similar to: How to print a variable with in double quotes

Dear R-helpers: I am an R newbie and have a question related to writing functions that accept unlimited number of input arguments. (I tried to peek into functions such as paste and cbind, but failed, I cannot see their codes..) Can someone kindly show me through a summation example? Say, we have input scalar, 1 2 3 4 5 then the ideal function, say sum.test, can do

Hi, I need to draw a line segment between two points on different plots in the same multigraph.I've tried looking at the zoominplot function in plotrix but havent understood much.any help is appreciated ~Aks [[alternative HTML version deleted]]

Dear all, I am trying to read in and assign data from 50 tables in an automated fashion. I have the following code, which I created with the help of textbooks and the internet, but it only seems to read in the final data file over and over again. For example, when I type:> table_1951 I get the same values in the table as when I type> table_2000 despite the values in the source tables

Hi, I wonder whether there is a way to generate a polygon (a triangle in my case) with color gradient using grid.polygon() in package grid? I tried something like library(grid) grid.polygon(x=c(0, 0.5, 1), y=c(0.5, 1, 0.5), gp=gpar(col=NA, fill=colorRampPalette(c("green", "lightgray"), space="Lab")(200))) But am only

Hi R, When I use the below to write the text file try=data.frame(rep("a",5), rep("b",5)) write.table(try,"z:\\try.txt",row.names=F,col.names=F,sep="\t") the output contains two columns with quotes! Is there a way to write without quotes? I tried try[,1]=noquote(try[,1]) try[,2]=noquote(try[,2]) Thank you, Regards, Ravi Shankar

Hi, I've used the following command in qplot qplot(a$V1,geom="histogram",binwidth=0.15,fill = factor(a$V2),ylab="Frequency",xlab="Rate"); but the title in the legend shows up as factor(a$V2)...how can i change this? -- Rajesh.J [[alternative HTML version deleted]]

Hi, I have a graph with seven parallel horizontal lines. Is it possible to shade the area between two adjacent lines? Thank you in advance, Hannu

Dear R-helpers, I have a list containing 5000 elements, each element is a dataframe containing one ID column (identical over the 5000 dataframes) and 9 numeric variables, e.g. ID VAR1 VAR2 VAR3 ... VAR9 I would like to create a new dataframe containing the ID column and the mean values of the 9 numeric variables. So, the structure of this new dataframe would be identical to the structure of the

Dear list, I've written a function that plots subjects. Something like: myplot <- function(subject) { plot(subject) } Subjects are vectors, e.g. ... s1 <- c(200,200,190,180) ... and plotting them works fine, e.g. ... myplot(s1) Now I want to have "s1" etc appear in the plot title, but I don't know how to refer to this generically (the object "name"? I tried

Dear R-community, using R.2.9.1, how do you put reference lines or grids into the _background_ of a plot? For example: barplot(3:1) abline(h = seq(0.5, 2.5 ,0.5), col = "red", lty = "dashed") -> The lines are before the bars (and the line parts going through the bars might be considered as "chart junk", I suppose). Now, in this toy example I could just do the

Hi I have the following function which should return the name of FUN: myFUN <- function(FUN) { return( THE_NAME_OF_FUN(FUN)) } Is it possible? What do I have tio use here instead of THE_NAME_OF_FUN? Thanks, Rainer -- Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch University, South Africa [[alternative HTML version deleted]]

/Dear all,/// /I am very thankful, if you could tell what is the right way to write: mtext(paste(expression("R"^2),round(marco2[1,i],digits=3)," N? of proteins:",marco3[i]),side=4,cex=.6) in this case the output is: "R"^2 I tried also in this way: mtext(paste(expression(paste("R"^2)),round(marco2[1,i],digits=3)," N? of

Dear R-Help subsribers, upon running into a wonderful ggplot2 package by accident, I abruptly encountered another problem. Almost every command run with ggplot2 results in some sort of error. The one below is far the most common one. Kind people from ggplot2 mailing list couldn't manage to solve the problem, so I'm re-posting it here to try my luck. I will recommend myself for any tips

Hi, I have the following data that looks like this: > names(dat) [1] "(-2329,-2319]" "(-1399,-1389]" "(-669.4,-659.4]" How can I modify those names into just this? [1] -2329 -1399 -669.4 - Gundala Viswanath Jakarta - Indonesia

Hi, x=matrix(1:60,nr=6) I can refer the last 2 rows by x[5:6,] If I don't know the total number of rows is 6, is there a way to refer the last 2 rows? Regards, Peng

Hello R Folks... Using the example below, I¹d like two of the panels to be plotted with type = ³p² but the third to be done with type = ³h². I can¹t use type = c(³p², ³p², ³h²) because this syntax applies all given types to every panel. I don¹t think I can use groups and distribute.type because these are intended for different styles of plotting within a single panel. As you can see, I tried

Dirk, I came upon your message below in searching for a different Sweave topic. Have you pursued this? I'm new to R (9 months). Recently have been doing data analysis in R and have started pondering the same question, but don't have any ideas yet. I did do something related to this in SAS, the main purpose of which was to build a standardized file of statistical results to be

Dear R-users, I am trying to use the grid.text and expression functions to display several character strings and plotmath text on a viewport. Some strings can include a variable portion (PI.limits in the following example), which I thought could be implemented by combining the bquote and the expression functions. Unfortunately, my expressions do not seem to be evaluated. I would greatly

Dear All, I have a problem which *should* be pretty straightforward to resolve - but I can't work out how! I have a list of 3 coefficient estimates for 4 different datasets: Coefs<-list(c(1,0.6,0.5),c(0.98,0.65,0.4),c(1.05,0.55,0.45),c(0.99,0.50,0.47)) All I want to do is take the sum (or mean) of each coefficient across the 4 datasets. I can do this using a "for" loop, but

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