Create Range of Hours in Python

Question

So I'm looking to take input such as 7:00 AM as a start time, and say, 10:00 PM as an endtime, and then compare a set of times (also in the form of 8:00 AM, 3:00 PM, etc) to see if those time are or aren't in the range of time. Is there a good way to create a "day" that has a range of hours (the endtimes can be as late as 4:00 AM, so I was figuring I would go from 6:00 AM - 5:59 AM instead of midnight - 11:59 PM, if that's possible) that I can check times against?

For example: starttime = 8:00 AM, endtime = 3:00 AM, and checking time = 7:00 AM would return time as being out of the range.

Thanks!

bereal · Accepted Answer · 2012-04-11 18:51:58Z

2

Use datetime parsing and comparison capabilities:

from datetime import datetime, timedelta

def parse_time(s):
    ''' Parse 12-hours format '''
    return datetime.strptime(s, '%I:%M %p')

starttime = parse_time('8:00 AM')
endtime   = parse_time('3:00 AM')
if endtime < starttime:
   # add 1 day to the end so that it's after start
   endtime += timedelta(days=1)

checked_time = parse_time('7:00 AM')

# Can compare:
print starttime <= checked_time < endtime

answered Apr 11, 2012 at 18:51

bereal

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1 Comment

westbyb Over a year ago

Thanks, this should solve my problem of the 3:00 AM being after 8:00 AM.

Fred Larson · Accepted Answer · 2012-04-11 18:47:38Z

2

I'd probably use the datetime library. Use strptime() on your input strings, then you can compare the datetime objects directly.

answered Apr 11, 2012 at 18:47

Fred Larson

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Comments

Abhranil Das · Accepted Answer · 2012-04-11 18:42:37Z

1

Convert all times into a 24-hour format, taking minutes and seconds as fractions of hours, then just compare them as numbers to see if a time lies in an interval.

answered Apr 11, 2012 at 18:42

Abhranil Das

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user1202136 · Accepted Answer · 2012-04-11 18:47:46Z

1

Python's datetime.* classes already support comparison:

>>> import datetime
>>> datetime.datetime(2012,10,2,10) > datetime.datetime(2012,10,2,11)
False
>>> datetime.datetime(2012,10,2,10) > datetime.datetime(2012,10,2,9)
True

So all you have to do is declare a class TimeRange, with two members startTime and endTime. Then a "inside" function can be as simple as:

def inside(self, time):
    return self.startTime <= time and time <= self.endTime

answered Apr 11, 2012 at 18:47

user1202136

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1 Comment

Fred Larson Over a year ago

Actually, comparison is even easier than that in Python: self.startTime <= time <= self.endTime.

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