Yes, but either
drop(t(df[,1,drop=TRUE]))
or
t(df[,1,drop=TRUE])[1,]
does work. My minimal effort to check timings found that the first
version was a hair faster.
-- Bert
On Sat, Aug 18, 2012 at 9:01 AM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
A bit more general
nv - c(a=1,
Or to expand just a hair on Joshua's suggestion, is the following what you want:
x - 1:10
names(x) - letters[1:10]
x
a b c d e f g h i j
1 2 3 4 5 6 7 8 9 10
df - data.frame(x=x,y=LETTERS[1:10],row.names=names(x))
df
x y
a 1 A
b 2 B
c 3 C
d 4 D
e 5 E
f 6 F
g 7 G
Sorry! -- Change that to drop = FALSE !
drop(t(df[,1,drop=FALSE]))
t(df[,1,drop=FALSE])[1,]
-- Bert
On Sat, Aug 18, 2012 at 9:37 AM, Bert Gunter bgun...@gene.com wrote:
Yes, but either
drop(t(df[,1,drop=TRUE]))
or
t(df[,1,drop=TRUE])[1,]
does work. My minimal effort to check timings
On 2012-08-18 11:03, Martin Maechler wrote:
Today, I was looking for an elegant (and efficient) way
to get a named (atomic) vector by selecting one column of a data frame.
Of course, the vector names must be the rownames of the data frame.
Ok, here is the quiz, I know one quite cute/slick
On Sat, Aug 18, 2012 at 02:13:20PM -0400, Christian Brechb?hler wrote:
On 8/18/12, Martin Maechler maech...@stat.math.ethz.ch wrote:
On Sat, Aug 18, 2012 at 5:14 PM, Christian Brechb?hler wrote:
On Sat, Aug 18, 2012 at 11:03 AM, Martin Maechler
maech...@stat.math.ethz.ch wrote:
