Python - Working out if time now is between two ti

问题:

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- for example (just using simple examples):

onhour=23
onmin=30
offhour=4
offmin=15
timenow = datetime.datetime.now().time()

Doing a straight if START < NOW < END scenario won't work for this!

What I have currently is some code which evaluates if it's currently "NightTime", which looks like this:

def check_time(timenow, onhour, onmin, offhour, offmin, verbose):
    now = datetime.datetime.now()
    now_time = now.time()
    # If we're actually scheduling at night:
    if int(offhour) < int(onhour):
        # Check to see if we're in daylight times (ie. off schedule)
        if datetime.time(int(offhour),int(offmin)) <= now_time <= datetime.time(int(onhour),int(onmin)):
            if verbose == True:
                print("Day Time detected.")
            return False
        else:
            if verbose == True:
                print("Night Time detected.")
            return True
    else:
        if datetime.time(int(onhour),int(onmin)) <= now_time <= datetime.time(int(offhour),int(offmin)):
            if verbose == True:
                print("Night Time detected.")
            return True
        else:
            if verbose == True:
                print("Day Time detected.")
            return False

Apologies if the title doesn't sound like anything new, but having reviewed a few existing answers for similar problems such as:

• Difference between two time intervals?

I noticed that these don't seem to account for instances where the Start and End times occur over a day boundary.

In addition to this; any ideas surrounding adding Day based scheduling would be quite useful too! ie. "for Mon - Fri, turn on at 23:00, off a 04:00" - but managing on and off for a day either side (else; something will be turned on, on Friday, but not be turned off on the Saturday-- and yet, including Saturday means it gets turned back on again at 23!...)

I've considered doing a simple "Turn on at X, sleep for Y" to get around this... but if the script is started up during an "On" cycle, it won't be initiated until the next run... But it seems like the simplest option! :)

I'm hoping there's some sort of awesome module that does all this... :D

Compatibility of Python2.7 - 3.2 is pretty important to me too!

回答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.")

回答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")

回答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)