# [R] sapply behavior

Liaw, Andy andy_liaw at merck.com
Tue Sep 28 03:12:07 CEST 2004

```The problem is that temp2 is a data frame, and the function you are
sapply()ing to returns a row from a data frame.  A data frame is really a
list, with each variable corresponding to a component.  If you extract a row
of a data frame, you get another data frame, not a vector, even if all
variables are the same type.  sapply() can really `simplify' the right way
if it's given a vector (or matrix).  Consider:

> str(temp2)
`data.frame':   6 obs. of  4 variables:
\$ A1: int  5 2 4 6 1 3
\$ A2: int  5 2 4 6 1 3
\$ A3: int  5 2 4 6 1 3
\$ A4: int  5 2 4 6 1 3
> temp2 <- as.matrix(temp2)
> str(temp2)
int [1:6, 1:4] 5 2 4 6 1 3 5 2 4 6 ...
- attr(*, "dimnames")=List of 2
..\$ : chr [1:6] "1" "2" "3" "4" ...
..\$ : chr [1:4] "A1" "A2" "A3" "A4"
> str(sapply(1:6,function(x){xmat<-temp2[temp2[,1]==x,,drop=F]; xmat[1,]}))
int [1:4, 1:6] 1 1 1 1 2 2 2 2 3 3 ...
- attr(*, "dimnames")=List of 2
..\$ : chr [1:4] "A1" "A2" "A3" "A4"
..\$ : NULL

(The is.matrix() function probably just check whether the dim attribute is a
vector of length 2, and not a data frame (as it says in ?is.matrix).  The
newtemp2 object you get is a list with 24 components, each component is a
vector of one integer, and has a dim attribute of c(4, 6).  Not what I would
call a matrix.)

HTH,
Andy

> From: Elizabeth Purdom
>
> Hi,
>
> I use sapply very frequently, but I have recently noticed a
> behavior of
> sapply which I don't understand and have never seen before.
> Basically,
> sapply returns what looks like a matrix,  says it a matrix,
> and appears to
> let me do matrix things (like transpose). But it is also a
> list and behaves
> like a list when I subset it, not a vector (so I can't sort a row for
> instance). I don't know where this is coming from so as to
> avoid it, nor
> how to handle the beast that sapply is returning. I double
> checked my old
> version of R and apparently this same thing happens in 1.8.0,
> though I
> never experienced it. I had a hard time reproducing it, and I
> don't know
> what's setting it off, but the code below seems to do it for
> me. (I'm using
> R on Windows XP, either 1.8.0 or 1.9.1)
>
> Thanks for any help,
> Elizabeth Purdom
>
>
>  > temp2<-matrix(sample(1:6,6,replace=F),byrow=F,nrow=6,ncol=4)
>  > colnames(temp2)<-paste("A",as.character(1:4),sep="")
>  > temp2<-as.data.frame(temp2)
>  >
> newtemp2<-sapply((1:6),function(x){xmat<-temp2[temp2[,1]==x,,d
> rop=F];return(xmat[1,])})
>  > print(newtemp2) #looks like matrix
>     [,1] [,2] [,3] [,4] [,5] [,6]
> A1 1    2    3    4    5    6
> A2 1    2    3    4    5    6
> A3 1    2    3    4    5    6
> A4 1    2    3    4    5    6
>  > is.matrix(newtemp2) #says it's matrix
> [1] TRUE
>  > class(newtemp2)
> [1] "matrix"
>  > is.list(newtemp2) #but also list
> [1] TRUE
>  > newtemp2[,1] #can't subset and get a vector back; same
> thing happens for
> rows.
> \$A1
> [1] 1
>
> \$A2
> [1] 1
>
> \$A3
> [1] 1
>
> \$A4
> [1] 1
>  > names(newtemp2)
> NULL
>  > dimnames(newtemp2)
> [[1]]
> [1] "A1" "A2" "A3" "A4"
>
> [[2]]
> NULL
>  > dim(newtemp2)
> [1] 4 6
>  > length(newtemp2)
> [1] 24
>
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