Convert timedelta to total seconds

265,484

Solution 1

Use timedelta.total_seconds().

>>> import datetime
>>> datetime.timedelta(seconds=24*60*60).total_seconds()
86400.0

Solution 2

You have a problem one way or the other with your datetime.datetime.fromtimestamp(time.mktime(time.gmtime())) expression.

(1) If all you need is the difference between two instants in seconds, the very simple time.time() does the job.

(2) If you are using those timestamps for other purposes, you need to consider what you are doing, because the result has a big smell all over it:

gmtime() returns a time tuple in UTC but mktime() expects a time tuple in local time.

I'm in Melbourne, Australia where the standard TZ is UTC+10, but daylight saving is still in force until tomorrow morning so it's UTC+11. When I executed the following, it was 2011-04-02T20:31 local time here ... UTC was 2011-04-02T09:31

>>> import time, datetime
>>> t1 = time.gmtime()
>>> t2 = time.mktime(t1)
>>> t3 = datetime.datetime.fromtimestamp(t2)
>>> print t0
1301735358.78
>>> print t1
time.struct_time(tm_year=2011, tm_mon=4, tm_mday=2, tm_hour=9, tm_min=31, tm_sec=3, tm_wday=5, tm_yday=92, tm_isdst=0) ### this is UTC
>>> print t2
1301700663.0
>>> print t3
2011-04-02 10:31:03 ### this is UTC+1
>>> tt = time.time(); print tt
1301736663.88
>>> print datetime.datetime.now()
2011-04-02 20:31:03.882000 ### UTC+11, my local time
>>> print datetime.datetime(1970,1,1) + datetime.timedelta(seconds=tt)
2011-04-02 09:31:03.880000 ### UTC
>>> print time.localtime()
time.struct_time(tm_year=2011, tm_mon=4, tm_mday=2, tm_hour=20, tm_min=31, tm_sec=3, tm_wday=5, tm_yday=92, tm_isdst=1) ### UTC+11, my local time

You'll notice that t3, the result of your expression is UTC+1, which appears to be UTC + (my local DST difference) ... not very meaningful. You should consider using datetime.datetime.utcnow() which won't jump by an hour when DST goes on/off and may give you more precision than time.time()

Solution 3

More compact way to get the difference between two datetime objects and then convert the difference into seconds is shown below (Python 3x):

from datetime import datetime
        
time1 = datetime.strftime('18 01 2021', '%d %m %Y')
    
time2 = datetime.strftime('19 01 2021', '%d %m %Y')

difference = time2 - time1

difference_in_seconds = difference.total_seconds()
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Updated on January 20, 2021

Comments

  • ripper234
    ripper234 almost 3 years

    I have a time difference

    import time
    import datetime
    
    time1 = datetime.datetime.fromtimestamp(time.mktime(time.gmtime()))
    ...
    time2 = datetime.datetime.fromtimestamp(time.mktime(time.gmtime()))
    diff = time2 - time1
    

    Now, how do I find the total number of seconds that passed? diff.seconds doesn't count days. I could do:

    diff.seconds + diff.days * 24 * 3600
    

    Is there a builtin method for this?

    • ripper234
      ripper234 over 12 years
      @RestRisiko - you're right. Still, it's useful to have the question on Stack Overflow, so the next time me, or someone else, Googles for it, he has a good answer as the top result.
  • ripper234
    ripper234 over 12 years
    Thanks for the clarification. For the purpose I'm using right now, a difference of even a few hours every now and then isn't important, but I'll be sure to check this out in the future when I write something more meaningful.
  • Uwe Geuder
    Uwe Geuder almost 10 years
    If somebody still needs to be compatible with 2.6: See stackoverflow.com/questions/3318348/… for how to extend datetime.timedelta with the new method yourself.
  • Russo
    Russo over 5 years
    datetime.timedelta.total_seconds(time2-time1) in Python3.6
  • snakecharmerb
    snakecharmerb almost 3 years
    This is no different to the 10 year old accepted answer.
  • taras
    taras about 2 years
    (time2-time1).total_seconds() in python 3

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