312

I have a bunch of datetime objects and I want to calculate the number of seconds since a fixed time in the past for each one (for example since January 1, 1970).

import datetime
t = datetime.datetime(2009, 10, 21, 0, 0)

This seems to be only differentiating between dates that have different days:

t.toordinal()

How does one convert a datetime object to seconds?

3

13 Answers 13

302

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.

15
  • 1
    Python now warns me: "TypeError: can't subtract offset-naive and offset-aware datetimes" What's the best solution to fix that?
    – Aaron Ash
    Commented Apr 13, 2013 at 0:47
  • 2
    @Charybdis, try datetime.datetime(1970,1,1,tzinfo=pytz.utc). Commented Apr 13, 2013 at 1:00
  • 13
    Consider using: datetime.datetime.utcfromtimestamp(0) I've used this to get the 'epoch' easily. Note that epoch is not always the same on all systems.
    – D. A.
    Commented Nov 5, 2013 at 20:04
  • 3
    Be very careful with timezones here. I'm at UTC+2, which means that the output of time.time() and datetime.now() - datetime(1970, 1, 1) differ by 7200 seconds. Rather use (t - datetime.datetime.fromtimestamp(0)).total_seconds(). Do not use utcfromtimestamp(0) if you want to convert a datetime in your local timezone.
    – Carl
    Commented Jul 9, 2015 at 9:18
  • 2
    @D.A.: Python does not support non-POSIX epochs. All systems where python works use the same Epoch: 1970-01-01 00:00:00 UTC
    – jfs
    Commented Jul 10, 2015 at 20:24
225

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.

4
  • 4
    This should be the accepted answer / the accepted answer should be updated accordingly. Commented Sep 12, 2016 at 12:49
  • 20
    Time to upgrade I guess ;) Commented Aug 1, 2017 at 11:42
  • is possible to calculate this distance from another starting date? (i mean to change the 1970-01-01 date)
    – Parsifal
    Commented Feb 25, 2020 at 14:20
  • 1
    @Parsifal, yep! You can subtract a new datetime object representing 1970-01-01 from dt; i.e. dt - datetime(1970, 1, 1). Where dt is datetime.today() from the answer above.
    – UrbanConor
    Commented Feb 15, 2023 at 14:52
137

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
4
  • 24
    be careful when using time.mktime for it's express of local time and it's platform-dependent Commented Jan 9, 2013 at 19:10
  • 9
    Be careful indeed. It bit me to my ass big time
    – Arg
    Commented Aug 9, 2013 at 7:29
  • it assumes that t is a local time. UTC offset for the local timezone may have been different in the past and if mktime() (C library) has no access to a historical timezone data on a given platform than it may fail (pytz is a portable way to access the tz database). Also, local time may be ambiguous e.g., during DST transitions -- you need additional info to disambiguate e.g., if you know that consecutive date/time values should be increasing in a log file
    – jfs
    Commented Jul 10, 2015 at 21:00
  • 1
    Be careful when using this with times that have fractions of a second. time.mktime(datetime.datetime(2019, 8, 3, 4, 5, 6, 912000).timetuple()) results in 1564819506.0, silently dropping the milliseconds, but datetime.datetime(2019, 8, 3, 4, 5, 6, 912000).timestamp() (Andrzej Pronobis' answer) results in 1564819506.912, the expected result.
    – Alex
    Commented Nov 5, 2019 at 23:40
33

Maybe off-the-topic: to get UNIX/POSIX time from datetime and convert it back:

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

# and back
>>> datetime.datetime.fromtimestamp(s)
datetime.datetime(2011, 10, 21, 0, 0)

Note that different timezones have impact on results, e.g. my current TZ/DST returns:

>>>  time.mktime(datetime.datetime(1970, 1, 1, 0, 0).timetuple())
-3600 # -1h

therefore one should consider normalizing to UTC by using UTC versions of the functions.

