Python time differences

I have two time objects.

Example

开发者_StackOverflowtime.struct_time(tm_year=2010, tm_mon=9, tm_mday=24, tm_hour=19, tm_min=13, tm_sec=37, tm_wday=4, tm_yday=267, tm_isdst=-1)

time.struct_time(tm_year=2010, tm_mon=9, tm_mday=25, tm_hour=13, tm_min=7, tm_sec=25, tm_wday=5, tm_yday=268, tm_isdst=-1)

I want to have the difference of those two. How could I do that? I need minutes and seconds only, as well as the duration of those two.

Time instances do not support the subtraction operation. Given that one way to solve this would be to convert the time to seconds since epoch and then find the difference, use:

>>> t1 = time.localtime()
>>> t1
time.struct_time(tm_year=2010, tm_mon=10, tm_mday=13, tm_hour=10, tm_min=12, tm_sec=27, tm_wday=2, tm_yday=286, tm_isdst=0)
>>> t2 = time.gmtime()
>>> t2
time.struct_time(tm_year=2010, tm_mon=10, tm_mday=13, tm_hour=4, tm_min=42, tm_sec=37, tm_wday=2, tm_yday=286, tm_isdst=0)

>>> (time.mktime(t1) - time.mktime(t2)) / 60
329.83333333333331

>>> t1 = time.mktime(time.strptime("10 Oct 10", "%d %b %y"))
>>> t2 = time.mktime(time.strptime("15 Oct 10", "%d %b %y"))
>>> print(datetime.timedelta(seconds=t2-t1))
5 days, 0:00:00

You can use time.mktime(t) with the struct_time object (passed as "t") to convert it to a "seconds since epoch" floating point value. Then you can subtract those to get difference in seconds, and divide by 60 to get difference in minutes.

There is another way to find the time difference between any two dates (no better than the previous solution, I guess):

 >>> import datetime
 >>> dt1 = datetime.datetime.strptime("10 Oct 10", "%d %b %y")
 >>> dt2 = datetime.datetime.strptime("15 Oct 10", "%d %b %y")
 >>> (dt2 - dt1).days
 5
 >>> (dt2 - dt1).seconds
 0
 >>>

It will give the difference in days or seconds or combination of that. The type for (dt2 - dt1) is datetime.timedelta. Look in the library for further details.