R help - Oct 2005 - factor : how does it work ?

Dear all,

I try for long to understand exactly what is the factor type and especially
how it works, but it seems too difficult for me....
I read paragraphs about it, and I understand quite well what it is (I think)
but I still can't figure how to deal with.
Especially these 2 mysteries (for me) :

1st when I make a dataframe (with the as.data.frame() or the data.frame()
commands) from vectors, it seems that some "columns" of the dataframe
(which
where vectors) are factors and some not, but I didn't find an explanation
for which become factor and which don't.
(I know I can use I() to avoid the factor transformaton but I think it is
not an optimal solution to avoid the factor type just because I don't kno
how to deal with)

2d I can't manage to deal with factors, so when I have some, I transform
them in vectors (with levels()), but I think I miss the power and utility of
the factor type ?

Any help GREATLY appreciated,
best regards,

Florence.

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On 10/6/2005 9:14 AM, Florence Combes wrote:
> Dear all,
> 
> I try for long to understand exactly what is the factor type and especially
> how it works, but it seems too difficult for me....
> I read paragraphs about it, and I understand quite well what it is (I
think)
> but I still can't figure how to deal with.
> Especially these 2 mysteries (for me) :
> 
> 1st when I make a dataframe (with the as.data.frame() or the data.frame()
> commands) from vectors, it seems that some "columns" of the
dataframe (which
> where vectors) are factors and some not, but I didn't find an
explanation
> for which become factor and which don't.
> (I know I can use I() to avoid the factor transformaton but I think it is
> not an optimal solution to avoid the factor type just because I don't
kno
> how to deal with)
This is described in the ?data.frame man page:  "Character variables 
passed to 'data.frame' are converted to factor columns unless protected 
by 'I'."
> 2d I can't manage to deal with factors, so when I have some, I
transform
> them in vectors (with levels()), but I think I miss the power and utility
of
> the factor type ?
levels() is not the conversion you want.  That lists all the levels, but 
it doesn't tell you how they correspond to individual observations.  For 
example,

 > df <- data.frame(x=1:3, y=c('a','b','a'))
 > df
   x y
1 1 a
2 2 b
3 3 a
 > levels(df$y)
[1] "a" "b"

If you need to convert back to character values, use as.character():

 > as.character(df$y)
[1] "a" "b" "a"

For many purposes, you can ignore the fact that your data is stored as a 
factor instead of a character vector.  There are a few differences:

  1. You can't compare the levels of a factor unless you declared it to 
be ordered:

 > df$y[1] > df$y[2]
[1] NA
Warning message:
 > not meaningful for factors in: Ops.factor(df$y[1], df$y[2])

but

 > df$y <- ordered(df$y)
 > df$y[1] > df$y[2]
[1] FALSE

However, you need to watch out here: the comparison is done by the order 
of the factors, not an alphabetic comparison of their names:

 > levels(df$y) <- c("before", "after")
 > df
   x      y
1 1 before
2 2  after
3 3 before
 > df$y[1] > df$y[2]
[1] FALSE


  2. as.integer() works differently on factors:  it gets the position in 
the levels vector.  For example,

 > as.integer(df$y)
[1] 1 2 1
 > as.integer(as.character(df$y))
[1] NA NA NA
Warning message:
NAs introduced by coercion

There are other differences, but these are the two main ones that are 
likely to cause you trouble.

Duncan Murdoch

> > > 2d I can't manage to deal with factors, so when I have some,
I
> transform
> > > them in vectors (with levels()), but I think I miss the power and
> utility
> > of
> > > the factor type ?
> >
> > levels() is not the conversion you want.

in fact I use
'as.numeric(levels(f))[f]'
(from the ?factor description)

That lists all the levels, but
> > it doesn't tell you how they correspond to individual
observations. For
> > example,
> >
> > > df <- data.frame(x=1:3,
y=c('a','b','a'))
> > > df
> > x y
> > 1 1 a
> > 2 2 b
> > 3 3 a
> > > levels(df$y)
> > [1] "a" "b"
> >
> > If you need to convert back to character values, use as.character():
> >
> > > as.character(df$y)
> > [1] "a" "b" "a"

got it.

> > 1. You can't compare the levels of a factor unless you declared it
to
> > be ordered:
> >
> > > df$y[1] > df$y[2]
> > [1] NA
> > Warning message:
> > > not meaningful for factors in: Ops.factor(df$y[1], df$y[2])
> >
> > but
> >
> > > df$y <- ordered(df$y)
> > > df$y[1] > df$y[2]
> > [1] FALSE
> >
> > However, you need to watch out here: the comparison is done by the
order
> > of the factors

I am sorry I don't understand this.
here you compare the position of a in the factor and the position of b in
the factor ?

, not an alphabetic comparison of their names:
> >
> > > levels(df$y) <- c("before", "after")
> > > df
> > x y
> > 1 1 before
> > 2 2 after
> > 3 3 before
> > > df$y[1] > df$y[2]
> > [1] FALSE

best regards,

florence.

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