Python - Working out if time now is between two times

Question

I\'m trying to find the cleanest/most pythonic way of evaluating if \"now\" is between two times; However; the Start/End times may, or may not, fall across a day boundary- f

Answer 1

Your code is a bit chaotic. I would do something like this:

import datetime

DAY, NIGHT = 1, 2
def check_time(time_to_check, on_time, off_time):
    if on_time > off_time:
        if time_to_check > on_time or time_to_check < off_time:
            return NIGHT, True
    elif on_time < off_time:
        if time_to_check > on_time and time_to_check < off_time:
            return DAY, True
    elif time_to_check == on_time:
        return None, True
    return None, False


on_time = datetime.time(23,30)
off_time = datetime.time(4,15)
timenow = datetime.datetime.now().time()
current_time = datetime.datetime.now().time()

when, matching = check_time(current_time, on_time, off_time)

if matching:
    if when == NIGHT:
        print("Night Time detected.")
    elif when == DAY:
        print("Day Time detected.")

Answer 2

To find out whether a given time (no date) is in between given start, end times (the end is not included):

def in_between(now, start, end):
    if start <= end:
        return start <= now < end
    else: # over midnight e.g., 23:30-04:15
        return start <= now or now < end

Example:

from datetime import datetime, time

print("night" if in_between(datetime.now().time(), time(23), time(4)) else "day")

Answer 3

def is_hour_between(start, end, now):
    is_between = False

    is_between |= start <= now <= end
    is_between |= end < start and (start <= now or now <= end)

    return is_between

test with:

assert is_hour_between(6, 10, 6)
assert not is_hour_between(6, 10, 4)
assert is_hour_between(17, 20, 17)
assert not is_hour_between(17, 20, 16)