It really depends on what you want to do with them but one possibility
might be to represent them as chron dates and use a time of 0 for true dates
and noon for missing dates replacing the missing day with 01 or 15 or some
other day:
> library(chron)
> x <- c("01/00/05", "01/22/06")
> no.day <- regexpr("/00/", x) > 0
> as.chron(ifelse(no.day, sub("/00/", "/15/", x), x )) + no.day/2
[1] (01/15/05 12:00:00) (01/22/06 00:00:00)
We can then tell which have the incomplete dates with as.numeric(xx) %% 1 > 0
On Sun, Feb 24, 2008 at 9:00 PM, Anupa Fabian <[EMAIL PROTECTED]> wrote:
> I have a data frame which contains some valuable date information. But for a
> few of the dates, the day information missing .
>
> Viz:
> > interesting.data$date
> [1] "1/22/93" "1/22/93" "1/23/93" "1/00/93" "1/28/93" "1/31/93" "1/12/93"
>
> i.e. for dates where the day info is missing, the "%d" part of the
> "%m/%d/%yy" format is simply represented as "00".
>
> When I apply as.Date to the date information, the dates which don't contain
> exact day information are converted to "NA".
>
> Viz:
> > as.Date(interesting.data$date)
> [1] "1993-01-22" "1993-01-22" "1993-01-23" NA "1993-01-28" "1993-01-31"
> "1993-01-12"
>
> Is there a way of using the as.Date function when I only have partial dates
> (eg missing day information which is represented as "00", as above) such
> that the date isn't represented as NA?
>
> Thanks,
> Anupa
>
>
>
>
>
>
>
>
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