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Quiz: How to get a "named column" from a data frame

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Martin Maechler

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

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Joshua Ulrich

Re: Quiz: How to get a "named column" from a data frame

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
<[hidden email]> 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
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel

______________________________________________
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Rui Barradas

Re: Quiz: How to get a "named column" from a data frame

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
> <[hidden email]> 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
>>
>> ______________________________________________
>> [hidden email] mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel

______________________________________________
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Bert Gunter

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Joshua Ulrich

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
h 8 H
i 9 I
j 10 J
> y <- t(df[,1,drop=FALSE])[1,]
> y
a b c d e f g h i j
1 2 3 4 5 6 7 8 9 10
> identical(x,y)
[1] TRUE

Cheers,
Bert

On Sat, Aug 18, 2012 at 8:16 AM, Joshua Ulrich <[hidden email]> wrote:

> 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
> <[hidden email]> 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
>>
>> ______________________________________________
>> [hidden email] mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel

--

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel

Martin Maechler

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Martin Maechler

On Sat, Aug 18, 2012 at 5:14 PM, Christian Brechbühler .... wrote:

> On Sat, Aug 18, 2012 at 11:03 AM, Martin Maechler
> <[hidden email]> 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

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Martin Maechler

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Joshua Ulrich

>>>>> Joshua Ulrich <[hidden email]>
>>>>> 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)))
> [1] TRUE

Yes, that's definitely shorter,
congratulations!

One gotta is that I'd want a solution that also works when the
df has more columns than just one...

Your idea to use t(.) is nice and "perfect" insofar as it
coerces the data frame to a matrix, and that's really the clue:

Where as df[,1] is losing the names,
the matrix indexing is not.
So your solution can be changed into

t(df)[1,]

which is even shorter...
and slightly less efficient, at least conceptually, than mine, which has
been

as.matrix(df)[,1]

Now, the remaining question is: Shouldn't there be something
more natural to achieve that?
(There is not, currently, AFAIK).

Martin

> Best,
> --
> Joshua Ulrich | about.me/joshuaulrich
> FOSS Trading | www.fosstrading.com

> On Sat, Aug 18, 2012 at 10:03 AM, Martin Maechler
> <[hidden email]> 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
>>
>> ______________________________________________
>> [hidden email] mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel

______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel

Bert Gunter

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Rui Barradas

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 <[hidden email]> wrote:

> 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
>> <[hidden email]> 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
>>>
>>> ______________________________________________
>>> [hidden email] mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-devel
>>
>> ______________________________________________
>> [hidden email] mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel
>
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel

Bert Gunter

Re: Quiz: How to get a "named column" from a data frame

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 <[hidden email]> 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 that the first
> version was a hair faster.
>
> -- Bert
>
> On Sat, Aug 18, 2012 at 9:01 AM, Rui Barradas <[hidden email]> wrote:
>> 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
>>> <[hidden email]> 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
>>>>
>>>> ______________________________________________
>>>> [hidden email] mailing list
>>>> https://stat.ethz.ch/mailman/listinfo/r-devel
>>>
>>> ______________________________________________
>>> [hidden email] mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-devel
>>
>>
>> ______________________________________________
>> [hidden email] mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel
>
>
>
> --
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
>
> Internal Contact Info:
> Phone: 467-7374
> Website:
> http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

Joshua Wiley-2

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Martin Maechler

On Sat, Aug 18, 2012 at 9:33 AM, Martin Maechler
<[hidden email]> wrote:

>>>>>> Joshua Ulrich <[hidden email]>
>>>>>> 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)))
> > [1] TRUE
>
> Yes, that's definitely shorter,
> congratulations!
>
> One gotta is that I'd want a solution that also works when the
> df has more columns than just one...
>
> Your idea to use t(.) is nice and "perfect" insofar as it
> coerces the data frame to a matrix, and that's really the clue:
>
> Where as df[,1] is losing the names,
> the matrix indexing is not.
> So your solution can be changed into
>
> t(df)[1,]
>
> which is even shorter...
> and slightly less efficient, at least conceptually, than mine, which has
> been
>
> as.matrix(df)[,1]
>
> Now, the remaining question is: Shouldn't there be something
> more natural to achieve that?
> (There is not, currently, AFAIK).

