similar to: Covariance calculation gives different answer than Excel (PR#13720)

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

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 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"

This is the patch from Amos Waterland for the xenconsole to use /dev/xvc0 instead of taking over ttys. I''ve fixed a few places which needed to check for XVC mode in addition to serial mode. Also, until LANANA responds with an official minor, I''ve adjusted it to use char 250/187 (in the experimental range) as opposed to 204/187. (Should be identical to this patch from last

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),

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

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`

This is the patch from Amos Waterland for the xenconsole to use /dev/xvc0 instead of taking over ttys. I''ve fixed a few places which needed to check for XVC mode in addition to serial mode. Also, until LANANA responds with an official minor, I''ve adjusted it to use char 250/187 (in the experimental range) as opposed to 204/187 Signed-off-by: Jeremy Katz

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

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

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 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

> 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

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

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

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 was wondering if someone can help me interpret the results of running weibreg. I run the following and get the following R output. > weibreg(Surv(time, censor)~covar) fit$fail = 0 Call: weibreg(formula = Surv(time, censor)~covar) Covariate Mean Coef Rel.Risk L-R p Wald p covar 319.880 -0.002 0.998 0.000 log(scale) 0.000 8.239

In the first line, use the dist function, found in library mva, to get the distance between each pair of rows. From this calculate an incidence matrix for which element i,j is true if row i in dat equals row j in dat (and false elsewhere). In the second line, for each row calculate the indices of the matching rows and take the minimum of those as the key. incid <-

I need to play games with an expression similar to the following one: print(xyplot(DepVar ~ Group|Covar, groups=Othergroup, data=mydf, pch = 18 ,main="Testcase",auto.key = TRUE)) The problem is that the formula argument (the first argument) an the groups argument are passed over from another program as strings. The formula argument does not pose problems, I can do

Can someone help with a fairly simple task? I have a data set where I would like to insert a 0 time event between individuals: what I have: VAR DATE TIME CONC COVAR 1 NOV2 0.25 10 group1 1 NOV2 0.5 20 group1 1 NOV2 1 5 group1 1 NOV2 2 1 group1 1 NOV2 3 0.1 group1 2 NOV2 0.25 10