R help - Sep 2009 - How to assess object names within a function in lapply or l_ply?

Dear All,

to produce output of several columns of a data frame, I tried to use 
lapply and also l_ply. In both cases, I would like to print a header 
line containing also the name of the respective column in the data frame.

For example, I would like the following

lapply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x))))

to produce:
[1] "a"
[1] "b"

and not, what it actually does:
[1] "X[[1L]]"
[1] "X[[2L]]"
$a
[1] "X[[1L]]"

$b
[1] "X[[2L]]"

or with l_ply (plyr package)
l_ply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x))))

to produce:
[1] "a"
[1] "b"

and not, what it actually does:
[1] ".data[[i]]"
[1] ".data[[i]]"

Is this possible?

Thanks,
Heinz

You can use names insteed:

DF <- data.frame(a=1:3, b=2:4)
lapply(names(DF), function(x){
				print(x)
				DF[x]
			})

On Mon, Sep 28, 2009 at 8:22 AM, Heinz Tuechler <tuechler at gmx.at>
wrote:
> Dear All,
>
> to produce output of several columns of a data frame, I tried to use lapply
> and also l_ply. In both cases, I would like to print a header line
> containing also the name of the respective column in the data frame.
>
> For example, I would like the following
>
> lapply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x))))
>
> to produce:
> [1] "a"
> [1] "b"
>
> and not, what it actually does:
> [1] "X[[1L]]"
> [1] "X[[2L]]"
> $a
> [1] "X[[1L]]"
>
> $b
> [1] "X[[2L]]"
>
> or with l_ply (plyr package)
> l_ply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x))))
>
> to produce:
> [1] "a"
> [1] "b"
>
> and not, what it actually does:
> [1] ".data[[i]]"
> [1] ".data[[i]]"
>
> Is this possible?
>
> Thanks,
> Heinz
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


-- 
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

Heinz,

Try this:

lapply(DF, function(x)names(DF)[as.numeric(gsub("[^0-9]",
"",
deparse(substitute(x))))])

On Mon, Sep 28, 2009 at 8:43 AM, Heinz Tuechler <tuechler at gmx.at>
wrote:
> Thank you, Henrique,
>
> my example was simplified. In a more complexe function I want to use the
> objects, not just their names. In your solution, I have to adapt the
> function itself, depending on the name of the data.frame, which I would
like
> to avoid.
>
> Thanks,
> Heinz
>
>
> At 13:36 28.09.2009, Henrique Dallazuanna wrote:
>>
>> You can use names insteed:
>>
>> DF <- data.frame(a=1:3, b=2:4)
>> lapply(names(DF), function(x){
>> ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?print(x)
>> ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?DF[x]
>> ? ? ? ? ? ? ? ? ? ? ? ?})
>>
>> On Mon, Sep 28, 2009 at 8:22 AM, Heinz Tuechler <tuechler at
gmx.at> wrote:
>> > Dear All,
>> >
>> > to produce output of several columns of a data frame, I tried to
use
>> > lapply
>> > and also l_ply. In both cases, I would like to print a header line
>> > containing also the name of the respective column in the data
frame.
>> >
>> > For example, I would like the following
>> >
>> > lapply(data.frame(a=1:3, b=2:4), function(x)
>> > print(deparse(substitute(x))))
>> >
>> > to produce:
>> > [1] "a"
>> > [1] "b"
>> >
>> > and not, what it actually does:
>> > [1] "X[[1L]]"
>> > [1] "X[[2L]]"
>> > $a
>> > [1] "X[[1L]]"
>> >
>> > $b
>> > [1] "X[[2L]]"
>> >
>> > or with l_ply (plyr package)
>> > l_ply(data.frame(a=1:3, b=2:4), function(x)
>> > print(deparse(substitute(x))))
>> >
>> > to produce:
>> > [1] "a"
>> > [1] "b"
>> >
>> > and not, what it actually does:
>> > [1] ".data[[i]]"
>> > [1] ".data[[i]]"
>> >
>> > Is this possible?
>> >
>> > Thanks,
>> > Heinz
>> >
>> > ______________________________________________
>> > R-help at r-project.org mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> > http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>>
>>
>>
>> --
>> Henrique Dallazuanna
>> Curitiba-Paran?-Brasil
>> 25? 25' 40" S 49? 16' 22" O
>
>
>


-- 
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

> or with l_ply (plyr package)
> l_ply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x))))

The best way to do this is to supply both the object you want to
iterate over, and its names.  Unfortunately it's slightly difficult to
create a data structure of the correct form to do this with m_ply.

df <- data.frame(a=1:3, b=2:4)
input <- list(x = df, name = names(df))
inputdf <- structure(input,
    class = "data.frame",
    row.names = seq_along(input[[1]]))

m_ply(inputdf, function(x, name) {
  cat(name, "---------\n")
  print(x)
})

I'll think about how to improve this for a future version.

Hadley


-- 
http://had.co.nz/

> many thanks for your answer and for the enormous work you put into plyr, a
> really powerful package.
> For now, I will solve my problem with a variable label attribute, I usually
> attach to columns in data frames. I asked the list, because I thought, I am
> overlooking something trivial, since lapply itself apparently
"knows" the
> object names, as it labels the output by them. It just does not supply them
> to the function it calls.
lapply knows the names - the calling function doesn't - it takes the
output add then fixes up the names after it's run.

Hadley


-- 
http://had.co.nz/