Re: [R] Multiple expressions, when using substitute()
Yes, I did get a very helpful reply from Marc Schwartz. I have had substitute() working in legend(), when the legend argument has length one. The challenge was to find some way to do the equivalent of substitute() when several expressions appear in parallel, as may be required for legend(). The trick is to use bquote() to do the substitution. The resulting quoted expression (of mode call) can then be an element in a list, along with other quoted (or bquoted) expressions. The list elements, when passed to expression() via the args argument of do.call(), become unquoted expressions. Note that bquote() uses a syntax for the substitution of variables that is different from that used by substitute(). It would be useful to include some such example as below on the help page for bquote(): library(DAAG) Acmena - subset(rainforest, species=Acmena) plot(wood~dbh, data=Acmena) Acmena.lm - lm(log(wood) ~ log(dbh), data=Acmena) b - round(coef(Acmena.lm), 3) arg1 - bquote(italic(y) == .(A) * italic(x)^.(B), list(A=b[1], B=b[2])) arg2 - quote(where * italic(y) * =wood; * italic(x) * =dbh) legend(topleft, legend=do.call(expression, c(arg1, arg2)), bty=n) John Maindonald. On 11 Oct 2005, at 11:41 AM, Spencer Graves wrote: Have you received a reply to this post? I couldn't find one, and I couldn't find a solution, even though one must exist. I can get the substitute to work in main but not legend: B - 2:3 eB - substitute(y==a*x^b, list(a=B[1], b=B[2])) plot(1:2, 1:2, main=eB) You should be able to construct it using mtext, but I couldn't get the desired result using legend. hope this helps. spencer graves John Maindonald wrote: expression() accepts multiple expressions as arguments, thus: plot(1:2, 1:2) legend(topleft, expression(y == a * x^b, where * paste(y==wood; , x==dbh))) Is there a way to do this when values are to be substituted for a and b? i.e., the first element of the legend argument to legend() becomes, effectively: substitute(y == a * x^b, list(a = B[1], b=B[2])) John Maindonald email: [EMAIL PROTECTED] phone : +61 2 (6125)3473fax : +61 2(6125)5549 Centre for Bioinformation Science, Room 1194, John Dedman Mathematical Sciences Building (Building 27) Australian National University, Canberra ACT 0200. __ 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 -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 John Maindonald email: [EMAIL PROTECTED] phone : +61 2 (6125)3473fax : +61 2(6125)5549 Centre for Bioinformation Science, Room 1194, John Dedman Mathematical Sciences Building (Building 27) Australian National University, Canberra ACT 0200. __ 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
Re: [R] Multiple expressions, when using substitute()
Have you received a reply to this post? I couldn't find one, and I couldn't find a solution, even though one must exist. I can get the substitute to work in main but not legend: B - 2:3 eB - substitute(y==a*x^b, list(a=B[1], b=B[2])) plot(1:2, 1:2, main=eB) You should be able to construct it using mtext, but I couldn't get the desired result using legend. hope this helps. spencer graves John Maindonald wrote: expression() accepts multiple expressions as arguments, thus: plot(1:2, 1:2) legend(topleft, expression(y == a * x^b, where * paste(y==wood; , x==dbh))) Is there a way to do this when values are to be substituted for a and b? i.e., the first element of the legend argument to legend() becomes, effectively: substitute(y == a * x^b, list(a = B[1], b=B[2])) John Maindonald email: [EMAIL PROTECTED] phone : +61 2 (6125)3473fax : +61 2(6125)5549 Centre for Bioinformation Science, Room 1194, John Dedman Mathematical Sciences Building (Building 27) Australian National University, Canberra ACT 0200. __ 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 -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ 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
[R] Multiple expressions, when using substitute()
expression() accepts multiple expressions as arguments, thus: plot(1:2, 1:2) legend(topleft, expression(y == a * x^b, where * paste(y==wood; , x==dbh))) Is there a way to do this when values are to be substituted for a and b? i.e., the first element of the legend argument to legend() becomes, effectively: substitute(y == a * x^b, list(a = B[1], b=B[2])) John Maindonald email: [EMAIL PROTECTED] phone : +61 2 (6125)3473fax : +61 2(6125)5549 Centre for Bioinformation Science, Room 1194, John Dedman Mathematical Sciences Building (Building 27) Australian National University, Canberra ACT 0200. __ 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
Re: [R] Multiple expressions, when using substitute()
On Sat, 2005-10-01 at 20:32 +1000, John Maindonald wrote: expression() accepts multiple expressions as arguments, thus: plot(1:2, 1:2) legend(topleft, expression(y == a * x^b, where * paste(y==wood; , x==dbh))) Is there a way to do this when values are to be substituted for a and b? i.e., the first element of the legend argument to legend() becomes, effectively: substitute(y == a * x^b, list(a = B[1], b=B[2])) John, Try this: a - 5 b - 3 L - list(bquote(y == .(a) * x^.(b)), where y = wood; x = dbh) plot(1:2, 1:2) legend(legend = do.call(expression, L), topleft) Note the creation of the list 'L', which uses bquote() and then the .(Var) construct, where 'Var' are your variables to be replaced. Then in the legend() call, the use of do.call() to apply expression() to the elements of list 'L'. HTH, Marc Schwartz __ 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
