In Python, how do you convert a `datetime` object to seconds?

To get the Unix time (seconds since January 1, 1970):

>>> import datetime, time
>>> t = datetime.datetime(2011, 10, 21, 0, 0)
>>> time.mktime(t.timetuple())
1319148000.0

For the special date of January 1, 1970 there are multiple options.

For any other starting date you need to get the difference between the two dates in seconds. Subtracting two dates gives a timedelta object, which as of Python 2.7 has a total_seconds() function.

>>> (t-datetime.datetime(1970,1,1)).total_seconds()
1256083200.0

The starting date is usually specified in UTC, so for proper results the datetime you feed into this formula should be in UTC as well. If your datetime isn't in UTC already, you'll need to convert it before you use it, or attach a tzinfo class that has the proper offset.

As noted in the comments, if you have a tzinfo attached to your datetime then you'll need one on the starting date as well or the subtraction will fail; for the example above I would add tzinfo=pytz.utc if using Python 2 or tzinfo=timezone.utc if using Python 3.

Starting from Python 3.3 this becomes super easy with the datetime.timestamp() method. This of course will only be useful if you need the number of seconds from 1970-01-01 UTC.

from datetime import datetime
dt = datetime.today()  # Get timezone naive now
seconds = dt.timestamp()

The return value will be a float representing even fractions of a second. If the datetime is timezone naive (as in the example above), it will be assumed that the datetime object represents the local time, i.e. It will be the number of seconds from current time at your location to 1970-01-01 UTC.