R devel - Aug 2012 - Quiz: How to get a "named column" from a data frame

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

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
<maechler at 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
>
> ______________________________________________
> R-devel at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel

On Sat, Aug 18, 2012 at 5:14 PM, Christian Brechb?hler ....
wrote:
> On Sat, Aug 18, 2012 at 11:03 AM, Martin Maechler
> <maechler at 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)?
>
>
>> identical(nv, df[,1])
> [1] TRUE
>
>> In my solution, the above '.......' consists of 17 letters.
>
>
> I count 6 in mine
But it is not a solution in a current version of R!
though it's still interesting that   df[,1]  worked in some incantation of
R.

What's your sessionInfo()?
Martin
>
> /Christian

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 <- function() t(df)[1,]
f3 <- function() as.matrix(df)[,1]

r <- microbenchmark(f1(), f2(), f3(), times=1000)
r
# Unit: microseconds
#   expr    min      lq median      uq      max
# 1 f1() 14.589 17.0315 18.608 19.3220   89.388
# 2 f2() 68.057 70.8735 72.240 75.8065 3707.012
# 3 f3() 58.153 61.2600 62.521 65.0380  238.483

-Winston



On Sat, Aug 18, 2012 at 10:03 AM, Martin Maechler
<maechler at 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
>
> ______________________________________________
> R-devel at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel

On Sat, Aug 18, 2012 at 10:03 AM, Martin Maechler
<maechler at 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.
But aren't you making life difficult for yourself by not using I ?

df <- data.frame(VAR = I(nv))
str(df[[1]])

(which isn't quite identically because it now has the AsIs class)

Hadley

-- 
Assistant Professor
Department of Statistics / Rice University
http://had.co.nz/

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"
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
For this purpose my private function library has a function withnames():

withnames(): Extract from data frame as a named vector

Description: Extracts data from a data frame; if the result is a vector
(i.e. we extracted a single column and did not specify 'drop=FALSE')
it is assigned names derived from the row names of the data frame.

Usage: withnames(expr)

Arguments: expr: R expression.

Details: 'expr' is evaluated in an environment in which the extractor
functions '$.data.frame', '[.data.frame', and
'[[.data.frame' are
replaced by versions that attach the data frame's row names to an
extracted vector.

Value: 'expr', evaluated as described above.

## Code

withnames<-function(expr) {
   eval(substitute(expr),
   list(
     `[.data.frame` = function(x,i,...) {
       out<-x[i,...]
       if (is.null(dim(out))) names(out)<-row.names(x)[i]
       return(out)},
     `[[.data.frame` = function(x,...) {
       out<-x[[...]]
       if (is.null(dim(out))) names(out)<-row.names(x)
       return(out)},
     `$.data.frame` = function(x,name) {
       out<-x[[name, exact=FALSE]]
       if (is.null(dim(out))) names(out)<-row.names(x)
       return(out)}
     ),
   enclos=parent.frame())
}

## Examples

dd <- data.frame(aa=1:6, bb=letters[c(1,3,2,3,3,1)],
   row.names=LETTERS[1:6])
dd
dd$aa                          # Unnamed vector
withnames(dd$aa)               # Named vector
withnames(dd[["aa"]])          # Named vector
withnames(dd[2:4,"aa"])        # Named vector
withnames(dd$bb)               # Factor with names
withnames(outer(dd$a,dd$a))    # Both dimensions have names

## But now I am looking for a version that will play nicely with with():

withnames(with(dd, aa))  # No names!
with(dd, withnames(aa))  # No names!