similar to: simple index question

Thanks Dimitris!!! That's much clearer now. Still have a lot of work to do this weekend to understand every bit but your code will prove very useful. Cheers, Aziz -----Original Message----- From: Dimitrios Rizopoulos [mailto:Dimitris.Rizopoulos at med.kuleuven.be] Sent: May 12, 2006 4:35 PM To: Chaouch, Aziz Subject: RE: [R] Maximum likelihood estimate of bivariate

Hi all, My question is not really urgent. I can write a loop and solve the problem. But I know that I'll be in a similar situation many more times so it would be useful to find out the answer Is there a fast way to perform linear fit to all the columns of a matrix? (or in the one dimension of a multi-dimensional array.) I'm talking about many single linear fits, not about a multiple fit.

Hi all. I have a numerical function f(x), with x being a vector of generic size (say k=4), and I wanna take the numerically computed gradient, using deriv or numericDeriv (or something else). My difficulties here are that in deriv and numericDeric the function is passed as an expression, and one have to pass the list of variables involved as a char vector... So, it's a pure R programming

Thanks a lot to Demitris for a prompt answer some minutes ago on another tread (see below). To avoid excess mails on the list, I move onto next question: I have another small plotting problem that confuses me. I want to plot results from a field trial series, using the numbers of the trials as symbols in the plot. pch = as.character(trial_no) works fine, but truncates the trial number to the

Dear R-devels, I'd like to create a plot method for a class of objects that passes the '...' argument to both plot() and legend(), e.g., x <- list(data = rnorm(1000)) class(x) <- "foo" plot.foo <- function(x, legend = FALSE, cx = "topright", cy = NULL, ...){ dx <- sort(x$data) plot(dx, dnorm(dx), type = "l", ...) if (legend)

Hi R-users, How do I calculate a number of NA's in a row of every second column in my data frame? As a starting point: dfr <- data.frame(sapply(x, function(x) sample(0:x, 6, replace = TRUE))) dfr[dfr==0] <- NA So, I would like to count the number of NA in row one, two, three etc. of columns X1, X3, X5 etc. Thanks in advance Lauri [[alternative HTML version deleted]]

I have a vector that has 1,974 elements and each element is one of the following (B, F, N, Y). How do I recreate that vector accept in the place of N put 0 and in the place of B, F or Y put a 1? Thanks, Jacob [[alternative HTML version deleted]]

Dear R-users, Is there a way to get p-values for a one-sided hypothesis test about a poisson mean? Thanks, Jan Wijffels University Center for Statistics W. de Croylaan 54 3001 Heverlee Belgium tel: +32 (0)16 322784 fax: +32 (0)16 322831 <http://www.kuleuven.be/ucs> http://www.kuleuven.be/ucs Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm [[alternative HTML version

Hi Everyone, I have some data that predicts both a nominal and ordinal response variable. I was wondering what packages in R would help me analyze the data? I was also curious if anyone could recomend me some textbooks that would help with the analysis of such data? I have the 5th edition of "Applied Multivariate Statistical Analysis" by Richard A. Johnson and Dean W. Wichern

Hi everyone, I am new to R, but it's really great and helped me a lot! But now I have 2 questions. It would be great, if someone can help me: 1. I want to integrate a normal distribution, given a median and sd. The integrate function works great BUT the first argument has to be a function so I do integrate(dnorm,0,1) and it works with standard m. and sd. But I have the m and sd given.

Hello there! My operating system is Windows XP, my version of R is the latest (R-2.5.0). Recently I have downloaded the package "mvtnorm" and a problem with the command "pmvnorm" occured. Trying to enter the lines ... A <- diag(3) A[1,2] <-0.5 A[1,3] <- 0.25 A[2,3] <- 0.5 pvmnorm(lower=c(-Inf,-Inf,-Inf), upper=c(2,2,2),mean = c(0,0,0), corr=A) I got the following

Hello R-ologists, Imagine you have a list "list" like so: >list [[1]] [1] "IPI00776145.1" "IPI00776187.1" [[2]] [1] "Something" "IPI00807764.1" "IPI00807887.1" [[3]] [1] "IPI00807764.1" [[4]] [1] "Somethingelse" What I need to achieve is a filtered list "list2" like so: >list2 [[1]] [1]

I use matplot to get the density curve and then I use abline(v=g$V2, col = 3 ) to get the vertical line. Goal : I want very small lines at the bottom on the x axis , if possible in the arrow forms instead of vertical lines on the whole graph. Thanks a lot. [[alternative HTML version deleted]]

Dear All, I am producing a figure with many curves on it. How do I make the legends for all those curves smaller so that it can fit the figure itself? The commands I used for ploting are: plot(x1,y1,col=1,lty=1) lines(x2,y2,col=2,lty=2) ... legend(0.3,0.4,c("name1","name2",...),col=1:20,lty=1:20) Any tips for making the legend fit the figure will very welcome! Thanks! Best,

Dear, I am looking for the simplification of a formula to improve the calculation speed of my program. Therefore I want to simplify the following formula: H = sum{i=0..n-1 , [ sum {j=0..m-1 , sqrt ( (Ai - Bj)^2 + (Ci - Dj)^2) } ] } where: A, C = two vectors (with numerical data) of length n B, D = two vectors (with numerical data) of length m sqrt = square root Ai = element of A with index

As part of carrying out a complicated maximum likelihood estimation, I am trying to learn to program likelihoods in R. I started with a simple probit model but am unable to get the code to work. Any help or suggestions are most welcome. I give my code below: ************************************ mlogl <- function(mu, y, X) { n <- nrow(X) zeta <- X%*%mu llik <- 0 for (i in 1:n) { if

for example: > a$tic[1:10] [1] "AIR " "ABCB " "ABXA " "ACMR " "ADCT " "ADEX " [7] "ABM " "AFCE " "AG " "ATG " Can anyone please tell me how to strip the white spaces from a$tic? Thanks, Roger [[alternative HTML version deleted]]

Hi, How do I retrieve the var.test() function code? I had a similar problem once before with another function but getMethods() solved the problem then. Now I tried several combinations for var.test() without success. Regards benjamin -- Benjamin Otto Universitaetsklinikum Eppendorf Hamburg Institut fuer Klinische Chemie Martinistrasse 52 20246 Hamburg [[alternative HTML

ng, I have a matrix (x) with binary content. Each row of the matrix holds exactly one 1, and the rest of the row is zeros. The thing is that I need to 'collapse' the matrix to one column where each row holds the original column index of the 1's (y). Sometimes, the matrix is quite large, so I have a perfomance problem. x <- matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0,

Hello all Is There a fuction that return a index of a element in vector? like this semantic example: vector = c( 100, 200, 300 ) getINDEX( vector, value = 200 ) Thanks in advance for your attention. Cleber Borges