How to add delta to python datetime.time?

Question

From:

http://docs.python.org/py3k/library/datetime.html#timedelta-objects

A timedelta object represents a duration, the difference between two

Answer 1

A datetime.time object can be split into separate integer components that you can add to. No need for timedelta eg:

from datetime import datetime, time
    time_now = datetime.now().time()
    twelve_hours_time =  time(time_now.hour + 12, time_now.minute)

Answer 2

datetime.time objects do not support addition with datetime.timedeltas.

There is one natural definition though, clock arithmetic. You could compute it like this:

import datetime as dt
now = dt.datetime.now()
delta = dt.timedelta(hours = 12)
t = now.time()
print(t)
# 12:39:11.039864

print((dt.datetime.combine(dt.date(1,1,1),t) + delta).time())
# 00:39:11.039864

dt.datetime.combine(...) lifts the datetime.time t to a datetime.datetime object, the delta is then added, and the result is dropped back down to a datetime.time object.

Answer 3

Here is a function that adds a time to a timedelta:

def time_plus(time, timedelta):
    start = datetime.datetime(
        2000, 1, 1,
        hour=time.hour, minute=time.minute, second=time.second)
    end = start + timedelta
    return end.time()

This will provide the expected result so long as you don't add times in a way that crosses a midnight boundary.

Answer 4

How would this work? datetime.datetime.now().time() returns only hours, minutes, seconds and so on, there is no date information what .time() returns, only time.

Then, what should 18:00:00 + 8 hours return?
There's not answer to that question, and that's why you can't add a time and a timedelta.

In other words:

18:28:44, Sep. 16, 2012 + 8 hours #makes sense: it's 2:28:44, Sep. 17, 2012
18:28:44 + 5 hours # Doesn't make sense.

Answer 5

All the solutions above are too complicated, OP had already shown that we can do calculation between datetime.datetime and datetime.timedelta, so why not just do:

(datetime.now() + timedelta(hours=12)).time()