How to check if the current time is in range in python?

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

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, and calendar 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