How to find difference in years between two dates?

[John Machin]

Bruno Desthuilliers wrote:
> thebjorn wrote:
> > For the purpose of finding someone's age I was looking for a way to
> > find how the difference in years between two dates, so I could do
> > something like:
> >
> >   age = (date.today() - born).year
> >
> > but that didn't work (the timedelta class doesn't have a year
> > accessor).
> >
> > I looked in the docs and the cookbook, but I couldn't find anything, so
> > I came up with:
> >
> >   def age(born):
> >       now = date.today()
> >       birthday = date(now.year, born.month, born.day)
> >       return now.year - born.year - (birthday > now and 1 or 0)
> >
> > i.e. calculate the "raw" years first and subtract one if he hasn't had
> > his birthday yet this year... It works, but I'd rather use a standard
> > and generic approach if it exists...?
>
> You may want to have a look at mxDatetime, which has a RelativeDateTime
> type that seems to do what you want:
> http://www.egenix.com/files/python/mxDateTime.html
>

Which pieces of the following seem to be working to you?

>>> import mx.DateTime
>>> f = mx.DateTime.RelativeDateTimeDiff
>>> d = mx.DateTime.Date
>>> f(d(2000, 2, 29), d(2001, 2, 28))
<RelativeDateTime instance for 'YYYY-(-11)-(-28) HH:MM:SS' at 0xaee170>
>>> f(d(2000, 2, 29), d(2001, 3, 1))
<RelativeDateTime instance for '(-0001)-MM-(-01) HH:MM:SS' at 0xb06530>
>>> f(d(2001, 1, 31), d(2001, 2, 28))
<RelativeDateTime instance for 'YYYY-MM-(-28) HH:MM:SS' at 0xaee170>
>>> g = lambda x, y: f(y, x)
>>> g(d(2000, 2, 29), d(2001, 2, 28))
<RelativeDateTime instance for 'YYYY-(+11)-(+30) HH:MM:SS' at 0xb06580>
>>> g(d(2000, 2, 29), d(2001, 3, 1))
<RelativeDateTime instance for '(+0001)-MM-DD HH:MM:SS' at 0xaee170>
>>> g(d(2001, 1, 31), d(2001, 2, 28))
<RelativeDateTime instance for 'YYYY-MM-(+28) HH:MM:SS' at 0xb06580>
>>>

and going the other way, adding one month to 31 January gives you some
date in March which is ludicrous.