[R] value of strptime in R 1.8.0

[Prof Brian Ripley]

On Wed, 12 Nov 2003, RINNER Heinrich wrote:

> I am using R 1.8.0, under Windows XP.
> 
> What I want to do is a date conversion of a character column of a data frame
> and assign the result as a new column.
> 
> Simple example:
> > x <- data.frame(a=c("yesterday","today","tomorrow"), b=I(c("20031111",
> "20031112", "20031113")))
> # convert x$b from character to date:
> > strptime(x$b, format="%Y%m%d")
> [1] "2003-11-11" "2003-11-12" "2003-11-13"
> # trying to make this a new column in x:
> > x$c <- strptime(x$b, format="%Y%m%d")
> Error in "$<-.data.frame"(`*tmp*`, "c", value = strptime(x$b, format =
> "%Y%m%d")) : 
>         replacement has 9 rows, data has 3
> 
> I have done this before (in R 1.7.0), and I am pretty sure this has been
> working then.

It was an undetected error then.

> What am I doing wrong here (has something changed concerning the value of
> strptime() in R 1.8.0)?

You need as.POSXIct, just as you always have.  strptime gives a list of 
length 9, and that does not go into a dataframe of 3 rows.

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595