Here is yet another approach using aggregate(), which internally,
basically does what my first solution did:
> aggregate(z[, 2], list(z[, 1]), "[", 1)
Group.1 x
1 1 -1.2006469
2 2 -0.1614918
3 3 -0.5717729
4 4 -0.2398887
5 5 1.1690564
See ?aggregate
Note that you get a data frame as a result, rather than a matrix.
Also, you could 'collapse' the split() and sapply() part of my first
solution using tapply():
> tapply(z[, 2], z[, 1], "[", 1)
1 2 3 4 5
-1.2006469 -0.1614918 -0.5717729 -0.2398887 1.1690564
As has been said by fortune("Yoda"):
Evelyn Hall: I would like to know how (if) I can extract some of the
information from the summary of my nlme.
Simon Blomberg: This is R. There is no if. Only how.
-- Evelyn Hall and Simon 'Yoda' Blomberg
R-help (April 2005)
HTH,
Marc Schwartz
On Fri, 2007-09-28 at 16:12 -0600, Matthew Keller wrote:
> Is this easier?
>
> x.index <- duplicated(x.sample)==FALSE
> cbind(x.sample[x.index],y[x.index])
>
>
> - Matt
>
> On 9/28/07, Marc Schwartz <[EMAIL PROTECTED]> wrote:
> > On Fri, 2007-09-28 at 17:48 -0400, Brian Perron wrote:
> > > Hello all,
> > >
> > > An elementary question that I am sure can be easily cracked by an R
> > > enthusiast. Let's say I have multiple scores (y) on subjects (x.sample).
> > > Some subjects have a few more scores than others. Can somebody suggest
> > > some
> > > code that will select the first score for each subject?
> > >
> > > For example, the following code generates scores for 5 subjects:
> > >
> > > > x <- c(1:5)
> > > > x.sample <- sample(x, 20, replace = TRUE)
> > > > x.sample <- sort(x.sample)
> > > > y <- rnorm(20)
> > > > z <- cbind(x.sample, y)
> > > > z
> > >
> > > x.sample y
> > > [1,] 1 -1.2006469
> > > [2,] 1 0.7615261
> > > [3,] 1 -0.1287516
> > > [4,] 1 - 1.1796474
> > > [5,] 1 -1.2902519
> > > [6,] 2 -0.1614918
> > > [7,] 2 -0.1464773
> > > [8,] 2 -0.8875417
> > > [9,] 2 0.3062891
> > > [10,] 2 0.4398530
> > > [11,] 3 -0.5717729
> > > [12,] 3 - 0.2938118
> > > [13,] 4 -0.2398887
> > > [14,] 4 0.8425419
> > > [15,] 4 2.5269801
> > > [16,] 4 -0.3643613
> > > [17,] 5 1.1690564
> > > [18,] 5 -0.7644521
> > > [19,] 5 1.4178982
> > > [20,] 5 - 0.8198921
> > >
> > > I am only interested in extracting the first score (y) for each unique
> > > subject (x.sample). So, I would like to generate the following output.
> > >
> > > x.sample y
> > > [1,] 1 -1.2006469
> > > [2,] 2 -0.1614918
> > > [3,] 3 -0.5717729
> > > [4,] 4 -0.2398887
> > > [5,] 5 1.1690564
> > >
> > > Any assistance would be greatly appreciated.
> > >
> > > Regards,
> > > Brian
> >
> > See ?split, ?sapply and ?unique.
> >
> > Then try this:
> >
> > > cbind(unique(z[, 1]), sapply(split(z[, 2], z[, 1]), "[", 1))
> > [,1] [,2]
> > 1 1 -1.2006469
> > 2 2 -0.1614918
> > 3 3 -0.5717729
> > 4 4 -0.2398887
> > 5 5 1.1690564
> >
> >
> > The key part of that is:
> >
> > > split(z[, 2], z[, 1])
> > $`1`
> > [1] -1.2006469 0.7615261 -0.1287516 -1.1796474 -1.2902519
> >
> > $`2`
> > [1] -0.1614918 -0.1464773 -0.8875417 0.3062891 0.4398530
> >
> > $`3`
> > [1] -0.5717729 -0.2938118
> >
> > $`4`
> > [1] -0.2398887 0.8425419 2.5269801 -0.3643613
> >
> > $`5`
> > [1] 1.1690564 -0.7644521 1.4178982 -0.8198921
> >
> >
> > which splits 'z' by the values in the first column.
> >
> > Then we use sapply() to go through the list and subset the first element
> > in each vector:
> >
> > > sapply(split(z[, 2], z[, 1]), "[", 1)
> > 1 2 3 4 5
> > -1.2006469 -0.1614918 -0.5717729 -0.2398887 1.1690564
> >
> >
> > Then we cbind() that result to the unique values in the first column.
> >
> > HTH,
> >
> > Marc Schwartz
> >
> > ______________________________________________
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
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