similar to: PDF output problem

Note: I am almost certain that this has been asked and answered here before, so my apologies for the redundant query. I also know that there are several packages that will do this, but I wish to do it using base R functions only (see below). The query: Suppose I have a character vector of names like this: somenames <- c("Heigh", "Ho", "Silver", "Away")

Gabor, Duncan, et. al. 1. Thank you for your great comments and solutions. This is what I was hoping for! 2. Duncan: I completely agree with your criticisms. In fact, I realized the for() loop only needed the <- assignment, but your comment is important to note. However, I didn't like the for() loop either; I *much* preferred your Reduce() solution which is exactly the sort of elegant

Another solution. reformulate + substitute + as.formula: substitute(~ (.)^2, list(. = reformulate(somenames)[[2]])) |> as.formula() On Sat, Mar 29, 2025 at 5:31?PM Bert Gunter <bgunter.4567 at gmail.com> wrote: > > Note: I am almost certain that this has been asked and answered here > before, so my apologies for the redundant query. > > I also know that there are several

> somenames <- c("Heigh", "Ho", "Silver", "Away") > as.formula(paste("~(",paste(somenames, collapse="+"),")^2")) ~(Heigh + Ho + Silver + Away)^2 > > On Mar 29, 2025, at 14:30, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > somenames <- c("Heigh", "Ho", "Silver",

As always, I would like to thank all who responded for their insights and suggestions. I have learned from them. Thus far, my own aesthetic preference -- and therefore not to be considered in any sense as a "best" approach -- is to use Duncan's suggestion to produce the call directly with call() rather than substitute in my simple for() loop; i.e. somenames <-

Thanks, Rich. I thought of that, too, but it violates the spirit of my restraints (to avoid character strings), which I unfortunately did not clearly articulate. So my apologies for that failure. My concern is that with more complex model formula, using as.formula, etc. to parse/convert character strings can get a bit hairy. But in most cases, as here maybe, it may be perfectly fine. So think of

my take of the assignment was to avoid 'parse' specifically. we start with a character vector, so avoiding characters is not possible. i was dealing with the fortune "if parse is the answer, you have the wrong question" Sent from my iPhone On Mar 29, 2025, at 15:39, Bert Gunter <bgunter.4567 at gmail.com> wrote: ? Thanks, Rich. I thought of that, too, but it violates

I am confused. Richard's answer that Bert did not like did not use parse explicitly. Richard pasted together a string that a function like lm() will have to parse to run the analysis. However, the answers so far do not use parse(). In the reply to Richard, Bert indicated we cannot use strings. Even if I pass a vector where R can assume that the first variable is the dependent variable and all

Hello, I thought of answering "reformulate can solve the problem" but how do you create quadratic terms with reformulate? ~(Heigh + Ho + Silver + Away)^2 is still a problem with no solution that I know of but paste/as.formula. Or Bert's bquote or substitute. Rui Barradas ?s 23:18 de 29/03/2025, Ebert,Timothy Aaron escreveu: > The general formula is y ~ a + b + c + ... >

The general formula is y ~ a + b + c + ... There is this approach: formula <- reformulate(independent_vars, response = "y") model <- lm(formula, data = mydata) summary(model) It does not generate a string object, but the formula is still a string even if it is of class formula. Also, in this approach you only get + and if you want interactions or such you will need to code them

Hello >From the following script I received a grafic output when I called: - xyplot.test( 'green3' ) - call.xyplot.test( 'blue3' ) I did NOT receive a grafic output when I called: - loop.xyplot.test( 'red3' ) What's the Problem? NB: I am using R 1.4.1 on Linux. --------- START OF SCRIPT ---------------- n <- 1000 x <- seq( 1, n ) y <- rnorm( n )

Hello, I think this might be a beginner question, but I couldn't find the answer in the manual... http://www.nabble.com/file/p14618947/at-modality.png I created this image with R by using the following code: modality <- read.table("results.table", header=TRUE, sep=",") color <-

Dear List, I am trying to get a basic plot to show a continuous range of fill colors. It is probably easiest to demonstrate. I would like a legend like in the following example: Satellite.Palette <-colorRampPalette(c("blue3","cyan","aquamarine","yellow","orange","red")) require(fields) image.plot(volcano, col = Satellite.Palette

http://www.nabble.com/file/p14664173/at-modality.png I created the above image with R and I have one problem left: Some of the labels of the axes do not show up, probably because there's not enough space. I use the following code to create the plot: modality <- read.table("results.table", header=TRUE, sep=",") color <-

> Dear expeRts, > > Currently, my colors are as follows: > mycol <- > c("blue1","blue2","blue3","blue4","black","yellow4","yellow3","yellow2","y > ellow1") > heatmap(snp, Rowv=NA, Colv=NA, col=mycol) > > However, I would like to have the following colors: > bright blue ->

I need help with using graphics in Word 2007 that will later be converted into a pdf document.? I have tried several formats and found that?I get the best quality of graphics using .wmf, .eps format, but when I convert it to .pdf I get a bunch of lines across the figures.? I also tried .tiff and .png but they give me much lower quality.? The best quality that converts to pdf appears to be

Hi, I have processed measurements of a rough surface to a heigh-height correlation plot. What the meaning of this exactly is, is not important. Only that it is a plot that had two (almost ) linear parts when plotted on a logaritmic scale. In this plot, I want to draw the best fitting lines for these linear parts but I just can't get it done. It is easy when the scales are linear but as you

I want to write a function mimic the function of select.list(), here is my preliminary version. select <- function(x,multiple=TRUE,...){ ans<-new.env() g <- gwindow(title=title,wid=200,heigh=500) x1<-ggroup(FALSE,con=g) x2<-gtable(x,multiple=multiple,con=x1,expand=TRUE) gbutton("OK",con=x1,handler=function(h,...){ value <- svalue(x2) if (length(value)==0)

Dear List, I am trying to plot a color.legend() in the right outer margin of my device region. I have read multiple threads on the subject and still can't get it right. I have stolen an example from one of the threads to demonstrate my problem. I have extended the outer margin using par(oma()), and have used par(xpd=NA) to tell it to plot in the device region. I can get the legend to plot

Dear all, I am trying (!!!) to generate pdfs that have 8 plots on one page: df = data.frame( day = c(1,2,3,4), var1 = c(1,2,3,4), var2 = c(100,200,300,4000), var3 = c(10,20,300,40000), var4 = c(100000,20000,30000,4000), var5 = c(10,20,30,40), var6 = c(0.001,0.002,0.003,0.004), var7 = c(123,223,123,412), var8 = c(213,123,234,435), all = as.factor(c(1,1,1,1)))