What is the standard way to add N seconds to datetime.time?

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

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