ripper234
ripper234

Reputation: 230038

Convert timedelta to total seconds

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 built-in method for this?

Upvotes: 360

Views: 398998

Answers (6)

user1438233
user1438233

Reputation: 1233

Python 3:

import datetime
start_time = datetime.datetime.now()
# do something
total_seconds = (datetime.datetime.now() - start_time).total_seconds()

Upvotes: 1

ericbn
ericbn

Reputation: 10958

You also can divide a timedelta by another timedelta for a specific unit. As the documentation states:

timedelta.total_seconds()

Return the total number of seconds contained in the duration. Equivalent to td / timedelta(seconds=1). For interval units other than seconds, use the division form directly (e.g. td / timedelta(microseconds=1)).

Upvotes: 1

Atif Bashir
Atif Bashir

Reputation: 359

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()

Upvotes: 1

user2665694
user2665694

Reputation:

Use timedelta.total_seconds().

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

Upvotes: 650

gopisubramni
gopisubramni

Reputation: 23

You can use mx.DateTime module

import mx.DateTime as mt

t1 = mt.now() 
t2 = mt.now()
print int((t2-t1).seconds)

Upvotes: -7

John Machin
John Machin

Reputation: 82934

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()

Upvotes: 12

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