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

Question

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?

Answer 1

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()

Answer 2

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()).

Answer 3

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")

Answer 4

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()

Answer 5

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")

Answer 6

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:

• datetime.date.timetuple
• time.mktime
• datetime.datetime.fromtimestamp
• introduction in time module explains POSIX time, 1970 epoch, UTC, TZ, DST ...

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

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.

Answer 13

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