How to find difference in years between two dates?

[thebjorn]

John Machin wrote:
> thebjorn wrote:
> > John Machin wrote:
> > > thebjorn wrote:
[...]
> > > Holy code bloat, Batman! Try this:
> > >
> > >     return now.year - born.year - (birthday > now)
> >
> > yuck :-)
>
> But this:
>     return now.year - born.year - (birthday > now and 1 or 0) is not yuck???

Correct.

> > [...]
> > > It's the irregular-size months that cause the problems. If you can work
> > > out the months difference, then just floor_div by 12 to get the years
> > > difference.
> >
> > I don't agree that the irregular sized months cause a problem in this
> > case. They do cause a problem if you're asking "when is today + one
> > month?", i.e. there isn't an unambiguous answer to that question in
> > general (e.g. if today was January 31). We're asking a different kind
> > of question though: "has it been at least one month since January 31?",
> > the answer would be no on Feb 29 and yes on Mar 1.
>
> If a bank were paying you interest on a monthly basis, and you
> deposited money on Jan 31 and pulled it out on the last day of
> February, that would count as one month. This is what I call the "today
> - yesterday == 1" rule. For computing things like duration of employee
> service, you need the "today - yesterday == 2" rule -- on the
> assumption that service counts from start of business yesterday to
> close of business today. So hire date of 1 Feb to fire date of (last
> day of Feb) would count as one month.

You give a good argument that the concept of a month is fuzzy, I still
don't believe that it makes the concept of a year fuzzy (or that the
fuzziness of month needs to be accounted for when computing years).

> Sorry, I don't understand. Why are you speechless? What would I want to
> use the calendar module for? Apart from the leap() function and the
> table of days in a month, the calendar module doesn't have any of the
> functionality that one would expect in a general-purpose date class.

Well, I thought replacing a 4 line function with 31 lines, 13 of which
duplicated functionality in the standard library was overkill... I came
up with this yesterday which seems sufficient?

    def yeardiff(a, b):
      y = a.year - b.year
      if (a.month, a.day) < (b.month, b.day): # tuple comparison
          y -= 1
      return y 

-- bjorn