What you see is the result of 'printing' something of the class 'date'.
> x <- as.date(c("1jan1960"))
> x
[1] 1Jan60
> str(x)
Class 'date' int 0
> unclass(x)
[1] 0
> getAnywhere('print.date')
A single object matching print.date was found
It was found in the following places
registered S3 method for print from namespace survival
namespace:survival
with value
function (x, quote, prefix, ...)
{
fun <- options()$print.date
if (is.null(fun))
x <- date.ddmmmyy(x)
else x <- get(fun)(x)
if (missing(quote))
quote <- FALSE
invisible(print(x, quote = quote))
}
<environment: namespace:survival>
>
Notice that if you 'unclass' x you get the value 0 which is the number of
days from 1/1/1960 of your input.
On Thu, Apr 30, 2009 at 2:16 AM, utkarshsinghal <
utkarsh.sing...@global-analytics.com> wrote:
> I never understood that why is the value returned by as.date function in
> the library(survival) never matches with the description given in the help
> file:
> Following is the extract from ?as.date
>
> Description:
> Converts any of the following character forms to a Julian date:
> 8/31/56, 8-31-1956, 31 8 56, 083156, 31Aug56, or August 31 1956.
>
> Usage:
> as.date(x, order = "mdy", ...)
>
> Value:
> For each date, the number of days between it and January 1, 1960.
> The date will be missing if the string is not interpretable.
>
> But here is what happens when you actually use it:
>
> > as.date(c("1jan1960"))
> [1] 1Jan60
>
> Can somebody explain this?
> Also please suggest me the best function in R to convert from
> as.Date("2009-08-13") to Julian and vice versa.
>
> Regards
> Utkarsh
>
> ______________________________________________
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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