[Python-Dev] datetime issues

[Gustavo Niemeyer]

> The mxDateTime code comes under the eGenix Public License, so
> reuse is permittted under the terms of the license. That said,
> I doubt that you want to maintain a date/time parser in the
> Python core -- I certainly don't :-)

Thanks. I've got the relativedelta type ready. It's based on your
specification (thanks for it), but uses no code from your module (you'll
notice that clearly when looking at the code). Indeed, it has even a
different algorithm, which solves some of the current "problems" with
your implementation (like getting 03/03 when adding one month to 31/01).
You're free to adopt it, if you feel like so. I'll be releasing that
code under the PSF license.

I'll probably work on a generic parser, once I get some more time.
Depending on the results, I'll submit it for review and possible
inclusion as well.

Here is the documentation of relativedelta, FWIW:

"""
The relativedelta type is based on the specification of the excelent
work done by M.-A. Lemburg in his mx.DateTime extension. However,
notice that this type does *NOT* implement the same algorithm as
his work. Do *NOT* expect it to behave like mx.DateTime's counterpart.

There's two different ways to build a relativedelta instance. The
first one is passing it two date/datetime classes:

    relativedelta(datetime1, datetime2)

And the other way is to use the following keyword arguments:

    year, month, day, hour, minute, seconds, microseconds:
        Absolute information.

    years, months, weeks, days, hours, minutes, seconds, microseconds:
        Relative information, may be negative.

    weekday:
        Tuple with (wday, nth), specifying the nth relative weekday.

Here is the behavior of operations with relativedelta:

1) Calculate the absolute year, using the 'year' argument, or the
   original datetime year, if the argument is not present.

2) Add the relative 'years' argument to the absolute year.

3) Do steps 1 and 2 for month/months.

4) Calculate the absolute day, using the 'day' argument, or the
   original datetime day, if the argument is not present. Then,
   subtract from the day until it fits in the year and month
   found after their operations.

5) Add the relative 'days' argument to the absolute day. Notice
   that the 'weeks' argument is multiplied by 7 and added to
   'days'.

6) Do steps 1 and 2 for hour/hours, minute/minutes, second/seconds,
   microsecond/microseconds.

7) If the 'weekday' argument is present, calculate the weekday,
   with the given (wday, nth) tuple. wday is the index of the
   weekday (0-6, 0=Mon), and nth is the number of weeks to add
   forward or backward, depending on its signal. Notice that if
   the calculated date is already Monday, for example, using
   (0, 1) or (0, -1) won't change the day.
"""

And here is a little demo:

 >>> from dtutil import *; from datetime import *

 >>> relativedelta(datetime.now(),datetime(1983,9,10,10,00))
 relativedelta(years=+20, days=+6, minutes=+25, seconds=+4,
 microseconds=+231402)

 >>> datetime(1983,9,10,10,00)+relativedelta(years=+20, days=+6,
 minutes=+25, seconds=+4, microseconds=+231402)
 datetime.datetime(2003, 9, 16, 10, 25, 4, 231402)

 >>> relativedelta(datetime(2000,2,29),datetime(2001,3,29))
 relativedelta(years=-1, months=-1)
 >>> datetime(2001,3,29)+relativedelta(years=-1, months=-1)
 datetime.datetime(2000, 2, 29, 0, 0)
 >>> relativedelta(datetime(2000,2,29),datetime(2001,3,30))
 relativedelta(years=-1, months=-1)
 >>> datetime(2001,3,30)+relativedelta(years=-1, months=-1)
 datetime.datetime(2000, 2, 29, 0, 0)

 >>> relativedelta(datetime(2000,2,29),datetime(2001,2,28))
 relativedelta(months=-11, days=-28)
 >>> relativedelta(datetime(2000,2,28),datetime(2001,2,28))
 relativedelta(years=-1)
 >>> relativedelta(datetime(2000,2,27),datetime(2001,2,28))
 relativedelta(years=-1, days=-1)

 >>> relativedelta(datetime(2001,2,28),datetime(2000,2,29))
 relativedelta(years=+1)
 >>> relativedelta(datetime(2001,2,28),datetime(2000,2,28))
 relativedelta(years=+1)
 >>> relativedelta(datetime(2001,2,28),datetime(2000,2,27))
 relativedelta(years=+1, days=+1)

 >>> import calendar
 >>> date.today()+relativedelta(weekday=(calendar.MONDAY, +2))
 datetime.date(2003, 9, 29)
 >>> date.today()+relativedelta(weekday=(calendar.MONDAY, -1))
 datetime.date(2003, 9, 15)

-- 
Gustavo Niemeyer
http://niemeyer.net