On 2/9/06, Berton Gunter <[EMAIL PROTECTED]> wrote: > > Folks: > > R 2.2.0 on Windows. > > I find the following somewhat puzzling: > > > a<-1; x<-0:1; y<-x > > ## following works fine: > > plot(x,y ,main= bquote(n[1] == .(a) )) > > ## following produces an error: > > plot(y~x ,main= bquote(n[1] == .(a) )) > Error in paste(n[1] == 1, " and ", n[2] == 2) : object "n" not found > > ******************************* > > Note 1: I assume that this is due to the following documented behavior of > plot.formula(): > > "Both the terms in the formula and the ... arguments are evaluated in data > enclosed in parent.frame() if data is a list or a data frame." > > Nevertheless, the behavior seems inconsistent to me. Am I missing something > (including the "I assume ..." comment)? > > Note 2: If one uses substitute() instead, it works fine: > > plot(y~x ,main= substitute(bquote(n[1] == a),list(a=a))) >
Regarding your note 2, the key thing that seems to be necessary is not really substitute vs. bquote but just doing it twice. In your example above you did not replace bquote with substitute but did a substitute and a bquote so now its two levels deep. But if we just did two bquotes or two substitutes that would be ok too. For example, we can just apply bquote twice and then it works: plot(y~x ,main = bquote(bquote(n[1] == .(a)))) The various calls must be stripping off one layer so that you have to protect it twice so that the underlying code does not get evaluated. ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
