从Python中的epoch(unix time)开始,如何将datetime对象转换为毫秒?

时间:2011-08-09 16:43:09

标签: python datetime epoch

我想要转换为unix时间的Python datetime对象,或者自1970年代以来的秒/毫秒。

我该怎么做?

13 个答案:

答案 0 :(得分:412)

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

答案 1 :(得分:103)

在Python 3.3中,添加新方法。

datetime.timestamp()

https://docs.python.org/3.3/library/datetime.html#datetime.datetime.timestamp

答案 2 :(得分:93)

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

或时间模块的帮助(没有日期格式化):

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

http://pleac.sourceforge.net/pleac_python/datesandtimes.html

的帮助下回答

文档:

答案 3 :(得分:14)

你可以使用Delorean在空间和时间旅行!

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

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

答案 4 :(得分:14)

答案 5 :(得分:12)

我就是这样做的:

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

答案 6 :(得分:7)

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))

答案 7 :(得分:3)

>>> 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

答案 8 :(得分:1)

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

答案 9 :(得分:1)

这是另一种形式的解决方案,用于规范您的时间对象:

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

答案 10 :(得分:1)

这是我根据以上答案做出的功能

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

您可以像这样包装返回的值:str(int(res)) 要返回它而没有十进制值用作字符串或仅返回int(不带str)

答案 11 :(得分:0)

一些熊猫代码:

import pandas

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

答案 12 :(得分:-9)

另一个解决方案是将隐藏日期时间改为unixtimestampmillis。

private static readonly DateTime UnixEpoch = new DateTime(1970, 1, 1, 0, 0, 0, DateTimeKind.Utc);

    public static long GetCurrentUnixTimestampMillis()
    {
        DateTime localDateTime, univDateTime;
        localDateTime = DateTime.Now;          
        univDateTime = localDateTime.ToUniversalTime();
        return (long)(univDateTime - UnixEpoch).TotalMilliseconds;
    }