Python Cookbook 2Nd Edition Jun 1002005 [Electronic resources] نسخه متنی

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.

See Also

dateutil documentation at DateUtil, and
datetime documentation in the Library
Reference. mx.DateTime can be found at
http://www.egenix.com//image/library/english/10241_python/mxDateTimel.
mx.DateTime can be found at http://www.egenix.com//image/library/english/10241_python/mxDateTimel.