How to check if the current time is in range in python?
Date/time is trickier than you think
Calculations involving date/time can be very tricky because you must consider timezone, leap years, day-light-savings and lots of corner cases. There is an enlightening video from the talk by Taavi Burns at PyCon2012 entitled "What you need to know about datetimes":
What you need to know about datetimes:
time
,datetime
, andcalendar
from the standard library are a bit messy. Find out: what to use where and how (particularly when you have users in many timezones), and what extra modules you might want to look into.Event: PyCon US 2012 / Speakers: Taavi Burns / Recorded: March 10, 2012
Use timezone-aware datetime for calculations
The concept of a datetime.time
for tomorrow is void, because datetime.time
lacks any date information. You probably want to convert everything to timezone-aware datetime.datetime
before comparing:
def time_in_range(start, end, x):
today = timezone.localtime().date()
start = timezone.make_aware(datetime.datetime.combine(today, start))
end = timezone.make_aware(datetime.datetime.combine(today, end))
x = timezone.make_aware(datetime.datetime.combine(today, x))
if end <= start:
end += datetime.timedelta(days=1) # tomorrow!
if x <= start
x += datetime.timedelta(days=1) # tomorrow!
return start <= x <= end
The Python solution is going to be much, much shorter.
def time_in_range(start, end, x):
"""Return true if x is in the range [start, end]"""
if start <= end:
return start <= x <= end
else:
return start <= x or x <= end
Use the datetime.time
class for start
, end
, and x
.
>>> import datetime
>>> start = datetime.time(23, 0, 0)
>>> end = datetime.time(1, 0, 0)
>>> time_in_range(start, end, datetime.time(23, 30, 0))
True
>>> time_in_range(start, end, datetime.time(12, 30, 0))
False