R help - Mar 2010 - Why can't "apply" be used with "as.factor" on a data.frame ?

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



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The basic reason because apply works with matrices - it first turns
the input into a matrix, processes each column and then returns a
matrix.  See colwise in the plyr package for a function that works
column wise on a data frame, returning a data frame.

Hadley

On Sun, Mar 7, 2010 at 11:07 AM, Tal Galili <tal.galili at gmail.com>
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.
>


-- 
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/

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!

-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