Question

Given a datetime.time value in Python, is there a standard way to add an integer number of seconds to it, so that 11:34:59 + 3 = 11:35:02, for example?

These obvious ideas don’t work:

>>> datetime.time(11, 34, 59) + 3 TypeError: unsupported operand type(s) for +: 'datetime.time' and 'int' >>> datetime.time(11, 34, 59) + datetime.timedelta(0, 3) TypeError: unsupported operand type(s) for +: 'datetime.time' and 'datetime.timedelta' >>> datetime.time(11, 34, 59) + datetime.time(0, 0, 3) TypeError: unsupported operand type(s) for +: 'datetime.time' and 'datetime.time' 

In the end I have written functions like this:

def add_secs_to_time(timeval, secs_to_add):     secs = timeval.hour * 3600 + timeval.minute * 60 + timeval.second     secs += secs_to_add     return datetime.time(secs // 3600, (secs % 3600) // 60, secs % 60) 

I can’t help thinking that I’m missing an easier way to do this though.

Related

• python time + timedelta equivalent

Answer 1

You can use full datetime variables with timedelta, and by providing a dummy date then using time to just get the time value.

For example:

import datetime a = datetime.datetime(100,1,1,11,34,59) b = a + datetime.timedelta(0,3) # days, seconds, then other fields. print(a.time()) print(b.time()) 

results in the two values, three seconds apart:

11:34:59 11:35:02 

You could also opt for the more readable

b = a + datetime.timedelta(seconds=3) 

if you’re so inclined.

---

If you’re after a function that can do this, you can look into using addSecs below:

import datetime  def addSecs(tm, secs):     fulldate = datetime.datetime(100, 1, 1, tm.hour, tm.minute, tm.second)     fulldate = fulldate + datetime.timedelta(seconds=secs)     return fulldate.time()  a = datetime.datetime.now().time() b = addSecs(a, 300) print(a) print(b) 

This outputs:

 09:11:55.775695  09:16:55