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

I need to check if the current time is in timerange. The most simple case time_end > time_start:

if time(6,0) <= now.time() <= time(12,00): print '1'

But troubles begin when user enters a time range when the end time is smaller than the start time, e.g. “23:00 – 06:00”. A time like ’00:00′ will be in this range. About 5 years ago I wrote this PHP function:

function checkInterval($start, $end)
  {    
    $dt = date("H:i:s");    

    $tstart = explode(":", $start);
    $tend =   explode(":", $end);
    $tnow =   explode(":", $dt);

    if (!$tstart[2])
      $tstart[2] = 0;

    if (!$tend[2])
      $tend[2] = 0;  

    $tstart = $tstart[0]*60*60 + $tstart[1]*60 + $tstart[2];
    $tend   = $tend[0]*60*60   + $tend[1]*60   + $tend[2];
    $tnow   = $tnow[0]*60*60   + $tnow[1]*60   + $tnow[2];

    if ($tend < $tstart)
      {
        if ($tend - $tnow > 0 && $tnow > $tstart)
          return true;
        else if ($tnow - $tstart > 0 && $tnow > $tend)
          return true;
        else if ($tend > $tnow && $tend < $tstart && $tstart > $tnow)
          return true;
        else return false;
      } else
      {
        if ($tstart < $tnow && $tend > $tnow)
          return true;
        else
          return false;
      }

Now I need to do the same thing, but I want to make it good looking. So, what algorithm should I use to determine if the current time ’00:00′ is in reversed range e.g. ['23:00', '01:00']?

Answers:

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Method 1

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

Method 2

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

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0