Re: Subtraction Broken?
> Perhaps I'm just approaching this all wrong. I'm just looking for a simple way to compute some ones age today. What you want is to normalize the values of a duration relative to some fixed point in time. I agree this is something that we need to do. Patches are welcome. :) So if I read this correctly the answer is at this time what I'm after is not possible with DateTime directly. But, what I request is something you all desire and hope to have working when someone has the time. I actually wouldn't mind taking a look to see if I could contribute but at this point if: 1968-06-28 2003-08-17 Years:0 Months:0 Days:2 Delta Months:0 Delta Days:12833 is viewed as correct in representing the duration between those two dates I feel I'm missing a little to much to grasp some of the basics to contribute. I really don't see what the years months days methods are all about and if the delta's are simply just supposed to all add up to the duration then I can make since of the Delta's but it seems like I should have some methods representing the deltas in other forms that would also include years. Do you have some other design specs for the DateTime::Duration module that might help me to understand the above output better. However if on the other hand some part of the output above although perhaps correct not exactly what is intended or just stubs for what is desired then if I new what was intended/desired I probably could take a look and see if the effort to produce what I desire from DateTime is worth it. If years really should produce 35 and simply isn't because that method is a stub I would be happy to see what I can produce.
Re: Subtraction Broken?
Try: print $age->deltas, "\n"; If the output from that doesn't look right to you please send it on to the list. Code (DateTime is the current version available from CPAN): #!/usr/bin/perl use DateTime; $birth=DateTime->new(year=>1968,month=>6,day=>28); print $birth->ymd."\n"; $today=DateTime->today; print $today->ymd."\n"; $age=$today-$birth; print "Years:".$age->years."\n"; print "Months:".$age->months."\n"; print "Days:".$age->days."\n"; print "Deltas:".$age->deltas."\n"; print "Delta Months:".$age->delta_months."\n"; print "Delta Days:".$age->delta_days."\n"; Output: 1968-06-28 2003-08-17 Years:0 Months:0 Days:2 Deltas:10 Delta Months:0 Delta Days:12833 Again I fail to see the logic or even value in the DateTime::Duration behaving as above. But, I'm sure I'm probably just missing something important. The only one that does makes since is Deltas but only because it is returning a hash that has 10 elements in it. The Delta Days is sort of interesting but its seems like a lot of work for me to figure out the number of years from that especially when I think the point of this is to take into account all the strange ness that goes on between missing days seconds etc that go on over time. Perhaps I'm just approaching this all wrong. I'm just looking for a simple way to compute some ones age today.
Re: Subtraction Broken?
> Again I fail to see the logic or even value in the DateTime::Duration > behaving as above. But, I'm sure I'm probably just missing something > important. Durations are independent of dates and times. > The only one that does makes since is Deltas but only > because it is returning a hash that has 10 elements in it. The Delta If you would have used my example it would have printed all the hash keys and values. Your use of the concatenation operator forced scalar context so you ended up with the '10'. > Days is sort of interesting but its seems like a lot of work for me > to figure out the number of years from that especially when I think > the point of this is to take into account all the strange ness that > goes on between missing days seconds etc that go on over time. Years relative to what? Years are _not_ all the same length. Nor are months, days, hours, or even minutes. This is what makes this such a complicated problem. I [we] understand your confusion but DT::Duration is correct. > Perhaps I'm just approaching this all wrong. I'm just looking for a > simple way to compute some ones age today. What you want is to normalize the values of a duration relative to some fixed point in time. I agree this is something that we need to do. Patches are welcome. :) -J --
Re: DateTime::Set - Number of elements in set
On Mon, 2003-08-18 at 02:37, [EMAIL PROTECTED] wrote:
> > Flavio, is there already a method to determine
> > the number of elements in a DateTime::Set?
> >
> > If not, I can currently make it as_list, and then
> > get the length of the list, but it would be
> > cleaner to have a method to do this.
>
> What would be a name for this method?
$set->length (sound like it could be a DT::Duration between the first
and last)
$set->elements (sounds like a clone of as_list)
$set->count (possible, though could it be a count of something else?)
$set->number_of_elements (long, but self explainitory)
So what does it return? I figure it returns
1 + $#{$set->{set}{list}}
when it can, and it returns
$DateTime::INF
when it can't.
It would also be good if it could accept a span like as_list does.
Then if there is an infinite set (Say a set of Friday 13ths), we can get
the number of them in a particular span. Without the span parameter it
returns the count of the whole set or it returns INF.
Cheers!
Rick
Re: Subtraction Broken?
Hi Mattew, > $birth=DateTime->new(year=>1968,month=>6,day=>28); > print $birth->ymd."\n"; > $today=DateTime->today; > print $today->ymd."\n"; > $age=$today-$birth; > print $age->years."\n"; $age is a DateTime::Duration object. Unfortunately math with these object can be a little non-intuitive. > I get this output: > > 1968-06-28 > 2003-08-17 > 0 Try: print $age->deltas, "\n"; If the output from that doesn't look right to you please send it on to the list. > That zero sure doesn't seem correct to me. It means zero year 'units' - that doesn't preclude other 'units' adding up to year+ values. Cheers, -J --
Re: DateTime::Set - Number of elements in set
--- Rick Measham <[EMAIL PROTECTED]> wrote: > Flavio, is there already a method to determine the > number of elements in > a DateTime::Set? > > If not, I can currently make it as_list, and then > get the length of the > list, but it would be cleaner to have a method to do > this. There is an internal method. Would you suggest a name for it? $set->cardinality $set->n $set->??? - Flavio S. Glock __ Do you Yahoo!? Yahoo! SiteBuilder - Free, easy-to-use web site design software http://sitebuilder.yahoo.com
Re: DateTime::Set - Number of elements in set
> Flavio, is there already a method to determine
> the number of elements in a DateTime::Set?
>
> If not, I can currently make it as_list, and then
> get the length of the list, but it would be
> cleaner to have a method to do this.
What would be a name for this method?
There is no accessor yet. The number of
elements is stored as:
$set->{set}{too_complex} -- if set is infinite
1 + $#{$set->{set}{list}} -- number of elements
- Flavio S. Glock
PS: sorry if this mail is being re-sent,
I had a problem with my webmail
Subtraction Broken?
I'm not sure if I'm doing something really wrong or if things are broke but with this code: #!/usr/bin/perl use DateTime; $birth=DateTime->new(year=>1968,month=>6,day=>28); print $birth->ymd."\n"; $today=DateTime->today; print $today->ymd."\n"; $age=$today-$birth; print $age->years."\n"; I get this output: 1968-06-28 2003-08-17 0 That zero sure doesn't seem correct to me.
[ANNOUNCE] DateTime::Event::Easter 1.02
The new Event-Easter has been uploaded to CPAN and should be available shortly. This release fixes the issue raised by Ron Hill concerning sets. Basically sets weren't being tested so the problem never showed its head. The sets test was just a duplicate of the lists test! All fixed now though! Cheers! Rick
DateTime::Set - Number of elements in set
Flavio, is there already a method to determine the number of elements in a DateTime::Set? If not, I can currently make it as_list, and then get the length of the list, but it would be cleaner to have a method to do this. Cheers! Rick
