python: convert year/month/day/hour/min/second to # seconds since Jan 1 1970

I know how to do it in C and Java, but I don't know a quick way of converting year/month/day/hour/min/second to the # of seconds since the Jan 1 1970 epoch.

Can someone help me?

So far I've figured out how to create a datetime object but I can't seem to get the elapsed # seconds since the epoch.

(edit: my question is the inverse of this other one: Python: Seconds since epoch to relative date)

Answers


Use timetuple or utctimetuple method to get time tuple and convert it to timestamp using time.mktime

>>> import datetime
>>> dt = datetime.datetime(2011, 12, 13, 10, 23)
>>> import time
>>> time.mktime(dt.timetuple())
1323793380.0

There is a nice bug related to it http://bugs.python.org/issue2736, this is interesting read and anybody trying to convert to timestamp should read this. According to that thread correct way is

timestamp = (dt - datetime(1970, 1, 1)) / timedelta(seconds=1)

You can use datetime.datetime(1970, 1, 1) as a reference and get the total amount of seconds from a datetime.timedelta object as follows:

from datetime import datetime

delta = your_date - datetime(1970, 1, 1)
delta.total_seconds()

import calendar

calendar.timegm(datetime_object.utctimetuple())

These lines can return a float number representing seconds since epoch.

import time

time.time()

To convert a datetime object (broken-downtime time: year/month/day/hour/min/second) to seconds since the Epoch (POSIX time):

seconds_since_epoch = datetime_object.timestamp()

Note: POSIX Epoch is "00:00:00 GMT, January 1, 1970".

If datetime_object has no timezone info then .timestamp() method uses a local timezone.

datetime.timestamp() method is introduced in Python 3.3; for code that works on older Python versions, see Converting datetime.date to UTC timestamp in Python.


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