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)

>>> 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

在我看来，最简单的方法是

import datetime

epoch = datetime.datetime.utcfromtimestamp(0)

def unix_time_millis(dt):
    return (dt - epoch).total_seconds() * 1000.0

Python 2.7文档对time模块的建议

一段熊猫代码:

import pandas

def to_millis(dt):
    return int(pandas.to_datetime(dt).value / 1000000)

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