[Python-Dev] Issue5434: datetime.monthdelta (original) (raw)

Robert Kern robert.kern at gmail.com
Fri Apr 17 00:29:24 CEST 2009

On 2009-04-16 17:17, Paul Moore wrote:

2009/4/16 Robert Kern<robert.kern at gmail.com>:

from dateutil.relativedelta import * dt = relativedelta(months=1) dt relativedelta(months=+1) from datetime import datetime datetime(2009, 1, 15) + dt datetime.datetime(2009, 2, 15, 0, 0) datetime(2009, 1, 31) + dt datetime.datetime(2009, 2, 28, 0, 0) dt.months 1 datetime(2009, 1, 31) + relativedelta(years=-1) datetime.datetime(2008, 1, 31, 0, 0) Yes, but given r = relativedelta(d1, d2) how do I determine the number of months between d1 and d2, and the "remainder" - what monthmod gives me.

Oops! Sorry, I read too quickly and misread "monthmod" as "monthdelta".

From the code, r.months looks like it works, but it's not documented, and I'm not 100% sure if it's always computed.

The result of relativedelta(d1, d2) is the same thing as if it were explicitly constructed from the years=, months=, etc. keyword arguments. From this example, I think this is something that can be relied upon:

""" It works with dates too.

relativedelta(TODAY, johnbirthday) relativedelta(years=+25, months=+5, days=+11, hours=+12) """

The use case I'm thinking of is converting the difference between 2 dates into "3 years, 2 months, 5 days" or whatever. I've got an application which needs to get this right for one of the dates being 29th Feb, so I really get to exercise the corner cases :-)

I believe relativedelta() is intended for this use case although it may resolve ambiguities in a different way than you were hoping.

-- Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco

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