[R] substitute, expression and factors

[Duncan Murdoch]

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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