RE: [R] sapply behavior
On Mon, 27 Sep 2004, Liaw, Andy wrote:
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.)
That *is* a matrix, though, and is useful for lists of length greater than
one. A matrix in R is just a vector with a dim attribute (and length the
product of the dims's), so as well as any of the atomic vectors it can be
a generic vector aka list.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] sapply behavior
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,,drop=F];return(xmat[1,])})
print(newtemp2) #looks like matrix
[,1] [,2] [,3] [,4] [,5] [,6]
A1 123456
A2 123456
A3 123456
A4 123456
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
#other things about it:
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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RE: [R] sapply behavior
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 123456
A2 123456
A3 123456
A4 123456
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
#other things about it:
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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Re: [R] sapply behavior
Elizabeth Purdom wrote:
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)
It is this coercion to the data frame that is injecting a list-like
property into the result. Try your script without that line and it will
work as you expect.
newtemp2-sapply((1:6),function(x){xmat-temp2[temp2[,1]==x,,drop=F];return(xmat[1,])})
print(newtemp2) #looks like matrix
[,1] [,2] [,3] [,4] [,5] [,6]
A1 123456
A2 123456
A3 123456
A4 123456
The best thing to do in a situation like this is to use the str function
to see the details of the structure of the object.
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Re: [R] sapply behavior
Elizabeth Purdom epurdom at stanford.edu writes:
:
: 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,,drop=F];return
(xmat[1,])})
: print(newtemp2) #looks like matrix
: [,1] [,2] [,3] [,4] [,5] [,6]
: A1 123456
: A2 123456
: A3 123456
: A4 123456
: 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
: #other things about it:
: names(newtemp2)
: NULL
: dimnames(newtemp2)
: [[1]]
: [1] A1 A2 A3 A4
:
: [[2]]
: NULL
: dim(newtemp2)
: [1] 4 6
: length(newtemp2)
: [1] 24
The problem is that your function is returning a one row data frame
and when sapply tries to simplify the resulting list of 6 data frames
that gives a list with dimensions rather what you were expecting
which is a vector with dimensions.
Let us call the original anonymous function in your post (i.e. the one
passed to sapply there), f. We can modify it to produce f2 which is like
f except that we wrap the return expression in c() to turn it into a
vector:
f2 - function(x){xmat-temp2[temp2[,1]==x,,drop=F];return(c(xmat[1,]))}
sapply(1:6, f2)
If you really do want to return a one row data frame then
use rbind to bind the data frames together rather than sapply:
do.call(rbind, lapply(1:6, f))
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