Add time afterwards; you can do so with the datetime.replace() method to produce a new datetime object:
my_time = datetime.datetime.strptime('07/05/15', '%m/%d/%y') my_time = my_time.replace(hour=23, minute=59)
The datetime.strptime() sets the hour and minute values to the default, 0. Note that for a two-digit year (like 15) you'd use %y, not %Y, which is for a four-digit year.
You could also use the datetime.combine() class method to pair up a date and a time object:
my_time = datetime.datetime.strptime('07/05/15', '%m/%d/%y') my_time = datetime.datetime.combine(my_time.date(), datetime.time(23, 59))
If you feel you must use a timedelta(), take into account that adding it will again produce a new datetime object. You could use augmented assignment to add it 'in-place':
my_time = datetime.datetime.strptime('07/05/15', '%m/%d/%y') my_time += datetime.timedelta(hours=23, minutes=59)
Demo:
>>> import datetime >>> my_time = datetime.datetime.strptime('07/05/15', '%m/%d/%y') >>> my_time.replace(hour=23, minute=59) datetime.datetime(2015, 7, 5, 23, 59) >>> datetime.datetime.combine(my_time.date(), datetime.time(23, 59)) datetime.datetime(2015, 7, 5, 23, 59) >>> my_time + datetime.timedelta(hours=23, minutes=59) datetime.datetime(2015, 7, 5, 23, 59)