similar to: Defining factors in GLM

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

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.

Dear all, I would like to Boostrap the stepAIC() procedure from package MASS for variety of model objects, i.e., fn <- function(object, data, B = 2){ n <- nrow(data) res <- vector(mode = "list", length = B) index <- sample(n, n * B, replace = TRUE) dim(index) <- c(n, B) for (i in 1:B) { up.obj <- update(object, data = data[index[, i], ])

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

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]]

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)

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

Hi. Can anyone point me to any module in R which implements "Little's Chi Square test" for MCAR. The problem is that i have around 60 behavioural variables on a 6 point categorical scale which i need to test for MCAR and MAR. What i can make out from preliminary analysis is that moderate (0.30 to 0.60) correlations may be present in several variable pairs leading me to suspect

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]]

Hi, I am an R beginner and having problems structuring my REML models. I have a model with y=weight x1=time x2=timesquared id=individual identity I need to structure the model such that in the random effects there is a constant intercept for all individuals but a separate individual slope for both x1 and x2 (a coefficient score for every individual).

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

Hello, I try to define a global variable. My example: R> a <- "old" R> test <- function () { a <- "new" } R> test() R> a # shoud be "new" This doesn't work. I would like to modify the variable "a" in a procedure. How can I do that. Thank you for helping. Sven Kn侟ppel (Germany-Berlin)

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.

R-help, Sorry for posting the same question (almost) once again. I'm using a lm model for some data. The code is below: fitData <- glm(pm ~ age + mw , family = binomial(link = logit), maxit = 30, data = inputModel1) "age" is a factor and "mw" is a continous variable. The summary function does not provide the parameter estimates for the first level of the

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]

Dear list, I am fitting a logistic multi-level regression model and need to test the difference between the ordinary logistic regression from a glm() fit and the mixed effects fit from glmer(), basically I want to do a likelihood ratio test between the two fits. The data are like this: My outcome is a (1,0) for health status, I have several (1,0) dummy variables RURAL, SMOKE, DRINK, EMPLOYED,

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

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]]