这是另一种形式的解决方案与规范化的时间对象:

def to_unix_time(timestamp):
    epoch = datetime.datetime.utcfromtimestamp(0) # start of epoch time
    my_time = datetime.datetime.strptime(timestamp, "%Y/%m/%d %H:%M:%S.%f") # plugin your time object
    delta = my_time - epoch
    return delta.total_seconds() * 1000.0

>>> import datetime
>>> # replace datetime.datetime.now() with your datetime object
>>> int(datetime.datetime.now().strftime("%s")) * 1000 
1312908481000

或者时间模块的帮助(没有日期格式):

>>> import datetime, time
>>> # replace datetime.datetime.now() with your datetime object
>>> time.mktime(datetime.datetime.now().timetuple()) * 1000
1312908681000.0

得到了来自:http://pleac.sourceforge.net/pleac_python/datesandtimes.html的帮助

文档:

time.mktime datetime.timetuple

import time
seconds_since_epoch = time.mktime(your_datetime.timetuple()) * 1000

from datetime import datetime
from calendar import timegm

# Note: if you pass in a naive dttm object it's assumed to already be in UTC
def unix_time(dttm=None):
    if dttm is None:
       dttm = datetime.utcnow()

    return timegm(dttm.utctimetuple())

print "Unix time now: %d" % unix_time()
print "Unix timestamp from an existing dttm: %d" % unix_time(datetime(2014, 12, 30, 12, 0))

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)

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。