Perhaps a data frame method for as.vector?

as.vector.data.frame <- function(x, ...) as.matrix(x)[,1]
as.vector(df[1])

or an additional argument to `[.data.frame` like keep.names, which
defaults to FALSE to maintain current behavior but can optionally be
TRUE.

Cheers,

Josh

>
> Martin
>
>
> > Best,
> > --
> > Joshua Ulrich | about.me/joshuaulrich
> > FOSS Trading | www.fosstrading.com
>
>
> > On Sat, Aug 18, 2012 at 10:03 AM, Martin Maechler
> > <[hidden email]> 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
> >>
> >> ______________________________________________
> >> [hidden email] mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-devel
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel

--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/

______________________________________________
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Quigi

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Martin Maechler

On 8/18/12, Martin Maechler <[hidden email]> wrote:
> On Sat, Aug 18, 2012 at 5:14 PM, Christian Brechbühler .... wrote:
>> On Sat, Aug 18, 2012 at 11:03 AM, Martin Maechler
>> <[hidden email]> wrote:

>>> 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

> 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.

My mistake! We disliked some quirks of indexing, so we've long had
our own patch for "[.data.frame" in place, which I used inadvertently.
In essence, it does this:

result <- base::"[.data.frame"(df,,1, drop=F)
if (drop && length(ncol(result) > 0) && ncol(result)==1) {
save.names <- dimnames(result)[[1]]
result <- result[[1]]
names(result) <- save.names
}

That obviously violated your constraint "no non-standard R packages".
I apologize.

Still, maybe the behavior of getting the named column would be
desirable in general?

/Christian

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Winston Chang

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Martin Maechler

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
<[hidden email]> wrote:

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Hadley Wickham-2

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Martin Maechler

On Sat, Aug 18, 2012 at 10:03 AM, Martin Maechler
<[hidden email]> wrote:

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/

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J. R. M. Hosking-3

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Martin Maechler

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!

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William Dunlap

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Winston Chang

That would have been essentially my suggestion as well. I prefer its clarity
(and speed). I didn't know if you wanted your solution to also apply
to matrices embedded in data.frames. In S+ rownames<-() works on vectors
(because it calls the generic rowId<-()) so the following works:
> f4 <- function(df, column) { tmp <- df[[column]] ; rownames(tmp) <- rownames(df) ; tmp}
> nv <- c(a=1,d=17,e=101)
> df <- data.frame(VAR=nv, Two=3^(1:3))
> f4(df, 2)
a d e
3 9 27
> df$Matrix <- matrix(1001:1006, ncol=2, nrow=3)
> f4(df, "Matrix")
[,1] [,2]
a 1001 1004
d 1002 1005
e 1003 1006

I forget if R has something like rowIds() (it is to names and rownames as
NROW is to length and nrow).

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com

> -----Original Message-----
> From: [hidden email] [mailto:[hidden email]] 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-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
> <[hidden email]> 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
> >
> > ______________________________________________
> > [hidden email] mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-devel
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel

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Andrew Piskorski

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Quigi

On Sat, Aug 18, 2012 at 02:13:20PM -0400, Christian Brechb?hler wrote:

> On 8/18/12, Martin Maechler <[hidden email]> 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
> >> <[hidden email]> wrote:
>
> >>> 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, df[,1])
> >> [1] TRUE
>
> > 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.
>
> My mistake! We disliked some quirks of indexing, so we've long had
> our own patch for "[.data.frame" in place, which I used inadvertently.

As I understand it, when when doing 'df[,1]' on a data frame, Bell
Labs S and all versions of S-Plus prior to 3.4 always retained the
data frame's row names as the names on the result vector. 'df[,1]'
gave you a named vector identical to your 'nv' above. Then in 1996
with S-Plus 3.4, Insightful broke that behavior, after which 'df[,1]'
returned a vector without any names. I believe R copied that
late-1990s S-Plus behavior, but I don't know why exactly.

When subscripting objects, R sometimes retains the object's dimnames
as names in the result, and sometimes not, which I find frustrating.
Personally, I think it would make much more sense if subscripting
ALWAYS retained any names it could, and worked as similarly as
possible across data frames, matrices, arrays, vectors, etc. After
all, explicitly dropping names afterwards is trivial, while adding
them back on is not.

