similar to: coxph and robust variance estimation

Hello, my problem is that I don't know how long must be an array of double while calling C from R. R-Code: > array <- c(1,1,1) > save <- .C ( "Ctest" , a = array ) C-Code: void Ctest ( double *array ) { ... array = (double*) realloc ( array , new_number * sizeof(double) ) ; ... } The length of "array" will be compute in C. At the end save$a has a length

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)

Dear R Helpers, I have noticed that when I use lmer to analyse data, the summary function gives different values for the AIC, BIC and log-likelihood compared with the anova function. Here is a sample program #make some data set.seed(1); datx=data.frame(array(runif(720),c(240,3),dimnames=list(NULL,c('x1','x2','y' )))) id=rep(1:120,2); datx=cbind(id,datx) #give x1 a

A time series text with a title that seems designed to hide its wide scope is: Forecasting with Exponential Smoothing The State Space Approach Hyndman, R.J., Koehler, A.B., Ord, J.K., Snyder, R.D. Springer 2009. This book is actually an excellent overview of time series theory, ARIMA as well as state space. It is of course, in part, a manual for the forecast and other packages in what has been

Dear R developers, I noticed a bug in the stats4 package, specifically in the confint method applied to ?mle? objects. In particular, when some ?fixed? parameters define the log likelihood, these parameters are stored within the mle object but they are not used by the ?confint" method, which retrieves their value from the global environment (whenever they still exist). Sample code: >

Hello dear R help members. I am trying to understand the anova.rq, and I am finding something which I can not explain (is it a bug?!): The example is for when we have 3 nested models. I run the anova once on the two models, and again on the three models. I expect that the p.value for the comparison of model 1 and model 2 would remain the same, whether or not I add a third model to be compared

I am trying to find methods for testing and visualizing calibration to Cox models with time-depended coefficients. I have read this nice article <http://journals.sagepub.com/doi/10.1177/0962280213497434>. In this paper, we can fit three models: fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) p <- log(predict(fit0, newdata = data1, type = "expected")) lp

Hi , I want to apply the following code fo my data with 400 predictors. I was wondering if there ia an alternative way instead of typing 400 predictors for the following code. I really appreciate your help. fit0<-lm(Y~1, data= mydata) fit.final<- lm(Y~X1+X2+X3+.....+X400, data=mydata) ??? step(fit0, scope=list(lower=fit0, upper=fit.final), data=mydata, direction="forward")

Colleagues, I have written an R function (see fully annotated code below), with which I want to process a dataframe within levels of the variable StepType. My program works, it processes the data within levels of StepType, but the usual headers that separate the output by levels of StepType are at the end of the listing rather than being used as separators, i.e. I get Regression results StepType

First, as others have said please obey the mailing list rules and turn of First, as others have said please obey the mailing list rules and turn off html, not everyone uses an html email client. Here is your code, formatted and with line numbers added. I also fixed one error: "y" should be "status". 1. fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) 2. p

windows xp R 2.8.1 I am trying to plot the -log(log(survival)) to visually test the proportional hazards assumption of a Cox regression. The plot, which should give two lines (one for each treatment) gives only one line and a warning message. I would appreciate help getting two lines, and an explanation of the warning message. My problem may the that I have very few events in one of my strata,

Here is an example, modified from the help page to use test="Cp": -------------------------------------------------------------------------------- > fit0 <- lm(sr ~ 1, data = LifeCycleSavings) > fit1 <- update(fit0, . ~ . + pop15) > fit2 <- update(fit1, . ~ . + pop75) > anova(fit0, fit1, fit2, test="Cp") Error in `[.data.frame`(table, , "Resid.

R 2.8.1 Windows XP I am trying to plot the results of a coxph using plot(survfit()). The plot should, I believe, show two lines one for survival in each of two treatment (Drug) groups, however my plot shows only one line. What am I doing wrong? My code is reproduced below, my figure is attached to this EMail message. John > #Create simple survival object >

Dear R-helpers, I want to try to extract residuals from a multi-level linear mixed effects model, to correlate with another variable. I need to know which residuals relate to which experimental units in the lme. I can show the labels that relate to the experimental units via the command ranef(fit0)$resid which gives: 604/1/0 -1.276971e-05 604/1/1 -1.078644e-03 606/1/0 -7.391706e-03 606/1/1

Folks: I'm having trouble doing a forward variable selection using step() First, I fit an initial model: fit0 <- glm ( est~1 , data=all, subset=c(n>=25) ) then I invoke step(): fit1 <- step( fit0 , scope=list(upper=est~ pcped + pchosp + pfarm ,lower=est~1)) I get the error message: Error in eval(expr, envir, enclos) : invalid 'envir' argument I looked at the

I'm trying to use mars{mda} to model functions that look fairly close to a sequence of straight line segments. Unfortunately, 'mars' seems to totally miss the obvious places for the knots in the apparent first order spline model, and I wonder if someone can suggest a better way to do this. The following example consists of a slight downward trend followed by a jump up after

On 26/12/2019 11:14 p.m., Spencer Graves wrote: > Hello, All: > > > ????? The default "simulate" method for lm and glm seems to ignore the > sampling variance of the parameter estimates;? see the trivial lm and > glm examples below.? Both these examples estimate a mean with formula = > x~1.? In both cases, the variance of the estimated mean is 1. That's how

Hello, All: ????? The default "simulate" method for lm and glm seems to ignore the sampling variance of the parameter estimates;? see the trivial lm and glm examples below.? Both these examples estimate a mean with formula = x~1.? In both cases, the variance of the estimated mean is 1. ??? ??????? * In the lm example with x0 = c(-1, 1), var(x0) = 2, and

On 2019-12-27 04:34, Duncan Murdoch wrote: > On 26/12/2019 11:14 p.m., Spencer Graves wrote: >> Hello, All: >> >> >> ? ????? The default "simulate" method for lm and glm seems to ignore the >> sampling variance of the parameter estimates;? see the trivial lm and >> glm examples below.? Both these examples estimate a mean with formula = >>

Dear Group members I was wondering whether there is any interface to use processing (www.processing.org) to visualize R analyses? If so could somebody point me to relevant ressources? If not, are there any attempts to build such an interface / package? Best Regards, Simon -- Nur bis 31.05.: GMX FreeDSL Komplettanschluss mit DSL 6.000 Flatrate und Telefonanschluss nur 17,95 Euro/mtl.!*