Convert timedelta to total seconds

Question

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

User · Answer

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

User · Answer

You can use mx DateTime module  import mx DateTime as mt  t1   mt now    t2   mt now   print int  t2-t1  seconds

User · Answer

Use timedelta total seconds      gt  gt  gt  import datetime  gt  gt  gt  datetime timedelta seconds 24 60 60  total seconds   86400 0

User · Answer

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   gt  gt  gt  import time  datetime  gt  gt  gt  t1   time gmtime    gt  gt  gt  t2   time mktime t1   gt  gt  gt  t3   datetime datetime fromtimestamp t2   gt  gt  gt  print t0 1301735358 78  gt  gt  gt  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  gt  gt  gt  print t2 1301700663 0  gt  gt  gt  print t3 2011-04-02 10 31 03     this is UTC 1  gt  gt  gt  tt   time time    print tt 1301736663 88  gt  gt  gt  print datetime datetime now   2011-04-02 20 31 03 882000     UTC 11  my local time  gt  gt  gt  print datetime datetime 1970 1 1    datetime timedelta seconds tt  2011-04-02 09 31 03 880000     UTC  gt  gt  gt  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

[python] Convert timedelta to total seconds

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