Recipe3.1.Calculating Yesterday and Tomorrow

Recipe 3.1. Calculating Yesterday and Tomorrow

Credit: Andrea Cavalcanti

Problem

You want to get today's date, then calculate yesterday's or tomorrow's.

Solution

Whenever you have to deal with a "change" or "difference" in time, think timedelta:

import datetime today = datetime.date.today( ) yesterday = today - datetime.timedelta(days=1) tomorrow = today + datetime.timedelta(days=1) print yesterday, today, tomorrow #emits: 2004-11-17 2004-11-18 2004-11-19

Discussion

This recipe's Problem has been a fairly frequent question on Python mailing lists since the datetime module arrived. When first confronted with this task, it's quite common for people to try to code it as yesterday = today - 1, which gives a TypeError: unsupported operand type(s) for -: 'datetime.date' and 'int'.

Some people have called this a bug, implying that Python should guess what they mean. However, one of the guiding principles that gives Python its simplicity and power is: "in the face of ambiguity, refuse the temptation to guess." Trying to guess would clutter datetime with heuristics meant to guess that you "really meant 1 day", rather than 1 second (which timedelta also supports), or 1 year.

Rather than trying to guess what you mean, Python, as usual, expects you to make your meaning explicit. If you want to subtract a time difference of one day, you code that explicitly. If, instead, you want to add a time difference of one second, you can use timedelta with a datetime.datetime object, and then you code the operation using exactly the same syntax. This way, for each task you might want to perform, there's only one obvious way of doing it. This approach also allows a fair amount of flexibility, without added complexity. Consider the following interactive snippet:

>>> anniversary = today + datetime.timedelta(days=365)          # add 1 year >>> print anniversary 2005-11-18 >>> t = datetime.datetime.today( )                               # get right now >>> t datetime.datetime(2004, 11, 19, 10, 12, 43, 801000) >>> t2 = t + datetime.timedelta(seconds=1)                      # add 1 second >>> t2 datetime.datetime(2004, 11, 19, 10, 12, 44, 801000) >>> t3 = t + datetime.timedelta(seconds=3600)                   # add 1 hour >>> t3 datetime.datetime(2004, 11, 19, 11, 12, 43, 801000)

Keep in mind that, if you want fancier control over date and time arithmetic, third-party packages, such as dateutil (which works together with the built-in datetime) and the classic mx.DateTime, are available. For example:

from dateutil import relativedelta  nextweek = today + relativedelta.relativedelta(weeks=1) print nextweek #emits: 2004-11-25

However, "always do the simplest thing that can possibly work." For simple, straightforward tasks such as the ones in this recipe, datetime.timedelta works just fine.

Recipe3.1.Calculating Yesterday and Tomorrow