R help - Dec 2010 - Remove 100 years from a date object

Hello,

I have some data that has dates in the form 27.02.37.  I convert them to
a date object as follows:
as.Date(data$date,format="%d.%m.%y")

But this gives me years such as 2037 when I would like them to be 1937.
 I thought of trying to take off some time i.e.
as.Date(camCD$DoB,format="%d.%m.%y") - 100*365
But that doesn't seem to work out correctly.  Any ideas how to do this?

Thanks

Dan
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On Fri, Dec 10, 2010 at 3:27 PM, Daniel Brewer <daniel.brewer at
icr.ac.uk> wrote:
> Hello,
>
> I have some data that has dates in the form 27.02.37. ?I convert them to
> a date object as follows:
> as.Date(data$date,format="%d.%m.%y")
>
> But this gives me years such as 2037 when I would like them to be 1937.
> ?I thought of trying to take off some time i.e.
> as.Date(camCD$DoB,format="%d.%m.%y") - 100*365
> But that doesn't seem to work out correctly. ?Any ideas how to do this?
Normally to adjust dates you can use as.difftime() and do arithmetic,
but a year is a variable thing (can be 365 or 366 days) so you cant
make a difftime of years. Days are variable things if you worry about
leap seconds...

Also, you could end up with an invalid date if you have 29-Feb-2000
and 29-Feb-1900. One wasn't a leap year...

A solution minus those caveats is to convert to POSIXlt and adjust the
$year element:

 > dob="27.02.37"
 > as.Date(dob,format="%d.%m.%y")
[1] "2037-02-27"
 > dobp = as.POSIXlt(as.Date(dob,format="%d.%m.%y"))
 > dobp$year = dobp$year - 100

 > dobp
[1] "1937-02-27 UTC"
 > as.Date(dobp)
[1] "1937-02-27"

although it might be easier to paste a '19' into your character variable

 > paste(substr(dob,1,6),"19",substr(dob,7,9),sep="")
[1] "27.02.1937"

and then do it the way you started. Assumes you have leading zeroes on
all fields though.

Barry

On Fri, Dec 10, 2010 at 10:27 AM, Daniel Brewer <daniel.brewer at
icr.ac.uk> wrote:
> Hello,
>
> I have some data that has dates in the form 27.02.37. ?I convert them to
> a date object as follows:
> as.Date(data$date,format="%d.%m.%y")
>
> But this gives me years such as 2037 when I would like them to be 1937.
> ?I thought of trying to take off some time i.e.
> as.Date(camCD$DoB,format="%d.%m.%y") - 100*365
> But that doesn't seem to work out correctly. ?Any ideas how to do this?
>
The easiest is just to use chron dates since it uses a cut.off of 30
by default.   That is, if yy is less than that then 2000+yy is used
and if greater than that then 1900+yy is used.

Thus try this:

library(chron)
d <- "27.02.37"
as.Date(dates(d, format = "d.m.y")) # "1937-02-27"
as.Date(d, format = "%d.%m.%y")  # "2037-02-27"


Also if that is not good enough and you want a different value for the
cut.off then note that the default in chron is to use the year.expand
function to expand two digit dates but you can change that via
something like this:

options(chron.year.expand = function(..., cut.off = 25)
year.expand(..., cut.off = cut.off))

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