Hello,
Thanks to all for your answers.
The solution given by R. Michael was perfect ! Thank you very much, that
helped a lot !
Thomas
On Tue, Feb 28, 2012 at 11:50 AM, Alemtsehai Abate wrote:
> Perhaps, the following does it as well.
>
> (d <- data.frame(x1=letters[2*1:4 - 1], x2=letters[2*1:4]))
Perhaps, the following does it as well.
(d <- data.frame(x1=letters[2*1:4 - 1], x2=letters[2*1:4]))
c(t(d))
Alemtsehai
>> Hello,
>> I am looking for a way to transform an array into a list (or a string).
>> My array has two columns 1 and 2, and I would like to create a list of
the
>> values
My apologies. The last line should have been
with(d, sort(c(as.character(x1), as.character(x2
Regards,
Jorge.-
On Sat, Feb 25, 2012 at 2:03 AM, Jorge I Velez <> wrote:
> Perhaps the following?
>
> d <- structure(list(x1 = structure(1:4, .Label = c("a", "c", "e",
> "g"), class = "factor"),
My apologies: I missed the order of the desired output: the easiest
thing to do is likely to use the same techniques given below (and by
others in this thread) with a transpose t() before.
Michael
On Sat, Feb 25, 2012 at 2:05 AM, R. Michael Weylandt
wrote:
> Your question is not well formed: do
Your question is not well formed: do you want a list or a string
(totally different things)? Or even more likely, a character vector?
What do you have now: is it really an array (=matrix) or is it the
data.frame it looks like?
If it's a matrix:
x <- matrix(letters[1:8], ncol = 2)
x <- as.vector(x
Perhaps the following?
d <- structure(list(x1 = structure(1:4, .Label = c("a", "c", "e",
"g"), class = "factor"), x2 = structure(1:4, .Label = c("b",
"d", "f", "h"), class = "factor")), .Names = c("x1", "x2"), class =
"data.frame", row.names = c("1",
"2", "3", "4"))
with(d, c(as.character(x1), as
Hello,
Try
(d <- data.frame(x1=letters[2*1:4 - 1], x2=letters[2*1:4]))
c(apply(d, 1, identity))
Note that you'll need the concatenation 'c()'.
Hope this helps,
Rui Barradas
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