vito muggeo wrote:
Dear all,
It appears that the function is.factor() returns different results when
used inside the apply() function: that is, is.factor() fails to
recognize a factor..
Where is the trick?
many thanks,
vito
df1-data.frame(y=1:10,x=rnorm(10),g=factor(c(rep(A,6),rep(B,4
is.factor(df1[,1])
[1] FALSE
is.factor(df1[,2])
[1] FALSE
is.factor(df1[,3])
[1] TRUE
is.factor(df1$g)
[1] TRUE
apply(df1,2,is.factor)
y x g
FALSE FALSE FALSE
Well, apply works on matrices and arrays, hence df1 is coerced to a
matrix, i.e. you invoked in fact the same as
apply(as.matrix(df1), 2, is.factor)
hence you get a matrix of character values which can be confirmed by
apply(df1, 2, is.character)
Anyway, what you really want is:
sapply(df1, is.factor)
Uwe Ligges
R.version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor 6.1
year 2007
month 11
day26
svn rev43537
language R
version.string R version 2.6.1 (2007-11-26)
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