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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(

I have been struggling with this problem for a while and I hope someone could help me. Or if someone could point me to a section in the manual I would be grateful. x <- "my" plot(1:10, main=expression(paste( x, Delta, "values" ))) Q : How do I get the title to say "my (triangle symbol) values" ? The following trial-and-error produced mainly errors :

Dear all, I am having trouble trying to install R-1.8.1 on a Sun Solaris 8 (Generic_108528-23 version) machine. The configuration was successful but make check fails. I traced the the problem to the plot() function. > 1 + 1 [1] 2 > capabilities() jpeg png tcltk X11 GNOME libz http/ftp sockets TRUE FALSE TRUE TRUE FALSE TRUE TRUE TRUE

Could some kindly tell me if I am supposed to be getting the same test statistic value with var.equal=TRUE and var.equal=FALSE in t.test ? set.seed(1066) x1 <- rnorm(50) x2 <- rnorm(50) t.test(x1, x2, var.equal=FALSE)$statistic # 0.5989774 t.test(x1, x2, var.equal=TRUE)$statistic # 0.5989774 ??? Here are my own calculations that shows that perhaps the result when var.equal=TRUE is

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

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, If the question below as been answered before I apologize for the posting. I would like to get the frequencies of occurrence of all values in a given variable in a multivariate dataset. In short for each variable (or field) a summary of values contained with in a value:frequency pair, there can be many such pairs for a given variable. I would like to do the same for several such variables.

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

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

There was a BioConductor thread today where the poster wanted to find pairwise difference between columns of a matrix. I suggested the slow solution below, hoping that someone might suggest a faster and/or more elegant solution, but no other response. I tried unsuccessfully with the apply() family. Searching the mailing list was not very fruitful either. The closest I got to was a cryptic chunk

Hi, I have a matrix with column names starting with a character in [0-9]. After some matrix operations (e.g. copy to another matrix), R seems to add a character 'X' in front of the column name. Is this a normal default behaviour of R? Why has it got this behaviour? Can it be changed? What would be the side effect? Thank you. Regards, Gilbert [[alternative HTML version deleted]]

You are using a method that needs to estimate the covariance matrix of all the variables. If you have 80 variables, there are (80+1)*80/2 = 3240 variances and covariances to estimate. How many data points do you think you need to do that? Some people assume the covariance matrix is diagonal (i.e., assuming all the variables are uncorrelated). Even then you still have 80 variances to estimate.

Dear R-users, I state that this is for beginners, so you may ignore this in order not to be irritated. By the way, "patience" is another important thing, together with "kindness", we should keep in mind when we teach students and our own children as Jim Lemon pointed out well in the context of the Socratic method. You may know that being kind does not mean giving

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

Hi all! I 'm beginner and i develop a bio-application with VB and i need some statistic functions! could i calculate StdError, CoeffOfVariance, SumSquared with R langage? if yes, what are functions to use? I need also to use ANOVA and t-test... Thanks for your help! Laurent Houdusse Analyste Programmeur

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

Dear all, I have a matrix X with 47 lines and say 500 columns - values are in {0,1}. I'd like to compare lines. For that, I first did: for (i in 1:(dim(X)[1]-1)) for (j in (i+1):dim(X)[1]) { Y <- X[i,]+Y[j,] etc. but, since it takes a long time, I would prefer avoding loops; for that, my first idea was to add this matrix: X1=X[,rep(1:46,46:1)] to this one: res=NULL for (i in

Hi all, My data genes [,1] [,2] [,3] [,4] [1,] 25 72 23 55 [2,] 34 53 41 33 [3,] 26 43 26 44 [4,] 36 64 64 22 [5,] 47 72 67 34 stu<-t.test(genes[,1:2],genes[,3:4]) > stu$p.value [1] 0.4198002 i get 1 pvalue for the entire col1:col2 Vs col3:col4. I am trying to get 5 p values for the 5 rows i have. I am trying to avoid a for loop coz my

Hi all, for example: > X<- rnorm(1000) > X0 <- 0.899 I want to draw a density distribution plot with a arrow pointer indicating the position of X0, meanwhile, giving out the p-value. any functions? Thanks very much.

Take a look at the manual page for "layout". The final example does pretty much what you're asking about (with a scatterplot and histograms, but the idea should be the same). Hope this helps, Matthew Wiener Applied Computer Science & Mathematics Dept. Merck Research Labs Rahway, NJ 07090 732-594-5303 -----Original Message----- From: Adaikalavan Ramasamy [mailto:ramasamy at