[R] Why can't "apply" be used with "as.factor" on a data.frame ?

[David Winsemius]

On Mar 7, 2010, at 3:20 PM, Don MacQueen wrote:

> And just a small followup. To find out what class each column is,  
> you wanted
>
>> lapply(a,class)
> $x1
> [1] "numeric"
>
> $x2
> [1] "factor"
>
> $x3
> [1] "factor"
>
> With regard to your solution, and why it works, it is my  
> understanding that data frames are in some sense actually lists,  
> each column corresponding to one element in a list.
>
> Hence, lapply() works column-wise on data frames.
>
> Also for this reason it's pretty easy to convert back and forth  
> between data frames and lists . Provided, of course, that each  
> element of the list has an appropriate structure; see this example:
>
>> data.frame( list(a=1:2, b=3:4) )
>  a b
> 1 1 3
> 2 2 4
>
>> data.frame( list(a=1:2, b=3:7) )
> Error in data.frame(a = 1:2, b = 3:7, check.names = FALSE,  
> stringsAsFactors = TRUE) :
>  arguments imply differing number of rows: 2, 5
>
>
> No doubt there are subtle details, but don't ask me to provide  
> details on what exactly the "some sense" is!

It's not that complicated:

 > class(dfrm)
[1] "data.frame"
 > is.list(dfrm)
[1] TRUE
 >
 > dput(dfrm)
structure(list(a = 1:2, b = 3:4), .Names = c("a", "b"), row.names =  
c(NA,
-2L), class = "data.frame")
# Let's do some violence to this dataframe ...
 > class(dfrm) <- "list"
 > dfrm
$a
[1] 1 2

$b
[1] 3 4

attr(,"row.names")
[1] 1 2
 > is.data.frame(dfrm)
[1] FALSE
 > is.data.frame(as.data.frame(dfrm))
[1] TRUE
 > dput(dfrm)
structure(list(a = 1:2, b = 3:4), .Names = c("a", "b"), row.names =  
c(NA,
-2L))
# Now let's restore it to its original data.frame-ish state:
 > class(dfrm) <- "data.frame"
 > dput(dfrm)
structure(list(a = 1:2, b = 3:4), .Names = c("a", "b"), row.names =  
c(NA,
-2L), class = "data.frame")

>
> -Don
>
> At 12:07 PM +0200 3/7/10, Tal Galili wrote:
>> Hi all,
>>
>> Let's say I have a data.frame and wants to turn each of it's  
>> columns into a
>> factor.
>> My instinct would be to use as.factor with apply. But this won't  
>> work, and
>> result with a data.frame of characters.
>> I found another solution for how to achieve this, but I would also  
>> like to
>> understand - *WHY* does it work this way?
>>
>> Here is an example script:
>> a <- data.frame(x1 = rnorm(100), x2 = sample(c("a","b"), 100,  
>> replace = T),
>> x3 = factor(c(rep("a",50) , rep("b",50))))
>> apply(a2, 2,class) # why is column 3 not a factor ?
>> a[,3]  # since it IS a factor.
>> a2 <- apply(a, 2,as.factor) # won't work - why not ?
>> a2[,3]  # Why was this just turned into a character ???
>> # A solution
>> a2 <- lapply(a, as.factor)
>> a3 <- as.data.frame(a2)
>> str(a3)
>>
>>
>> Thanks,
>> Tal
>>
>>
>>
>> ----------------Contact
>> Details:-------------------------------------------------------
>> Contact me: Tal.Galili at gmail.com |  972-52-7275845
>> Read me: www.*talgalili.com (Hebrew) | www.*biostatistics.co.il  
>> (Hebrew) |
>> www.*r-statistics.com (English)
>> ----------------------------------------------------------------------------------------------
>>
>> 	[[alternative HTML version deleted]]
>>
>> ______________________________________________
>> 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.
>
>
> -- 
> ---------------------------------
> Don MacQueen
> Lawrence Livermore National Laboratory
> Livermore, CA, USA
> 925-423-1062
> macq at llnl.gov
>
> ______________________________________________
> 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.

David Winsemius, MD
West Hartford, CT