On Tue, Aug 21, 2012 at 2:34 PM, Duncan Murdoch
wrote:
> On 12-08-18 12:33 PM, Martin Maechler wrote:
>>>
>>> Joshua Ulrich
>>> on Sat, 18 Aug 2012 10:16:09 -0500 writes:
>>
>>
>> > I don't know if this is better, but it's the most obvious/shortest
>> I
>> > could come
On 12-08-18 12:33 PM, Martin Maechler wrote:
Joshua Ulrich
on Sat, 18 Aug 2012 10:16:09 -0500 writes:
> 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.
On Sat, Aug 18, 2012 at 02:13:20PM -0400, Christian Brechb?hler wrote:
> On 8/18/12, Martin Maechler 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
> >> wrote:
>
> >>> Consider this toy example, where the dataf
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" an
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 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 found t
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
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 wrote:
> Hello,
>
> A bit more general
>
> nv <- c(a=1, d=17, e=101); nv
>
project.org [mailto:r-devel-boun...@r-project.org] On
> Behalf
> Of Winston Chang
> Sent: Saturday, August 18, 2012 11:54 AM
> To: Martin Maechler
> Cc: R. Devel List
> Subject: Re: [Rd] Quiz: How to get a "named column" from a data frame
>
> This isn't super-
On Sat, Aug 18, 2012 at 10:03 AM, 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 qu
This isn't super-concise, but has the virtue of being clear:
nv <- c(a=1, d=17, e=101)
df <- as.data.frame(cbind(VAR = nv))
identical(nv, setNames(df$VAR, rownames(df)))
# TRUE
It seems to be more efficient than the other methods as well:
f1 <- function() setNames(df$VAR, rownames(df))
f2 <- f
On 8/18/12, Martin Maechler wrote:
> On Sat, Aug 18, 2012 at 5:14 PM, Christian Brechbühler wrote:
>> On Sat, Aug 18, 2012 at 11:03 AM, Martin Maechler
>> wrote:
>>> Consider this toy example, where the dataframe already has only
>>> one column :
>>>
>>> > nv <- c(a=1, d=17, e=101); nv
>>>
On Sat, Aug 18, 2012 at 9:33 AM, Martin Maechler
wrote:
>> Joshua Ulrich
>> on Sat, 18 Aug 2012 10:16:09 -0500 writes:
>
> > 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
> >
> Joshua Ulrich
> on Sat, 18 Aug 2012 10:16:09 -0500 writes:
> 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)))
>
On Sat, Aug 18, 2012 at 5:14 PM, Christian Brechbühler wrote:
> On Sat, Aug 18, 2012 at 11:03 AM, 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 mu
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 i
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 Sa
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 b
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