similar to: Coplot graph size manipulation

Just out of curiosity, what is the difference between the terms for package and library ? Why are we loading a package with the library() command ? If this is a case of RTFM, I would be happy to do so if pointed in the right direction. I have searched the FAQ and mail archives and only came up with http://tolstoy.newcastle.edu.au/R/help/05/02/12162.html but this still does not explain what is

Dear R-friends, For the following data: > xy x y i 1 731 0.313 2 2 739 0.340 2 3 790 0.373 2 4 855 0.451 2 5 980 0.608 2 6 575 0.156 1 7 608 0.207 1 8 630 0.249 1 9 670 0.332 1 10 838 0.377 1 11 964 0.466 1 > coplot(y ~ x|i, data=xy) coplot gives 3 panels, rather than 2, namely one for i=1 and one for i=2. Futhermore, when I extand data fram xy to have i=3 as follows:

Dear all, I am trying to create a web interface using Perl-CGI to call R plots and to display them. The following codes works perfectly fine when I copy and paste into the console directly or if I save it into script.file and then R --no-save < script.file producing the graphs. jpeg("graph.jpeg", width=400, height=400) plot(rnorm(100)) dev.off() Now, I put the line system("R

Dear all, I am trying to compile the package e1071 (version 1.3-11) with R CMD INSTALL. I tried with R 1.7.0 on Redhat Linux 2.4.7-10 and R 1.6.2 on Linux 2.4.9-34smp but keep getting the same error message during configure : WARNING: g++ 2.96 cannot reliably be used with this package. Please use a different compiler. Can anyone help me with this or at least point me in the right direction ?

---1149173172-1804289383-983267779=:26068 Content-Type: TEXT/plain; charset=us-ascii Hello, and a big thank you for providing R! Please find attached a diff for coplot which you may want to consider for the next release. The diff is against R 1.2.2. The reasons for this patch are: 1. The boxes of coplot did not align very well with the panel graphs if applied to a factor 2. Putting the

Just out of curiosity, can some please explain the following return NA. x <- 6 var(x) y <- c( NA, NA, 10000 ) var(y, na.rm=T) Unless I am seriously misguided, I believe that the variance of a single number (i.e. a constant) should be zero. Thanks. Regards, Adai.

I am trying to write a wrapper for the last example in help(cv.glm) that deals with leave-one-out-cross-validation (LOOCV) for a logistic model. This wrapper will be used as part of a bigger program. Here is my wrapper funtion : logistic.LOOCV.err <- function( formu=NULL, data=NULL ){ cost.fn <- function(cl, pred) mean( abs(cl-pred) > 0.5 ) glmfit <- glm(

R 1.9.1 on Win2000 or Win98SE. I am using coplot as follows: coplot(AVG~LRPI| REGION) the output seems normal but I get: "Warning message: calling par(new=) with no plot" This is the only explanation that I have for being unable to use par() with coplot for changing the way the xlab and ylab appears. From within coplot I can change the text itself but not the font, fontsize, etc,

Sorry if this is a question more on regular expressions. I am dealing with several files which have been badly named. For example the files are given either the extensions txt, TXT or Txt. I wish to select all those files ending with 'txt' ignoring case. Here is how I would do it in bash (Redhat FC2) : touch a1.txt a2.TXT a3.Txt txt.control TXT.control ls -1 | grep -i

I am comparing two different algorithms in terms of speed and memory usage. I can calculate the processing time with proc.time() as follows but am not sure how to calculate the memory usage. ptm <- proc.time() x <- rnorm(1000000) proc.time() - ptm I would like to be within R itself since I will test the algorithm several hundred times and in batch mode. So manually looking up

Forgive me for I do not fully comprehend the idea of classes and methods but I was wondering if someone could help explain why the function args () behaves the way it does. Why does args(cut) show the simplified version instead of the more complete one as in help("cut"). This is true for few other functions (e.g. plot, rep). > args(cut) function (x, ...) Sometime I can get

Hi! I would like to know, how do you plot more than one data series when using "coplot"? I think I know the answer if "plot" is being used: x <- 1:10 y <- 1:10 y2 <- 1:10* 1.1 plot ( y ~ x, type="n" ) points( y ~ x, type = "b", col = "red" ) # plot the points and lines for the first series points( y2 ~ x, type = "b", col =

I am using R0.63.1 under Windows 95. Settings of <number> and <interval> when coplot is called are ignored, except for generating a warning. For example: > data(quakes) > coplot(long ~ lat | depth * mag, data = quakes, number=3) Warning: parameter "number" couldn't be set in high-level plot() function The warning is repeated ad nauseum. The following changes

Dear useRs, When I use coplot() and output to png/jpeg/bmp, the grid lines from the scatter plots disappear. If I output to pdf() the grid lines are there, however I can't use it - I have many points, and the resulting PDF file is large and very slow to open and scroll through. (By the way, if I click File-Save As-png/jpeg/bmp from Rgui.exe, the grid lines are preserved - but I need to use

Hello, I am trying to use self created functions in other scripts than the one where they are stored. For the moment I am using the following structure of commands to do that: 1. Load the text file with the functions in the current script: x=parse("path") 2. transform the tex in a function: f1=eval(x[1]), f2=eval(x[2]) if more than one function is stored in the text file 3. use the

Hi all, I could not find a way to add the dashed horizontal separator line to the top panel of a coplot (the one indicating which panel belongs to which range of the given variable a in y~x | a). This line should separate the ranges plotted in a row of panels in the panel matrix below (hmmm, hope this is reasonably clear). I like it because it enhances readability. I'll try a sketch: >

Greetings: I am unsuccessful in suppressing "Given : myvariable" from a coplot. There was such a question in the past but the thread breaks down. I am sure this is a "for dummies" question :-). I tried: coplot(myvar~myvar | myvar, show.given=FALSE, xlab="....", ylab="...", main=" ") and some other variations (including without main=" ")

Hi there, Suppose mu is constant, and error is normally distributed with mean 0 and fixed variance s. I need to find a statistics that: Y_i = mu + beta1* I1_i beta2*I2_i + beta3*I1_i*I2_i + +error, where I_i is 1 Y_i is from group A, and 0 if Y_i is from group B. It is large when beta1=beta2=0 It is small when beta1 and/or beta2 is not equal to 0 How can I find it by R? Thank you very much

Hello guys, I am wondering the default way of transferring arguments in R. Is it by value or by ref in default case, or could that be changed explicitly? Cheers, Xiaofan --- Xiaofan Li Department of Applied Mathematics and Theoretical Physics University of Cambridge

Hi, I calculating the output of a function when applied to pairs of row from a single matrix or dataframe similar to how cor() and pairs() work. This is the code that I have been using: pairwise.apply <- function(x, FUN, ...){ n <- nrow(x) r <- rownames(x) output <- matrix(NA, nc=n, nr=n, dimnames=list(r, r)) for(i in 1:n){ for(j