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
asked Sep 19, 2008 at 7:19
Paul StephensonPaul Stephenson
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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
answered Sep 19, 2008 at 7:54
paxdiablopaxdiablo
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As others here have stated, you can just use full datetime objects throughout:
from datetime import datetime, date, time, timedelta
sometime = time[8,00] # 8am
later = [datetime.combine[date.today[], sometime] + timedelta[seconds=3]].time[]
However, I think it's worth explaining why full datetime objects are required. Consider what would happen if I added 2 hours to 11pm. What's the correct behavior? An exception, because you can't have a time larger than 11:59pm? Should it wrap back around?
Different programmers will expect different things, so whichever result they picked would surprise a lot of people. Worse yet, programmers would write code that worked just fine when they tested it initially, and then have it break later by doing something unexpected. This is very bad, which is why you're not allowed to add timedelta objects to time objects.
answered Sep 19, 2008 at 13:47
Eli CourtwrightEli Courtwright
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One little thing, might add clarity to override the default value for seconds
>>> b = a + datetime.timedelta[seconds=3000]
>>> b
datetime.datetime[1, 1, 1, 12, 24, 59]
answered Sep 19, 2008 at 8:08
0
Thanks to @Pax Diablo, @bvmou and @Arachnid for the suggestion of using full datetimes throughout. If I have to accept datetime.time objects from an external source, then this seems to be an alternative add_secs_to_time[]
function:
def add_secs_to_time[timeval, secs_to_add]:
dummy_date = datetime.date[1, 1, 1]
full_datetime = datetime.datetime.combine[dummy_date, timeval]
added_datetime = full_datetime + datetime.timedelta[seconds=secs_to_add]
return added_datetime.time[]
This verbose code can be compressed to this one-liner:
[datetime.datetime.combine[datetime.date[1, 1, 1], timeval] + datetime.timedelta[seconds=secs_to_add]].time[]
but I think I'd want to wrap that up in a function for code clarity anyway.
answered Sep 19, 2008 at 9:40
Paul StephensonPaul Stephenson
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You cannot simply add number to datetime
because it's unclear what unit is used: seconds, hours, weeks...
There is timedelta
class for manipulations with date and time. datetime
minus datetime
gives timedelta
, datetime
plus timedelta
gives datetime
, two datetime
objects cannot be added although two timedelta
can.
Create
timedelta
object with how many seconds you want to add and add it to datetime
object:
>>> from datetime import datetime, timedelta
>>> t = datetime.now[] + timedelta[seconds=3000]
>>> print[t]
datetime.datetime[2018, 1, 17, 21, 47, 13, 90244]
There is same concept in C++: std::chrono::duration
.
answered Jan 18, 2018 at 7:30
user2387567user2387567
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If it's worth adding another file /
dependency to your project, I've just written a tiny little class that extends datetime.time
with the ability to do arithmetic. When you go past midnight, it wraps around zero. Now, "What time will it be, 24 hours from now" has a lot of corner cases, including daylight savings time, leap seconds, historical timezone changes, and so on. But sometimes you really do need the simple case, and that's what this will do.
Your example would be written:
>>> import datetime
>>> import nptime
>>> nptime.nptime[11, 34, 59] + datetime.timedelta[0, 3]
nptime[11, 35, 2]
nptime
inherits from datetime.time
, so
any of those methods should be usable, too.
It's available from PyPi as nptime
["non-pedantic time"], or on GitHub: //github.com/tgs/nptime
answered Jul 27, 2011 at 3:49
rescdskrescdsk
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For completeness' sake, here's the way to do it with arrow
[better dates and times for Python]:
sometime = arrow.now[]
abitlater = sometime.shift[seconds=3]
answered Oct 9, 2017 at 19:41
In a real world environment it's never a good idea to work solely with time
, always use datetime
, even better utc
, to avoid conflicts like overnight, daylight saving, different timezones between user and server etc.
So I'd recommend this approach:
import datetime as dt
_now = dt.datetime.now[] # or dt.datetime.now[dt.timezone.utc]
_in_5_sec = _now + dt.timedelta[seconds=5]
# get '14:39:57':
_in_5_sec.strftime['%H:%M:%S']
answered Jan 27, 2021 at 13:44
VengaVengaVengaVenga
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Try adding a datetime.datetime
to a datetime.timedelta
. If you only want the time portion, you can call the time[]
method on the resultant datetime.datetime
object to get it.
answered Sep 19, 2008 at 7:29
Nick JohnsonNick Johnson
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Old question, but I figured I'd throw in a
function that handles timezones. The key parts are passing the datetime.time
object's tzinfo
attribute into combine, and then using timetz[]
instead of time[]
on the resulting dummy datetime. This answer partly inspired by the other answers here.
def add_timedelta_to_time[t, td]:
"""Add a timedelta object to a time object using a dummy datetime.
:param t: datetime.time object.
:param td: datetime.timedelta object.
:returns: datetime.time object, representing the result of t + td.
NOTE: Using a gigantic td may result in an overflow. You've been
warned.
"""
# Create a dummy date object.
dummy_date = date[year=100, month=1, day=1]
# Combine the dummy date with the given time.
dummy_datetime = datetime.combine[date=dummy_date, time=t, tzinfo=t.tzinfo]
# Add the timedelta to the dummy datetime.
new_datetime = dummy_datetime + td
# Return the resulting time, including timezone information.
return new_datetime.timetz[]
And here's a really simple test case class [using built-in unittest
]:
import unittest
from datetime import datetime, timezone, timedelta, time
class AddTimedeltaToTimeTestCase[unittest.TestCase]:
"""Test add_timedelta_to_time."""
def test_wraps[self]:
t = time[hour=23, minute=59]
td = timedelta[minutes=2]
t_expected = time[hour=0, minute=1]
t_actual = add_timedelta_to_time[t=t, td=td]
self.assertEqual[t_expected, t_actual]
def test_tz[self]:
t = time[hour=4, minute=16, tzinfo=timezone.utc]
td = timedelta[hours=10, minutes=4]
t_expected = time[hour=14, minute=20, tzinfo=timezone.utc]
t_actual = add_timedelta_to_time[t=t, td=td]
self.assertEqual[t_expected, t_actual]
if __name__ == '__main__':
unittest.main[]
answered Aug 14, 2019 at 15:58
blthayerblthayer
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If you don't already have a timedelta object, another possibility would be to just initialize a new time object instead with the attributes of the old one and add values where needed:
new_time:time = time[
hour=curr_time.hour + n_hours,
minute=curr_time.minute + n_minutes,
seconds=curr_time.second + n_seconds
]
Admittedly this only works if you make a few assumptions about your values, since overflow is not handled here. But I just thought it was worth to keep this in mind as it can save a line or two
answered Mar 30 at 8:52