On 18/05/2010 4:36 PM, baptiste auguie wrote:> Dear list,
>
> I am puzzled by this,
>
> substitute(expression(x), list(x = factor(letters[1:2])))
> # expression(1:2)
>
> Why do I get back the factor levels inside the expression and not the
> labels?
As the documentation for substitute() says, there is no guarantee that
the result makes sense. Yours doesn't, and it confuses the deparser,
which is not displaying what you really have:
> y <- substitute(expression(x), list(x = factor(letters[1:2])))
> y
expression(1:2)
> str(y)
language expression(structure(1:2, .Label = c("a", "b"),
class = "factor"))
The problem is that expressions don't normally have attributes, and
factors have both .Label and class attributes. Put another way:
expressions don't normally include factors, they include names and calls
and atomic vectors. The deparser doesn't distinguish between the
language "1:2" and the atomic vector that is the value of that
expression, but it doesn't expect attributes, and doesn't go looking for
them.
Duncan Murdoch
> The following work as I expected,
>
> substitute(expression(x), list(x = letters[1:2]))
> # expression(c("a", "b"))
>
> substitute(x, list(x = factor(letters[1:2])))
> # [1] a b
> # Levels: a b
>
> bquote(.(factor(letters[1:2])))
> # [1] a b
> # Levels: a b
>
>
> All the best,
>
> baptiste
>
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