在Python 3.3中，添加了新的方法timestamp:

import datetime
seconds_since_epoch = datetime.datetime.now().timestamp()

你的问题说你需要毫秒，你可以得到这样的毫秒:

milliseconds_since_epoch = datetime.datetime.now().timestamp() * 1000

如果在naive datetime对象上使用时间戳，则假定该对象位于本地时区。如果您不希望发生这种情况，请使用时区感知的datetime对象。

下面是一个基于上面答案的函数

def getDateToEpoch(myDateTime):
    res = (datetime.datetime(myDateTime.year,myDateTime.month,myDateTime.day,myDateTime.hour,myDateTime.minute,myDateTime.second) - datetime.datetime(1970,1,1)).total_seconds()
    return res

你可以像这样包装返回值: 返回不带十进制值的字符串或只是int(不带str)

你可以用德罗宁去时空旅行!

import datetime
import delorean
dt = datetime.datetime.utcnow()
delorean.Delorean(dt, timezone="UTC").epoch

http://delorean.readthedocs.org/en/latest/quickstart.html  

>>> import datetime
>>> import time
>>> import calendar

>>> #your datetime object
>>> now = datetime.datetime.now()
>>> now
datetime.datetime(2013, 3, 19, 13, 0, 9, 351812)

>>> #use datetime module's timetuple method to get a `time.struct_time` object.[1]
>>> tt = datetime.datetime.timetuple(now)
>>> tt
time.struct_time(tm_year=2013, tm_mon=3, tm_mday=19, tm_hour=13, tm_min=0, tm_sec=9,     tm_wday=1, tm_yday=78, tm_isdst=-1)

>>> #If your datetime object is in utc you do this way. [2](see the first table on docs)
>>> sec_epoch_utc = calendar.timegm(tt) * 1000
>>> sec_epoch_utc
1363698009

>>> #If your datetime object is in local timeformat you do this way
>>> sec_epoch_loc = time.mktime(tt) * 1000
>>> sec_epoch_loc
1363678209.0

[1] http://docs.python.org/2/library/datetime.html#datetime.date.timetuple

[2] http://docs.python.org/2/library/time.html

Python 2.7文档对time模块的建议

我是这样做的:

from datetime import datetime
from time import mktime

dt = datetime.now()
sec_since_epoch = mktime(dt.timetuple()) + dt.microsecond/1000000.0

millis_since_epoch = sec_since_epoch * 1000