d\d):(\d\d):(\d\d)"
>>> m = re.match(pat,'0000-00-00 00:00:00')
>>> m.groups()
('0000', '00', '00', '00', '00', '00')
>>> 

一个更详细的选项,命名组,将字段名与值(用于 @skurpur )

>>> pat = "(?P<year>\d{4})-(?P<month>\d\d)-(?P<day>\d\d) (?P<hour>\d\d):(?P<min>\d\d):(?P<sec>\d\d)"
>>> m = re.match(pat,'0000-00-00 00:00:00')
>>> m.groupdict()
{'hour': '00', 'min': '00', 'month': '00', 'sec': '00', 'year': '0000', 'day': '00'}
>>>