Minutes left until midnight, take 2

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Duh… while driving home today, I realized “why do I need to figure out tomorrow?” (To obtain “midnight”.)  Midnight is always the same number of minutes from the start of the day.  I.e. 24 hrs * 60 min/hour = 1,440 minutes in a day.

So, let’s get the minute that we’re at “now”… and then just subtract that from 1440 to get the minutes that are left in the day.   Thus take 2:

from datetime import datetime

now = datetime.now()
minutes_left = 24 * 60 - (now.hour * 60 + now.minute)

print "As of " + str(now) + \
      ", there are %d minutes until midnight" % minutes_left

Learnings & TODOs

Realized I do need to go back and learn more of the basics, to get a grasp on correct teminology.  Found this post which gave some nice pointers on things to review in the replies:

I give advise you to read thoroughly the parts

3.1. Objects , values and types http://docs.python.org/reference/datamodel.html#the-standard-type-hierarchy

and

4.1. Naming and binding http://docs.python.org/reference/executionmodel.html#naming-and-binding

References:

  • http://docs.python.org/2/library/datetime.html
  • http://stackoverflow.com/questions/5103329/how-to-find-out-what-methods-properties-etc-a-python-module-possesses

 

Minutes left until midnight

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I often tend to do various things that involve date arithmetic, and thought, “hey, what’s a simple quicky problem to get my feet wet in exploring python’s date/time support?

How about, “how many minutes from now until midnight?”

Here’s the documentation for builtin module “datetime”.

A quick glance shows several types:  date, time, datetime and timedelta.  The timedelta looked interesting, and examples suggest I can do math with these types, too.

My thinking is do something like “tomorrow – now”, converted to minutes.

Messing around in IDLE, I got tripped up trying to perform timedelta operations with mixed types, and surprise, you can’t do that:

>>> import datetime
>>> date.today() + timedelta(1)
datetime.date(2013, 1, 6)
>>> datetime.now()
datetime.datetime(2013, 1, 5, 20, 12, 25, 131993)
>>> (date.today() + timedelta(1)) - datetime.now()

Traceback (most recent call last):
  File "<pyshell#285>", line 1, in 
    (date.today() + timedelta(1)) - datetime.now()
TypeError: unsupported operand type(s) for -: 'datetime.date' 
and 'datetime.datetime'
>>>

Ugh.

I gave up trying to use “date.today()”, and constructed ‘tomorrow’ as follows:

import datetime

def how_many_minutes_until_midnight():
     now = datetime.datetime.now()
     tomorrow = datetime.datetime(now.year, now.month, now.day) + \
                datetime.timedelta(1)
     return abs(tomorrow - now).seconds / 60

print how_many_minutes_until_midnight()

POSTSCRIPT

The code written above was the original solution I came up with. But, it’s a bit ugly I think.  Re-googling took me to these hits:

The latter whose comments revealed the use of time() and datetime.combine.  So the above example can be improved further, I think.  (Also note the different import statement.)

from datetime import *

def how_many_minutes_until_midnight():
    tomorrow = date.today() + timedelta(1)
    midnight = datetime.combine(tomorrow, time())
    now = datetime.now()
    return (midnight - now).seconds / 60

print how_many_minutes_until_midnight()

Learnings / TODOs

I did note the timedelta also has “total_seconds()” in addition to “seconds”, the latter which is only the seconds portion of the the overall delta which could span multiple days.  Since the nature of the calculation is within one day, either one is (mostly) equivalent for me:

    return (midnight - now).seconds / 60

and

    return (midnight - now).total_seconds() / 60

Testing these in IDLE showed that the latter returns a float, whereas the former is integer.  But both get the job done.

References