Hadley, Sunil, and list,
This is not quite doing what I wanted it to do (as far as I can tell). I
perhaps did not explain it thoroughly. It seems to be sampling one value
for each day leaving ~200 observations. I need for it randomly chose one hab
value for each bird if there is more than one value for a given day, I will
try and example below.
id,date,location2,hab
1,05/23/06,0,1
1,05/23/06,0,2
1,05/23/06,0,1
So in this case the animal was located 3 times on may 23rd but I only want
one of the locations and instead of arbitrarily choosing one I wanted to
randomly sample one.
I hope I did a better job explaining my issue. Thanks in advance.
jm
On Wed, Jul 1, 2009 at 3:38 PM, hadley wickham <h.wick...@gmail.com> wrote:
> On Wed, Jul 1, 2009 at 2:10 PM, Sunil
> Suchindran<sunilsuchind...@gmail.com> wrote:
> > #Highlight the text below (without the header)
> > # read the data in from clipboard
> >
> > df <- do.call(data.frame, scan("clipboard", what=list(id=0,
> > date="",loctype=0 ,haptype=0)))
> >
> > # split the data by date, sample 1 observation from each split, and rbind
> >
> > sampled_df <- do.call(rbind, lapply(split(df,
> > df$date),function(x)x[sample(1:nrow(x), 1),]))
>
> ddply from the plyr package (http://had.co.nz/plyr), makes this sort
> of operation a little simpler:
>
> ddply(df, "date", function(df) df[sample(nrow(df), 1), ])
>
> Hadley
>
>
> --
> http://had.co.nz/
>
--
James A. Martin
850-445-9773
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