At 12.08.2011 01:53 +0100, Heinz Tuechler wrote:
At 12.08.2011 11:05 +1200, Rolf Turner wrote:
On 12/08/11 11:34, Heinz Tuechler wrote:
At 12.08.2011 09:11 +1200, Rolf Turner wrote:
On 12/08/11 09:59, Heinz Tuechler wrote:
At 11.08.2011 21:50 +0300, Zeki Çatav wrote:
PrÅ, 2011-08-11 tarihinde 19:27 +0200 saatinde, Uwe Ligges yazdı:
>
> On 11.08.2011 19:22, David Winsemius wrote:
> >
> > On Aug 11, 2011, at 11:42 AM, Uwe Ligges wrote:
> >
> >>
> >>
> >> On 11.08.2011 16:10, zcatav wrote:
> >>> Hello R people,
> >>>
> >>> I have a "data.frame". Status variable has 3 values. 0->alive,
> >>> 1->dead and
> >>> 2->missed..........................
> .........................................
> As I understood the question, just how to rename the levels was the
> original question.
>
> Uwe
I don't want to rename levels or converting from numeric to string. I
want to add each corresponding levels value, a label, as in SPSS.
Level 0 labeled with alive,
level 1 labeled with dead and
level 2 labeled with missed.
This is not possible with a factor, because
factor levels can only be positive integers.
That is just plain (ridiculously) wrong. RTFM.
cheers,
Rolf Turner
So, how would you construct a factor with
levels 0, 1, 2 and labels alive, dead, and
missed, as the original post asked for?
Heinz
As I said, RTFM. But for completeness:
x <- sample(0:2,100,TRUE)
y <- factor(x,labels=c("alive","dead","missed"))
Duhhh.
cheers,
Rolf Turner
Maybe you would like to look at the structure.
str(y)
Factor w/ 3 levels "alive","dead",..: 3 1 2 2 3 3 2 1 1 1 ...
or
dput(y)
structure(c(3L, 1L, 2L, 2L, 3L, 3L, 2L, 1L, 1L, 1L, 2L, 1L, 3L,
1L, 2L, 3L, 2L, 2L, 1L, 3L, 1L, 3L, 1L, 3L, 1L, 2L, 1L, 2L, 1L,
1L, 3L, 1L, 2L, 3L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 3L,
2L, 1L, 2L, 3L, 3L, 2L, 1L, 1L, 2L, 3L, 3L, 2L, 1L, 3L, 1L, 1L,
2L, 2L, 1L, 2L, 1L, 2L, 3L, 3L, 3L, 2L, 3L, 3L, 1L, 2L, 3L, 2L,
3L, 1L, 3L, 1L, 1L, 2L, 2L, 1L, 2L, 2L, 3L, 2L, 2L, 1L, 1L, 1L,
2L, 3L, 3L, 2L, 1L, 2L, 3L), .Label = c("alive", "dead", "missed"
), class = "factor")
>
Anything else but positive integers?
Heinz
To be fair, you can construct a factor containing zeros.
b <- c(0L,0L,1L,1L,1L,2L,2L,2L,2L)
> b
[1] 0 0 1 1 1 2 2 2 2
> str(b)
int [1:9] 0 0 1 1 1 2 2 2 2
> table(b)
b
0 1 2
2 3 4
> levels(b) <- letters[1:3]
> str(b)
atomic [1:9] 0 0 1 1 1 2 2 2 2
- attr(*, "levels")= chr [1:3] "a" "b" "c"
> class(b) <- 'factor'
> str(b)
Factor w/ 3 levels "a","b","c": 0 0 1 1 1 2 2 2 2
But, if you print it, you get a warning.
> b
[1] a a a b b b b a a
Levels: a b c
Warning message:
In xx[] <- as.character(x) :
number of items to replace is not a multiple of replacement length
And table() gives a wrong result.
> table(b)
b
a b c
3 4 0
>
If you take a numeric, not explicitly integer vector, you are less lucky.
c <- c(0,0,1,1,1,2,2,2,2)
> str(c)
num [1:9] 0 0 1 1 1 2 2 2 2
> levels(c) <- letters[1:3]
> str(c)
atomic [1:9] 0 0 1 1 1 2 2 2 2
- attr(*, "levels")= chr [1:3] "a" "b" "c"
Assigning class "factor" is rejected with an error.
> class(c) <- 'factor'
Error in class(c) <- "factor" :
adding class "factor" to an invalid object
>
Heinz
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