Note that previous result can be used to calculate UTC offset of your current timezone. In this example this is +1h, i.e. UTC+0100.

References:

3
  • mktime() may fail. In general, you need pytz to convert local time to utc, to get POSIX timestamp.
    – jfs
    Commented Jul 10, 2015 at 21:07
  • calendar.timegm() seems to be the utc version of time.mktime()
    – frankster
    Commented Aug 5, 2015 at 10:56
  • Also beware: mktime is up to 10x slower than other approaches. Since it's not unlikely you're doing this in a hot path, it matters.
    – Remko
    Commented Oct 15, 2021 at 14:01
28

int (t.strftime("%s")) also works

3
  • 2
    Works for me in Python 2.7, with import datetime; t = datetime.datetime(2011, 10, 21, 0, 0) (as specified by OP). But really, I doubt %s is a recently-added time format.
    – dan3
    Commented Apr 5, 2014 at 10:30
  • 1
    @dan3: wrong. %s is not supported (it may work on some platforms iff t is a naive datetime object representing local time and if the local C library has access to the tz database otherwise the result may be wrong). Don't use it.
    – jfs
    Commented Jul 10, 2015 at 20:35
  • %s is not documented anywhere. I sow it in real code and was wondering what is this and look in the documentation and there is no such thing as %s. There is only with big S just for the seconds.
    – VStoykov
    Commented Oct 5, 2016 at 10:36
17

from the python docs:

timedelta.total_seconds()

Return the total number of seconds contained in the duration. Equivalent to

(td.microseconds + (td.seconds + td.days * 24 * 3600) * 10**6) / 10**6

computed with true division enabled.

Note that for very large time intervals (greater than 270 years on most platforms) this method will lose microsecond accuracy.

This functionality is new in version 2.7.

2
  • 1
    there is a small problem with the calculation it must be 106 instead of 10*6 ... td.microseconds + (td.seconds + td.days * 24 * 3600) * 106) / 10**6
    – sdu
    Commented Nov 27, 2012 at 10:22
  • @sdu great catch - the double asterisk is in my answer but stackoverflow consumes it, attempts to rectify only embolden the text.
    – Michael
    Commented Jan 18, 2013 at 19:01
4

I do not see this in all of the answers, although I guess it is the default need:

t_start = datetime.now()
sleep(2)
t_end = datetime.now()
duration = t_end - t_start
print(round(duration.total_seconds()))

If you do not use .total_seconds(), it throws: TypeError: type datetime.timedelta doesn't define __round__ method.

Example:

>>> duration
datetime.timedelta(seconds=53, microseconds=621861)
>>> round(duration.total_seconds())
54
>>> duration.seconds
53

Taking duration.seconds takes only the seconds, leaving aside the microseconds, the same as if you ran math.floor(duration.total_seconds()).

4

Comparing the 4 most common ways to do this, for accuracy:

Method 1: Manual Calculation

from datetime import datetime
total1 = int(datetimeobj.strftime('%S'))
total1 += int(datetimeobj.strftime('%M')) * 60
total1 += int(datetimeobj.strftime('%H')) * 60 * 60
total1 += (int(datetimeobj.strftime('%j')) - 1) * 60 * 60 * 24
total1 += (int(datetimeobj.strftime('%Y')) - 1970) * 60 * 60 * 24 * 365
print ("Method #1: Manual")
print ("Before: %s" % datetimeobj)
print ("Seconds: %s " % total1)
print ("After: %s" % datetime.fromtimestamp(total1))

Output:

Method #1: Manual
Before: 1970-10-01 12:00:00 
Seconds: 23630400 
After: 1970-10-01 16:00:00

Accuracy test: FAIL (time zone shift)

Method 2: Time Module

import time
from datetime import datetime
total2 = int(time.mktime(datetimeobj.timetuple()))
print ("Method #2: Time Module")
print ("Before: %s" % datetimeobj)
print ("Seconds: %s " % total2)
print ("After: %s" % datetime.fromtimestamp(total2))

