If there is a lot of names perhaps sprintf(format, 1:length(L), sep="")
where format is the appropriate format for sprintf so that it is easier to
read
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email [EMAIL PROTECTED]
Home [EMAIL
And if you have "thousands of names" like this:
names(L) <- paste ("names", 1:length(L), sep="")
Nael
On Sun, May 25, 2008 at 10:38 PM, Erin Hodgess <[EMAIL PROTECTED]>
wrote:
> try this:
>
> > names(L) <- c("name1","name2","name3")
> > L
> $name1
> [1] "Fred"
>
> $name2
> [1] "Mary"
>
> $name3
try this:
> names(L) <- c("name1","name2","name3")
> L
$name1
[1] "Fred"
$name2
[1] "Mary"
$name3
[1] "SAM"
>
>
HTH,
Sincerely,
Erin
On Sun, May 25, 2008 at 2:42 PM, joseph <[EMAIL PROTECTED]> wrote:
> Hi
> I have a character vector with thousands of names which looks like this:
>> V=c("Fred"
Hi
I have a character vector with thousands of names which looks like this:
> V=c("Fred", "Mary", "SAM")
> V
[1] "Fred" "Mary" "SAM"
> class(V)
[1] "character"
I would like to change it to a list:
> L=as.list(V)
> L
[[1]]
[1] "Fred"
[[2]]
[1] "Mary"
[[3]]
[1] "SAM"
but I need to name the compone
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