Back on 2005-10-19 with R 2.2.0, I gave a simple test of 15 cases; 4
of them dropped names during subscripting, the other 11 preseved them.
That's towards the end of the discussion here:

https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=8192

Contrary to the initial tone of my old 2005 "bug" report, current R
subscripting behavior is of course NOT a bug, as AFAIK it's working as
the R Core Team intended. However, I definitely consider the current
behavior a design infelicity.

Just now on stock R 2.15.1 (with --vanilla), I ran an updated version
of those same simple tests. Of 22 subscripting test cases, 7 lose
names and 15 preserve them. (If anyone's interested in the specific
tests, I can send them, or try to append them to that old 8192 feature
request.)

For what it's worth, at work, for years we ran various versions of
pre-namespace R using some ugly patches of "[" and "[.data.frame" to
force name retention during subscripting. Since we were not using
namespaces at all, those "keep names" subscripting hacks were
affecting ALL R code we ran, not just our own custom code which needed
and expected the names to be retained. Yet perhaps surprisingly, I
don't think I ever ran into a single case where the forced retention
of names broke any code. We of course ran only a tiny sample of the
huge amount of code on CRAN, but that experience suggests that most R
code which expects un-named objects doesn't mind at all if names are
present.

If anyone would genuinely like to add an option for name-preserving
subscripting to R, I'm willing to work on it, so please do let me know
your thoughts. So far though, I've never dug into the guts of the
.Primitive("[") and "[.data.frame" functions to see how/why they
sometimes keep and sometime discard names during subscripting.

--
Andrew Piskorski <[hidden email]>

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Duncan Murdoch-2

Re: Quiz: How to get a "named column" from a data frame

In reply to this post by Martin Maechler

On 12-08-18 12:33 PM, Martin Maechler wrote:

I've been offline, so I'm a bit late to this game, but the examples
above fail when df contains a character column as well as the desired
one, because everything gets coerced to a character matrix. You need to
select the column first, then convert to a matrix, e.g.

drop(t(df[,1,drop=FALSE]))

Duncan Murdoch

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Joshua Ulrich

Re: Quiz: How to get a "named column" from a data frame

On Tue, Aug 21, 2012 at 2:34 PM, Duncan Murdoch
<[hidden email]> wrote:

> On 12-08-18 12:33 PM, Martin Maechler wrote:
>>>>>>>
>>>>>>> Joshua Ulrich <[hidden email]>
>>>>>>> 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)))
>> > [1] TRUE
>>
>> Yes, that's definitely shorter,
>> congratulations!
>>
>> One gotta is that I'd want a solution that also works when the
>> df has more columns than just one...
>>
>> Your idea to use t(.) is nice and "perfect" insofar as it
>> coerces the data frame to a matrix, and that's really the clue:
>>
>> Where as df[,1] is losing the names,
>> the matrix indexing is not.
>> So your solution can be changed into
>>
>> t(df)[1,]
>>
>> which is even shorter...
>> and slightly less efficient, at least conceptually, than mine, which has
>> been
>>
>> as.matrix(df)[,1]
>>
>> Now, the remaining question is: Shouldn't there be something
>> more natural to achieve that?
>> (There is not, currently, AFAIK).
>
>
> I've been offline, so I'm a bit late to this game, but the examples above
> fail when df contains a character column as well as the desired one, because
> everything gets coerced to a character matrix. You need to select the
> column first, then convert to a matrix, e.g.
>
> drop(t(df[,1,drop=FALSE]))
>

That's true, but I was assuming a one-column data.frame, which can be
achieved via:
df <- data.frame(VAR=nv,CHAR=letters[1:3],stringsAsFactors=FALSE)
drop(t(df[1]))

That said, I prefer the setNames() solution for its efficiency.

Best,
Josh

> Duncan Murdoch
>
>
>>
>> Martin
>>
>>
>> > Best,
>> > --
>> > Joshua Ulrich | about.me/joshuaulrich
>> > FOSS Trading | www.fosstrading.com
>>
>>
>> > On Sat, Aug 18, 2012 at 10:03 AM, Martin Maechler
>> > <[hidden email]> 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
>> >>
>> >> ______________________________________________
>> >> [hidden email] mailing list
>> >> https://stat.ethz.ch/mailman/listinfo/r-devel
>>
>> ______________________________________________
>> [hidden email] mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel
>>
>

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