I am running R 1.2.1 under Windows 98SE.
termplot() currently fails when there are composite terms, thus:
> library(mass)
> data(hills)
> hills.lm <- lm(time ~ climb + poly(dist, 2), data = hills)
> termplot(hills.lm)
Hit <Return> to see next plot:
Error in eval(expr, envir, enclos) : Object "dist" not found
The call
> termplot(hills.lm,data=hills)
does run without generating an error message, i.e. there is a
problem with the default setting for the parameter 'data'.
However the x-axis for the term poly(dist,2) is then incorrectly
labelled as poly(dist,2), rather than dist
The label poly(dist,2) surely belongs as part of the y-axis label.
Here are suggested fixes
1. In the default parameters for the function call change
data = model.frame(model)
to
data = get(as.character(model$call$data))
[This does not work if the variable values are specified in the
function call. It is not clear to me how to fix that.]
2. A few lines from the beginning, replace
if (is.null(xlabs))
xlabs <- nmt
if (is.null(ylab))
ylab <- substitute(link(foo), list(foo = formula(model)[[2]]))
with
if (is.null(xlabs))
xlabs <- unlist(lapply(cn,function(x)
if(length(x)>1)as.character(x[[2]]) else as.character(x)))
if (is.null(ylab))
ylab <- paste("Partial for", nmt)
[The current y-label is "link(<yvar>)". If it is thought necessary
to specify this, it should be additional to e.g., in my example,
"Partial for poly(dist,2)"]
Replace ylab = ylab with
ylab = ylab[i]
in the two places (arguments to plot()) where it appears.
John Maindonald email : john.maindonald@anu.edu.au
Statistical Consulting Unit, phone : (6125)3998
c/o CMA, SMS, fax : (6125)5549
John Dedman Mathematical Sciences Building
Australian National University
Canberra ACT 0200
Australia
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
r-devel mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html
Send "info", "help", or "[un]subscribe"
(in the "body", not the subject !) To: r-devel-request@stat.math.ethz.ch
_._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
termplot fails for composite non-factor terms (PR#828)
2 messages · John Maindonald, Thomas Lumley
On Sat, 27 Jan 2001 john.maindonald@anu.edu.au wrote:
I am running R 1.2.1 under Windows 98SE. termplot() currently fails when there are composite terms, thus:
> library(mass) > data(hills) > hills.lm <- lm(time ~ climb + poly(dist, 2), data = hills) > termplot(hills.lm)
Hit <Return> to see next plot: Error in eval(expr, envir, enclos) : Object "dist" not found The call
> termplot(hills.lm,data=hills)
does run without generating an error message, i.e. there is a problem with the default setting for the parameter 'data'. However the x-axis for the term poly(dist,2) is then incorrectly labelled as poly(dist,2), rather than dist The label poly(dist,2) surely belongs as part of the y-axis label. Here are suggested fixes 1. In the default parameters for the function call change data = model.frame(model) to data = get(as.character(model$call$data)) [This does not work if the variable values are specified in the function call. It is not clear to me how to fix that.]
Also, it does not specify where to get() the data: what environment to use. The reason that the data= argument is present in termplot() is that we can't tell where to find the data in general. The situation is better as of R1.2, since we can look up variables in the environment of the model formula. This would fix not only termplot() but a number of other prediction-related problems.
2. A few lines from the beginning, replace
if (is.null(xlabs))
xlabs <- nmt
if (is.null(ylab))
ylab <- substitute(link(foo), list(foo = formula(model)[[2]]))
with
if (is.null(xlabs))
xlabs <- unlist(lapply(cn,function(x)
if(length(x)>1)as.character(x[[2]]) else as.character(x)))
if (is.null(ylab))
ylab <- paste("Partial for", nmt)
[The current y-label is "link(<yvar>)". If it is thought necessary
to specify this, it should be additional to e.g., in my example,
"Partial for poly(dist,2)"]
Replace ylab = ylab with
ylab = ylab[i]
in the two places (arguments to plot()) where it appears.
This looks like a good idea. -thomas -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-devel-request@stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._