Output:

Method #2: Time Module
Before: 1970-10-01 12:00:00 
Seconds: 23616000 
After: 1970-10-01 12:00:00

Accuracy test: PASS

Method 3: Calendar Module

import calendar
from datetime import datetime
total3 = calendar.timegm(datetimeobj.timetuple())
print ("Method #3: Calendar Module")
print ("Before: %s" % datetimeobj)
print ("Seconds: %s " % total3)
print ("After: %s" % datetime.fromtimestamp(total3))

Output:

Method #3: Calendar Module
Before: 1970-10-01 12:00:00
Seconds: 23616000
After: 1970-10-01 16:00:00

Accuracy test: FAIL (time zone shift)

Method 4: Datetime Timestamp

from datetime import datetime
total4 = datetimeobj.timestamp()
print ("Method #4: datetime timestamp")
print ("Before: %s" % datetimeobj)
print ("Seconds: %s " % total4)
print ("After: %s" % datetime.fromtimestamp(total4))

Output:

Method #2: Time Module
Before: 1970-10-01 12:00:00 
Seconds: 23616000 
After: 1970-10-01 12:00:00

Accuracy test: PASS

Conclusion

  • All 4 methods convert datetime to epoch (total seconds)
  • Both the Manual method and Calendar module method are time zone aware.
  • Both datetime.timestamp() and time.mktime() methods are time zone unaware.
  • Simplest method: datetime.timestamp()
2

To convert a datetime object that represents time in UTC to POSIX timestamp:

from datetime import timezone

seconds_since_epoch = utc_time.replace(tzinfo=timezone.utc).timestamp()

To convert a datetime object that represents time in the local timezone to POSIX timestamp:

import tzlocal # $ pip install tzlocal

local_timezone = tzlocal.get_localzone()
seconds_since_epoch = local_timezone.localize(local_time, is_dst=None).timestamp()

See How do I convert local time to UTC in Python? If the tz database is available on a given platform; a stdlib-only solution may work.

Follow the links if you need solutions for <3.3 Python versions.

1

I tried the standard library's calendar.timegm and it works quite well:

# convert a datetime to milliseconds since Epoch
def datetime_to_utc_milliseconds(aDateTime):
    return int(calendar.timegm(aDateTime.timetuple())*1000)

Ref: https://docs.python.org/2/library/calendar.html#calendar.timegm

1
  • it strips fractions of a second. It assumes that aDateTime is a UTC time
    – jfs
    Commented Jul 10, 2015 at 21:09
0

Python provides operation on datetime to compute the difference between two date. In your case that would be:

t - datetime.datetime(1970,1,1)

The value returned is a timedelta object from which you can use the member function total_seconds to get the value in seconds.

(t - datetime.datetime(1970,1,1)).total_seconds()
0
0
import datetime
import math


def getSeconds(inputDate):
    time = datetime.date.today().strftime('%m/%d/%Y')
    date_time = datetime.datetime.strptime(time, '%m/%d/%Y')
    msg = inputDate
    props = msg.split(".")
    a_timedelta = datetime.timedelta
    if(len(props)==3):
        a_timedelta = date_time - datetime.datetime(int(props[0]),int(props[1]),int(props[2]))
    else:
        print("Invalid date format")
        return
    seconds = math.trunc(a_timedelta.total_seconds())
    print(seconds)
    return seconds

Example getSeconds("2022.1.1")

-3

The standard way to find the processing time in ms of a block of code in python 3.x is the following:

import datetime

t_start = datetime.datetime.now()

# Here is the python3 code, you want 
# to check the processing time of

t_end = datetime.datetime.now()
print("Time taken : ", (t_end - t_start).total_seconds()*1000, " ms")

Not the answer you're looking for? Browse other questions tagged or ask your own question.