Python: Converting from `datetime.datetime` to `time.time`

Question

In Python, how do I convert a datetime.datetime into the kind of float that I would get from the time.time function?

Asked By: Ram Rachum

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Source

Answer 1

Given a datetime.datetime object dt, you could use

(dt - datetime.datetime.utcfromtimestamp(0)).total_seconds()

Example:

>>> dt = datetime.datetime.now(); t = time.time()
>>> t
1320516581.727343
>>> (dt - datetime.datetime.utcfromtimestamp(0)).total_seconds()
1320516581.727296

Note that the timedelta.total_seconds() method was introduced in Python 2.7.

Answered By: Sven Marnach

Answer 2

A combination of datetime.timetuple() and time.mktime():

>>> import datetime
>>> import time
>>> now = datetime.datetime.now()
>>> secondsSinceEpoch = time.mktime(now.timetuple())

Answered By: bgporter

Answer 3

time.mktime(dt_obj.timetuple())

Should do the trick.

Answered By: Amber

Answer 4

It’s not hard to use the time tuple method and still retain the microseconds:

>>> t = datetime.datetime.now()
>>> t
datetime.datetime(2011, 11, 5, 11, 26, 15, 37496)

>>> time.mktime(t.timetuple()) + t.microsecond / 1E6
1320517575.037496

Answered By: Raymond Hettinger

Answer 5

I know this is an old question, but in python 3.3+ there is now an easier way to do this using the datetime.timestamp() method:

from datetime import datetime
timestamp = datetime.now().timestamp()

Answered By: lsowen

Python: Converting from `datetime.datetime` to `time.time`

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