how to concatenate factor vectors?

hi Jorge,

* Jorge I Velez <wbetrvinairyrm at tznvy.pbz> [2012-10-18 16:43:58 +1100]:

a <- factor(5:1,levels=1:9)
b <- factor(9:1,levels=1:9)
lev <- sort(unique(f <- c(a, b)))
f <- factor(f, levels = lev)
str(f)

 Factor w/ 9 levels "1","2","3","4",..: 5 4 3 2 1 9 8 7 6 5 ...

is sort(unique()) really necessary?
I think
lev <- levels(a)
should be enough.

However, this does not quite do what I want.
I want a function which will _NOT_ have a non-factor vector as an
intermediate value because that would waste a LOT of memory in my case.
I want a function which will check that a and b have identical levels
(in Lisp lingo, the levels are EQ, not just EQUALP).

--8<---------------cut here---------------start------------->8---

a <- factor(letters[sample(1:10,20,replace=TRUE)],levels=letters)

[1] e e a b c e j d a b h i a e e g j a c e
Levels: a b c d e f g h i j k l m n o p q r s t u v w x y z

b <- factor(letters[sample(1:10,30,replace=TRUE)],levels=letters)

[1] d d f c j b d e j j g i g j j g g a j a b e d c b i i a b f
Levels: a b c d e f g h i j k l m n o p q r s t u v w x y z

c(a,b)

[1]  5  5  1  2  3  5 10  4  1  2  8  9  1  5  5  7 10  1  3  5  4  4  6  3 10
[26]  2  4  5 10 10  7  9  7 10 10  7  7  1 10  1  2  5  4  3  2  9  9  1  2  6

factor(letters[c(a,b)],levels=letters)

[1] e e a b c e j d a b h i a e e g j a c e d d f c j b d e j j g i g j j g g a
[39] j a b e d c b i i a b f
Levels: a b c d e f g h i j k l m n o p q r s t u v w x y z
--8<---------------cut here---------------end--------------->8---

however, this is not a "direct" way (unlike my unlist(list(...))):
there is an intermediate integer vector c(a,b) which is mapped to a
character vector via letters, which is converted back to integers
(==factors).

IIUC, a factor is an integer vector which knows that the integers refer
to levels.

c(a,b) creates such an integer vector.
How do I tell it that it is a factor?

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