[R] How do I use as.Date when day values are missing?

[Gabor Grothendieck]

In looking at this again here is a slight simplification.  Its now
only one line:

> library(chron)
> x <- c("01/00/05", "01/22/06")
> as.chron(sub("/00/", "/15/", x)) + (regexpr("/00/", x) > 0) / 2
[1] (01/15/05 12:00:00) (01/22/06 00:00:00)


On Sun, Feb 24, 2008 at 9:45 PM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
> 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 <anupa.fabian at yahoo.com> 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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