Re: [R] Automatic routine - NEW

[Spencer Graves]

 The documentation is good for things like this. In most (all recent?) 
versions of R, "help.start()" brings up a help page. Select "An 
Introduction to R", and go to "Vector Indices". Also look at "Arrays and 
Matrices: Array Indexing". There, you will find that "DF[3:5,]" might be 
what you want.

Actually, I'm not certain that your "c" below is what you want, because 
class(c[1,1]) is "list", and I could not do arithmetic on it. I 
discovered that when I tried all.equal(c, DF[3:5,]). {Also, "c" is the 
name of a very useful function; try "?c". It is usually wise to avoid 
using names of functions for matrices of data.frames. R will select the 
one you want from the context in most but not all cases.}

hope this helps.
spencer graves
Monica Palaseanu-Lovejoy wrote:
Hi Again,
First of all thank you for all the responses to my previous query. 
Your answers were very helpful and I did the job ;-). Now I hope you 
can answer as quick the following (sorry I am invading you with 
trivial questions):

Let’s use again the following data.frame example:
DF <- data.frame(x=rnorm(5), y=rnorm(5))
I want to obtain a new data.frame (or matrix) that contains only n 
rows (from the i rows DF has), and all the columns. If I have to do it 
step by step I would do something like that, for example:

a3 <- DF[3,]
a4 <- DF[4,]
a5 <- DF[5,]
b <- data.frame(a3, a4, a5)
c <- matrix(b, nrow=3, ncol=2, byrow=TRUE)
Now I want to do the same in one go, so I wrote:
for (i in 3:5)
{
        d[i] <- DF[i,]
        e <- data.frame(d[i])
        f <- matrix(e, ncol=2, nrow=3, byrow=TRUE)
}
Which of course gives me errors and the matrix f has all elements 
equal with DF[5,5]. If I don’t use [i] after d, the resulting f matrix is 
made up from the DF[5,] elements (which is quite normal since i 
replaces itself ...). So …. How is this done correctly?

I am really appreciating your time and effort to answer me,
Monica
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