Try:

format(d, "%a") or format(d, "%A") or as.POSIXlt(d)$wday

There is also day.of.week in chron.

On Sat, Jul 11, 2009 at 3:42 PM, Mark Knecht <markknecht@gmail.com> wrote:

> On Sat, Jul 11, 2009 at 12:05 PM, Gabor
> Grothendieck<ggrothendieck@gmail.com> wrote:
> > You want %Y, not %y.
> >
> > You might also want to look at the zoo package:
> >
> > library(zoo)
> > z <- read.zoo("Date1.txt", header = TRUE, sep = ",", format = "%m/%d/%Y")
> >
> > or using chron:
> >
> > library(zoo)
> > library(chron)
> > z <- read.zoo("Date1.txt", header = TRUE, sep = ",", FUN = as.chron)
> >
> > There are three vignettes that come with zoo that have more info.
> >
>
> Thanks Gabor. That solved the immediate problem. I appreciate the help.
>
> I'll take a look at zoo and chron. Any guidance on which package I
> should look at for getting what day of the week a certain date is?
> That's something I know I'll need.
>
> Cheers,
> Mark
>

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