as.name and namespaces
Here is an example problem: > mycall <- expression(lm(Y ~ x))[[1]] > mycall lm(Y ~ x) > newname <- "stats::lm" > desiredResult stats::lm(Y ~ x) I've solved the problem in the kludgy way of deparsing, fixing the string and then parsing. I like Duncan's third method, but it seems like it assumes the solution. Moving functions around is unappetizing for my use -- this is for testing and keeping things as faithful to real use is a good thing. Pat
On 23/04/2013 21:18, Duncan Murdoch wrote:
On 13-04-23 3:51 PM, Patrick Burns wrote:
Okay, that's a good reason why it shouldn't. Why it should is that I want to substitute the first element of a call to be a function including the namespace.
Three ways: 1. Assign the function from the namespace locally, then call the local one. 2. Import the function in your NAMESPACE (if you know the name in advance). 3. Construct an expression involving ::, and substitute that in. For example: substitute(foo(x), list(foo=quote(baz::bar))) Duncan Murdoch
Pat On 23/04/2013 18:32, peter dalgaard wrote:
On Apr 23, 2013, at 19:23 , Patrick Burns wrote:
'as.name' doesn't recognize a name with its namespace extension as a name:
as.name("lm")
lm
as.name("stats::lm")
`stats::lm`
as.name("stats:::lm")
`stats:::lm` Is there a reason why it shouldn't?
Any reason why it should? :: and ::: are operators. foo$bar is not the same as `foo$bar` either.
Patrick Burns pburns at pburns.seanet.com twitter: @burnsstat @portfolioprobe http://www.portfolioprobe.com/blog http://www.burns-stat.com (home of: 'Impatient R' 'The R Inferno' 'Tao Te Programming')