or use quote()... -- Bert
On Tue, May 29, 2012 at 10:48 AM, <cbe...@tajo.ucsd.edu> wrote: > Paul Johnson <pauljoh...@gmail.com> writes: > >> do.call( substitute, list( frm, list( x = as.name("z") ) ) ) ## or using quote() >>do.call( substitute, list( frm, list( x = quote(z)))) > y ~ log(z) * (w + u) > > > HTH, > > Chuck > > >> >> I have working code to finish that part of the problem, but it fails >> when the formula is more complicated. If the formula has log(x1) or >> x1:x2, the update code I'm testing doesn't get right. >> >> Here's the test code: >> >> ##PJ >> ## 2012-05-29 >> dat <- data.frame(x1=rnorm(100,m=50), x2=rnorm(100,m=50), >> x3=rnorm(100,m=50), y=rnorm(100)) >> >> m1 <- lm(y ~ log(x1) + x1 + sin(x2) + x2 + exp(x3), data=dat) >> m2 <- lm(y ~ log(x1) + x2*x3, data=dat) >> >> suffixX <- function(fmla, x, s){ >> upform <- as.formula(paste0(". ~ .", "-", x, "+", paste0(x, s))) >> update.formula(fmla, upform) >> } >> >> newFmla <- formula(m2) >> newFmla >> suffixX(newFmla, "x2", "c") >> suffixX(newFmla, "x1", "c") >> >> The last few lines of the output. See how the update misses x1 inside >> log(x1) or in the interaction? >> >> >>> newFmla <- formula(m2) >>> newFmla >> y ~ log(x1) + x2 * x3 >>> suffixX(newFmla, "x2", "c") >> y ~ log(x1) + x3 + x2c + x2:x3 >>> suffixX(newFmla, "x1", "c") >> y ~ log(x1) + x2 + x3 + x1c + x2:x3 >> >> It gets the target if the target is all by itself, but not otherwise. >> >> After messing with this for quite a while, I conclude that update was >> the wrong way to go because it is geared to replacement of individual >> bits, not editing all instances of a thing. >> >> So I started studying the structure of formula objects. I noticed >> this really interesting thing. the newFmla object can be probed >> recursively to eventually reveal all of the individual pieces: >> >> >>> newFmla >> y ~ log(x1) + x2 * x3 >>> newFmla[[3]] >> log(x1) + x2 * x3 >>> newFmla[[3]][[2]] >> log(x1) >>> newFmla[[3]][[2]][[2]] >> x1 >> >> So, if you could tell me of a general way to "walk" though a formula >> object, couldn't I use "gsub" or something like that to recognize each >> instance of "x1" and replace with "x1c"?? >> >> I just can't figure how to automate the checking of each possible >> element in a formula, to get the right combination of [[]][[]][[]]. >> See what I mean? I need to avoid this: >> >>> newFmla[[3]][[2]][[3]] >> Error in newFmla[[3]][[2]][[3]] : subscript out of bounds >> >> pj > > -- > Charles C. Berry Dept of Family/Preventive Medicine > cberry at ucsd edu UC San Diego > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.