问题:如何将日期时间对象转换为自epoch(unix时间)以来的毫秒数?
我有一个Python datetime
对象,我想将其转换为Unix时间,即自1970年以来的秒/毫秒。
我该怎么做呢?
I have a Python datetime
object that I want to convert to unix time, or seconds/milliseconds since the 1970 epoch.
How do I do this?
回答 0
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
It appears to me that the simplest way to do this is
import datetime
epoch = datetime.datetime.utcfromtimestamp(0)
def unix_time_millis(dt):
return (dt - epoch).total_seconds() * 1000.0
回答 1
在Python 3.3中,添加了新方法timestamp
:
import datetime
seconds_since_epoch = datetime.datetime.now().timestamp()
您的问题表明您需要毫秒,您可以这样获得:
milliseconds_since_epoch = datetime.datetime.now().timestamp() * 1000
如果您使用 timestamp
在朴素的datetime对象上使用,则假定它在本地时区中。如果这不是您打算发生的事情,请使用可识别时区的日期时间对象。
In Python 3.3, added new method timestamp
:
import datetime
seconds_since_epoch = datetime.datetime.now().timestamp()
Your question stated that you needed milliseconds, which you can get like this:
milliseconds_since_epoch = datetime.datetime.now().timestamp() * 1000
If you use timestamp
on a naive datetime object, then it assumed that it is in the local timezone. Use timezone-aware datetime objects if this is not what you intend to happen.
回答 2
>>> 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(不带日期格式):
>>> import datetime, time
>>> # replace datetime.datetime.now() with your datetime object
>>> time.mktime(datetime.datetime.now().timetuple()) * 1000
1312908681000.0
在以下网站的帮助下回答:http://pleac.sourceforge.net/pleac_python/datesandtimes.html
说明文件:
>>> import datetime
>>> # replace datetime.datetime.now() with your datetime object
>>> int(datetime.datetime.now().strftime("%s")) * 1000
1312908481000
Or the help of the time module (and without date formatting):
>>> import datetime, time
>>> # replace datetime.datetime.now() with your datetime object
>>> time.mktime(datetime.datetime.now().timetuple()) * 1000
1312908681000.0
Answered with help from: http://pleac.sourceforge.net/pleac_python/datesandtimes.html
Documentation:
回答 3
回答 4
这是我的方法:
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
This is how I do it:
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
回答 5
回答 6
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))
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))
回答 7
>>> 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
>>> 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
回答 8
这是时间对象标准化的另一种解决方案形式:
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
Here’s another form of a solution with normalization of your time object:
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
回答 9
一点熊猫代码:
import pandas
def to_millis(dt):
return int(pandas.to_datetime(dt).value / 1000000)
A bit of pandas code:
import pandas
def to_millis(dt):
return int(pandas.to_datetime(dt).value / 1000000)
回答 10
import time
seconds_since_epoch = time.mktime(your_datetime.timetuple()) * 1000
import time
seconds_since_epoch = time.mktime(your_datetime.timetuple()) * 1000
回答 11
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))要返回它而没有用作字符串的十进制值或仅用作int(不带str)
Here is a function I made based on the answer above
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
You can wrap the returned value like this : str(int(res))
To return it without a decimal value to be used as string or just int (without the str)
回答 12
这是unixtimestampmillis的秘密日期时间的另一种解决方案。
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;
}
This other solution for covert datetime to 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;
}