What is the standard way to add N seconds to datetime.time in Python?
datetime, math, python, time
Solution
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
Problem
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