Re: [R] Problem with very old dates

From: tom soyer <tom.soyer_at_gmail.com>
Date: Sun 29 Oct 2006 - 17:44:37 GMT

Thanks Marc and Gabor for the clarification.

On 10/29/06, Gabor Grothendieck <ggrothendieck@gmail.com> wrote:
>
> On 10/29/06, Marc Schwartz <MSchwartz@mn.rr.com> wrote:
> > On Sun, 2006-10-29 at 10:31 -0600, tom soyer wrote:
> > > Hi,
> > >
> > > I noticed that as.Date() could not convert date string to date type if
> the
> > > dates are very old. For example, if the date string is "1-Mar-50",
> then
> > > as.Date() would convert this to "2050-03-01", NOT "1950-03-01". This
> seems
> > > to be the behavior of as.Date() for dates older than 1969-1-1, and it
> is not
> > > documented in the R as.Date() documentation. It seems very strange
> that R
> > > would fail to convert old dates correctly. Does anyone know if this is
> the
> > > correct behavior? If so, then which method should one use to convert
> old

> > > dates?
> > >
> > > Thanks,
> > >
> > > Tom
> > >
> > > P.S., I am using R 2.4.0 for Windows.
> >
> > This is covered in ?strftime, which is also noted in the "See Also"
> > for ?as.Date, where it says:
> >
> > "Your system's help pages on strftime and strptime to see how to specify
> > their formats."
> >
> > In this case, the former help page in R indicates:
> >
> > %y
> > Year without century (0099). If you use this on input, which
> > century you get is system-specific. So don't! Often values up to
> > 69 (or 68) are prefixed by 20 and 70(or 69) to 99 by 19.
> >
> >
> > Thus on FC5 Linux, I get:
> >
> > > as.Date("1-Mar-50", format = "%d-%b-%y")
> > [1] "2050-03-01"
> >
> >
> > Ideally, you should change the representation of the Year component of
> > the dates you are working with to show a full four digit year and then
> > use (note %Y (capital 'Y') instead of %y):
> >
> > > as.Date("1-Mar-1950", format = "%d-%b-%Y")
> > [1] "1950-03-01"
> >
> > If this data was exported from another data source (ie. Excel) change
> > the format in that program prior to exporting.
> >
> > Otherwise, you could do something like this in R using sub():
> >
> > > sub("-([0-9]+)$", "-19\\1", "1-Mar-50")
> > [1] "1-Mar-1950"
> >
> > Which will change the two digit year ('50') to a four digit year
> > ('1950'). See ?sub and ?regexp for more information.
> >
> > HTH,
> >
> > Marc Schwartz
>
> As mentioned in the Help Desk article of Rnews 4-1, chron uses
> the chron.year.expand option with a default of year.expand to do
> the conversion from 2 digit to 4 digit. year.expand has a
> default cutoff of 30, i.e. years after 30 are regarded to be 19xx
> and ones before 30 are 20xx. Thus if that cutoff is ok for you:
>
> library(chron)
> as.Date(chron("1-Mar-50", format = "day-month-year"))
>
> If the cutoff of 30 is ok then we have a solution. Its possible to change
> that in chron although as discussed in the article but as mentioned
> it is not recommended that you change the chron.* options since it
> might interfere with making your software interoperable with other
> software.
>
> You could also do the 2 to 4 digit conversion yourself as suggested by
> Marc
> or using gsubfn like this (where this example uses a cutoff of 10):
>
> library(gsubfn)
> gsubfn("..$", ~ as.numeric(x) + 100*(as.numeric(x) < 10) + 1900,
> "1-Mar-50")
>
> This matches the last two digits and then adds 100+1900 if year <10
> or adds 1900 if year is greater replacing those digits with the new 4
> digit
> number. Then we can convert the output of gsubfn using as.Date
> unambiguously with the appropriate format argument.
>

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