[R] apply --> data.frame

[Bert Gunter]

?lapply

> z <-data.frame(a=1:3,b=letters[1:3])

> lapply(z,"[",1:2)
$a
[1] 1 2

$b
[1] a b
Levels: a b c

> data.frame(lapply(z,"[",1:2)) ## Is this not what you want?
  a b
1 1 a
2 2 b

You really should spend a little more time with the docs figuring out
what R _does_ and a little less complaining about what you think R
cannot do.

-- Bert

On Thu, Aug 30, 2012 at 9:44 AM, Sam Steingold <sds at gnu.org> wrote:
>> * Sam Steingold <fqf at tah.bet> [2012-08-30 08:56:17 -0400]:
>>
>> Is there a way for an apply-type function to return a data frame?
>> the closest thing I think of is
>>
>>   foo <- as.data.frame(t(sapply(...)))
>>   names(foo) <- c(....)
>
> alas, this has a problem of creating a "homogeneous" data frame, i.e.,
> all the columns are numbers or characters, because the function passed
> to sapply returns c(....) and
>> c(1,2,"a")
> [1] "1" "2" "a"
>
> e.g.,
> as.data.frame(t(sapply(c("a,1","b,2","c,3"),function (n) strsplit(n,",")[[1]])))
>     V1 V2
> a,1  a  1
> b,2  b  2
> c,3  c  3
>
> 'data.frame':   3 obs. of  2 variables:
>  $ V1: Factor w/ 3 levels "a","b","c": 1 2 3
>   ..- attr(*, "names")= chr  "a,1" "b,2" "c,3"
>  $ V2: Factor w/ 3 levels "1","2","3": 1 2 3
>   ..- attr(*, "names")= chr  "a,1" "b,2" "c,3"
>
> I wanted the V1 column to be a string, and V2 to be a number.
> (I know stringsAsFactors=FALSE would replace factors with strings, but I
> need a string and a number)
>
> I could, of course, do ret$V2 <- as.numeric(ret$V2) but this would mean
> a double conversion: from number to string first (by c()) and then back.
>
> thanks.
>
> --
> Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
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-- 

Bert Gunter
Genentech Nonclinical Biostatistics

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