similar to: ANCOVA in R, single CoVar, two Variables

Hi, Using package gclus in R, I have created some graphs that show the trends within subgroups of data and correlations among 9 variables (v1-v9). Being interested for more details on these data I have produced also the var-covar matrices. Question: From a pair of two subsets of data (with 9 variables each, I have two var-covar matrices for each subgroup, that differ for a treatment on one

> Date: Mon, 20 Feb 2006 16:43:55 -0600 > From: Aldi Kraja <aldi at wustl.edu> > > Hi, > Using package gclus in R, I have created some graphs that show the > trends within subgroups of data and correlations among 9 variables (v1-v9). > Being interested for more details on these data I have produced also the > var-covar matrices. > Question: From a pair of two

Hey I am trying to run /coeftest()/ using a heteroskedasticity-consistent estimation of the covariance matrix and i get this error: # packages >library(lmtest) >library(sandwich) #test > coeftest(*GSm_inc.pool*, vcov = vcovHC(*GSm_inc.pool*, method="arellano", > type="HC3")) /Fehler in 1 - diaghat : nicht-numerisches Argument f?r bin?ren Operator/ something like:

Is there an R function for computing a variance-covariance matrix that guarantees that it will have no negative eigenvalues? In my case, there is a *lot* of missing data, especially for a subset of variables. I think my tactic will be to compute cor(x, use="pairwise.complete.obs") and then pre- and post-multiply by a diagonal matrix of standard deviations that were computed based

Hi all, I want to construct relatedness among individuals and have a look at the following script. ######################### rm(list=ls()) N=5 id = c(1:N) dad = c(0,0,0,3,3) mom = c(0,0,2,1,1) sex = c(2,2,1,2,2) # 1= M and 2=F A=diag(nrow = N) for(i in 1:N) { for(j in i:N) { ss = dad[j] dd = mom[j] sx = sex[j] if( ss > 0

Dear all, I'm having a few problems trying to reshape a data frame. I tried with reshape{stats} and melt{reshape} but I was missing something. Any help is very welcome. Please find details below: ################################# # data in its original shape: indiv <- rep(c("A","B"),c(10,10)) level.1 <- rpois(20, lambda=3) covar.1 <- rlnorm(20, 3, 1) level.2

Hello! I have something like this: test1 <- data.frame(intx=c(4,3,1,1,2,2,3), status=c(1,1,1,0,1,1,0), x1=c(0,2,1,1,1,0,0), x2=c(1,1,0,0,2,2,0), sex=c(0,0,0,0,1,1,1)) and I can easily fit a cox model: library(survival) coxph(Surv(intx,status) ~ x1 + x2 + strata(sex),test1) However, I want to

Dear R users, I am trying to make this (3 by 10) matrix A --A---------------------------------------------------- 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0.5 0.5 0 0 0 0 0 0 0 ------------------------------------------------------- from "mass.func" --mass.func------------------------------------------- > mass.func $`00` prop 5 1 $`10`

Dear R users: I'm new to R and am trying to fit a mixed model Cox regression model with coxme function. I have one two-level factor (treat) and one covariate (covar) and 32 different groups (centers). I'd like to fit a random coefficients model, with treat and covar as fixed factors and a random intercept, random treat effect and random covar slope per center. I haver a couple of

I am new to R. I just tried to recreate in R (using sem package and the identical input data) a solution for a simple measurment model I have found before in LISREL. LISREL had no problems and converged in just 3 iterations. In sem, I got no solution, just the warning message: "Could not compute QR decomposition of Hessian. Optimization probably did not converge. in: sem.default(ram =

Dear All, This is not strictly R related (though I would implement the solution in R; besides, being this list so helpful for these kinds of stats questions...). I got a "strange" request from a colleage. He has a bunch (approx. 25000) subjects that belong to one of 12 possible classes. In addition, there are 8 covariates (factors) that can take as values either "absence"

Dear helpers, I wrote a code which estimates a multi-equation model with generalized least squares (GLS). I can use GLS because I know the covariance matrix of the residuals a priori. However, it is a bit slow and I wonder if anybody would be able to point out a way to make it faster (it is part of a bigger code and needs to run several times). Any suggestion would be greatly appreciated. Carlo

Hello, I finally run my measurement model in sem - successfully. Now, I am trying to print out the path diagram that is based on the results - but for some reason it's not working. Below is my script - but the problem is probably in my very last line: # ANALYSIS OF ANXIETY, DEPRESSION, AND FEAR - LISREL P.31 library(sem) # Creating the ANXIETY, DEPRESSION, AND FEAR intercorrelation matrix

Newsgroup members, I appreciate the help on this topic. David Duffy provided a solution (below) that was quite helpful, and came close to what I needed. It did a great job creating two vectors of dichotomous variables with a known correlation (what I referred to as a phi-coefficient). My situation is a bit more complicated and I'm not sure it is easily solved. The problem is that I must

Dear expeRts, I have a data.frame with certain covariate combinations ('group' and 'year') and corresponding values: set.seed(1) x <- data.frame(group = c(rep("A", 4), rep("B", 3)), year = c(2001, 2003, 2004, 2005, 2003, 2004, 2005), value = rexp(7)) My goal is essentially to

Hi, ? I am very new to R. So, pardon my dumb question. I was trying to write my own function to run a different model (perform an ordered logistic regression) using the example in website http://pngu.mgh.harvard.edu/~purcell/plink/rfunc.shtml But R returns a error `R Error in eval(expr, envir, enclos) : object 's' not found' when I run it. What am I doing wrong here? Here's

Hi I'm running QDA on some data and calculating the discriminant function. qda.res <- qda(type ~ npreg + glu + bp + skin + bmi + ped + age) ind_yes <- c(1:N)[type == "Yes"] > ind_no <- c(1:N)[type == "No"] > cov_yes <- cov(table[ind_yes, 1:7] ) > cov_no <- cov(table[ind_no, 1:7] ) > covar<-list(cov_no, cov_yes) qdf<- function(x,

Hola a todos, Soy nuevo en R y estoy intentando modelizar una serie de datos no estacionarios usand la distribucion Generalizada de Valores Extremos GEV. ¿Podriais indicarme como se modeliza una tendencia polinómica (cuadrática, por ejemplo) en alguno de los 3 parámetros (situación, escala o forma)? He encontrado documentación a cerca de modelización linear o exponencial, pero no acabo de

Hi all, I cant' wrap my head around an error from the coxph function (package survival). Here's an example: library(survival) n = 100; set.seed(1); time = rexp(n); event = sample(c(0,1), n, replace = TRUE) covar = data.frame(z = rnorm(n)); model = coxph(Surv(time, event)~ . , data = covar) R gives the following error: > model = coxph(Surv(time, event)~ . , data = covar) Error in

Dear R users. I'm trying to fit a generalised linear spatial mode using the geoRglm package. To do so, I'm preparing my data (geodata) as follow: geoData9093 = as.geodata(data9093, coords.col= 17:18, data.col=15,* covar.col=16*) where covar.col is a factor variable (years in this case 90-91-92-93)). Then I run the model as follow: / model.5 = list(cov.pars=c(1,1),