Hello,
A bit more general
nv <- c(a=1, d=17, e=101); nv
nv2 <- c(a="a", d="d", e="e")
df2 <- data.frame(VAR = nv, CHAR = nv2); df2
identical( nv, drop(t( df2[1] )) ) # TRUE
identical( nv, drop(t( df2[[1]] )) ) # FALSE
Rui Barradas
Em 18-08-2012 16:16, Joshua Ulrich escreveu:
I don't know if this is better, but it's the most obvious/shortest I
could come up with. Transpose the data.frame column to a 'row' vector
and drop the dimensions.
R> identical(nv, drop(t(df)))
[1] TRUE
Best,
--
Joshua Ulrich | about.me/joshuaulrich
FOSS Trading | www.fosstrading.com
On Sat, Aug 18, 2012 at 10:03 AM, Martin Maechler
<maech...@stat.math.ethz.ch> 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" answer, but was
wondering if there are obvious better ones, and
also if this should not become more idiomatic (hence "R-devel"):
Consider this toy example, where the dataframe already has only
one column :
nv <- c(a=1, d=17, e=101); nv
a d e
1 17 101
df <- as.data.frame(cbind(VAR = nv)); df
VAR
a 1
d 17
e 101
Now how, can I get 'nv' back from 'df' ? I.e., how to get
identical(nv, .......)
[1] TRUE
where ...... only uses 'df' (and no non-standard R packages)?
As said, I know a simple solution (*), but I'm sure it is not
obvious to most R users and probably not even to the majority of
R-devel readers... OTOH, people like Bill Dunlap will not take
long to provide it or a better one.
(*) In my solution, the above '.......' consists of 17 letters.
I'll post it later today (CEST time) ... or confirm
that someone else has done so.
Martin
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