>>>>> On Thu, 8 Sep 2005 15:01:23 -0400 (EDT), >>>>> Jonathan Dushoff (JD) wrote:
> As instructed, I have spent a long time searching the web for an answer > to this question. > I am trying to use Sweave to produce lecture slides, and have the > problem that I can't control the formatting of my R source. Setting > options(width), as recommended in this forum, works fine on the R > _output_, but seems to have unpredictable effects on the echoing of the > source code. > If I try setting options(width) directly in R, I note that it has _no_ > effect on echoed source code, whereas Sweave does sometimes break source > code, but not predictably, and not to the same width as output code. > I would be happy with any method of manually or automatically > controlling the line width of Sweave source, using R, Sweave or LaTeX > options. Making the font smaller does not count, though; I want to > break the lines. > Any help is appreciated. > An example of Sweave input and output is appended. > The last break is right, while the others are too late. The deparser of R only *tries* to break lines at the given cutoff (Sweave uses 0.75*getOption("width") for input lines), so you soemtimes have to play a little bit, and yes, results are somewhat unpredictable. After storing your code in file ex1.R I get: R> expr=parse("ex1.R") R> length(expr) [1] 5 R> for(n in 1:length(expr)) print(deparse(expr[[n]], width.cutoff=.75*55)) [1] "options(width = 55)" [1] "data(state)" [1] "data.frame(area = mean(state.area), pop = mean(state.pop), " [2] " hop = mean(state.area))" [1] "c(medianarea = median(state.area), medianpop = median(state.pop))" [1] "c(medianarea = median(median(state.area)), " [2] " medianpop = median(state.pop))" R> for(n in 1:length(expr)) print(deparse(expr[[n]], width.cutoff=.75*45)) [1] "options(width = 55)" [1] "data(state)" [1] "data.frame(area = mean(state.area), " [2] " pop = mean(state.pop), hop = mean(state.area))" [1] "c(medianarea = median(state.area), " [2] " medianpop = median(state.pop))" [1] "c(medianarea = median(median(state.area)), " [2] " medianpop = median(state.pop))" I know that's not exactly the answer you were looking for, but that's the way it is. I'll add a sentence or two to the FAQ. Best, -- ------------------------------------------------------------------- Friedrich Leisch Institut für Statistik Tel: (+43 1) 58801 10715 Technische Universität Wien Fax: (+43 1) 58801 10798 Wiedner Hauptstraße 8-10/1071 A-1040 Wien, Austria http://www.ci.tuwien.ac.at/~leisch ______________________________________________ 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