-----Original Message-----
From: [email protected] [mailto:[email protected]] On
Behalf Of Dan Murphy
Sent: Monday, January 23, 2012 10:31 PM
To: peter dalgaard
Cc: [email protected]
Subject: Re: [Rd] factor S4 class is NA when as.character method exists
Thank you for your reply, Peter. But that didn't work either. Continuing
the example:
setGeneric("unique")
setMethod("unique", "foo", function(x, incomparables = FALSE, ...){
y<- callNextMethod(x = getDataPart(x), incomparables = incomparables,
...)
new("foo", y)
})
unique(bar)
An object of class "foo"
[1] 12
factor(bar)
[1]<NA>
Levels: 12
Indeed I had tried stepping through the 'factor' call, but perhaps in an
unsophisticated manner -- I had copied the body of 'factor' to a local
version of the function:
myfactor<- function (x = character(), levels, labels = levels, exclude =
NA,
ordered = is.ordered(x))
{
if (is.null(x)) ...
etc.
And 'myfactor' worked as desired:
myfactor(bar)
[1] x= 12
Levels: x= 12
I hypothesized that there might be a deeper interaction of an S4
'as.character' method with base::factor, but, having exhausted my woeful
lack of expertise, I decided to write my original email.
Thanks for your consideration.
Dan
On Mon, Jan 23, 2012 at 8:25 AM, peter dalgaard<[email protected]> wrote:
On Jan 23, 2012, at 16:07 , Dan Murphy wrote:
Hello,
'factor' returns<NA> for my S4 object when the class is given an
"as.character" method. Here is a minimal example:
setClass("foo", contains="numeric")
bar<- new("foo", 12)
factor(bar)
[1] 12
Levels: 12
setMethod("as.character", "foo", function(x) paste("x=", [email protected]))
[1] "as.character"
as.character(bar)
[1] "x= 12"
factor(bar)
[1]<NA>
Levels: 12
I would like to 'aggregate' by my S4 objects, but 'factor' seems to be
getting in the way. Is there an 'as.character' implementation that works
better for S4 classes? I searched help.search("factor S4 class") and
help.search("factor S4 as.character") without success.
Single-stepping the factor call would have shown you that the real problem
is that you don't have a unique() method for your class:
unique(bar)
[1] 12
i.e., you are getting the default numeric method, which returns a numeric
vector, so the levels become as.character(unique(bar)) which is c("12") and
doesn't match any of the values of as.character(bar).
So, either provide a unique() method, or use factor(as.character(bar)).
Thank you.
Dan Murphy
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--
Peter Dalgaard, Professor
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: [email protected] Priv: [email protected]
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