similar to: Variable selection in Cox proportional hazards model?

Hello, I ran my simulations on the Unix verson of R1.9.1 and the Windows version of R2.0.1 on XP. I kept getting different values for the mean of the same column of the same matrix, and am perplexed. I would appreciate if anyone could help explain the difference? Here is a sample code: set.seed(7293) z1v <- rnorm(1000, mean=68, sd=13) z1v <- (z1v-mean(z1v))/sd(z1v) Using R1.9.1 on Unix,

Hello all I would like to visualize the hazard ratios of a cox proportional hazards model. I have seen some good examples in the New England Journal of Medicine: http://content.nejm.org/cgi/content/short/353/26/2747/F3 http://content.nejm.org/cgi/content/short/350/26/2654/F2 (I hope these are open access, but I am not sure) And something similar in the book 'R Graphics' by Paul

Hi useRs, Many studies of the link between red meat and colorectal cancer use Cox proportional hazards with (among other things) a gender covariate. If it is true that men eat more red meat, drink more alcohol and smoke more than women, and if it is also true that alcohol and tobacco are known risk factors then why does it make sense to "adjust" for gender? I would think that in this

Good afternoon, I am currently trying to do some work on survival analysis. - I hope to seek your advice re: 2 questions (1 general and 1 specific) (1) I'm trying to do a stratified Cox analysis and subsequently plot(survfit(object)). It seems to work for some strata, but not for others. I have tumor grade, which is a range of 1 - 4. When I divide this range of 1:4 into 2 groups, it

I know that mvrnorm from MASS (generously provided by Profs. Venables and Ripley) can be used to generate multivariable normal data that can be used in a linear regression with certain desired characteristics (e.g. a given mean for each variable as well as a given variance-covariance pattern). Is there any similar facility that can be used to generate data for (1) a logistic regression and (2) a

I would like to perform a permutation test for Cox proportional hazards regression model. I only find it for t-test and other tests (e.g. comparing two medians). Is there a way that I can perform a Cox PH model in R or SAS for the LR-test? I am doing the following B <- 1000; LRtestx <- rep(NA,B); Srv <- Surv(Time, Event); for(j in 1:B){ LRtestx[j] <-

Dear All, I am contemplating centering the covariates in my Cox model to reduce multicollinearity between the predictors and the interaction term and to render a more meaningful interpretation of the regression coefficient. Suppose I have two indicator variables, x1 and x2 which represent age categories (x1 is patients less than 16 while x2 is for patients older than 65). If I use the following

Hi. I used coxph(surv(start,end,event)~~event,data) to deal with interval censor data. Does anyone know similar samples using coxph(surv(start,end,event)~~event,data)? If you knows, can you tell me? I'll really appreciate it. Thank you very much R learner.

I am calculating cox propotional hazards models with the coxph function from the survival package. My data relates to failure of various types of endovascular interventions. I can successfully obtain the LR, Wald, and Score test p-values from the coxph.object, as well as the hazard ratio as follows: formula.obj = Surv(days, status) ~ type coxph.model = coxph(formula.obj, df) fit =

Hello, My advisor and I have been working on some survivorship analyses in R and we are hoping to get some feedback on a particular issue involving nesting. We are interested in patterns of food discovery by ant species. Our observations consist of time to discovery by an ant for three different food types, each of two different sizes. These data were collected at 6 plots located in each of

I'm getting unexpected results from the coxph function when using weights from counter-matching. For example, the following code produces a parameter estimate of -1.59 where I expect 0.63: d2 = structure(list(x = c(1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1), wt = c(5, 42, 40, 4, 43, 4, 42, 4, 44, 5, 38, 4, 39, 4, 4, 37, 40, 4, 44, 5, 45, 5, 44, 5), riskset =

Hi, I'm new to R and using it for Cox survival analysis. Thanks to this great forum I learned how to compute the HR with its confidence interval. My question would be: Is there any way to get the p-value for a hazard ratio in addition to the confidence interval? Thanks, Thierry -- Thierry Panje Visiting Student Researcher Department of Psychology Stanford

Hello Everyone,   I'm learning how to perform various statistical analyses in R. I'm checking my understanding by replicating examples from my SAS books. Below is an attempt to replicate a Cox Proportional Hazards model with a time-varying covariate. I think I'm doing this correctly but am not completely sure. I would appreciate it if someone could double-check my results. In case

Buenas. Abajo pongo la salida de un modelo de cox , dónde he estratificado por una variable de país (Countryb) y por otra (Q6). Además hay interacción entre la variable mobilityPDurG2 (es una variable 0,1, y 0 es la categoría de referencia) país. La categoría de referencia para país es "united kingdom". Mi duda surge si quiero calcular el hazard ratio para los que tienen un 1

Hello everyone, I am a new user of R. Does anybody know how hazard ratios are extracted for each factor level in a stratified Cox proportional hazard regression model? I have a cancer data set where the variable ?differentiation? is a factor with three levels: poor, intermediate and good. I would like to extract the hazard ratio for each grade level and relate it to another prognostic factor.

Hi all, Is there an equivalent to predict(...,type="linear") of a Proportional hazard model for a Cox model instead? For example, the Figure 13.12 in MASS (p384) is produced by: (aids.ps <- survreg(Surv(survtime + 0.9, status) ~ state + T.categ + pspline(age, df=6), data = Aidsp)) zz <- predict(aids.ps, data.frame(state = factor(rep("NSW", 83), levels =

Dear List, Is the test for proportional hazards valid when the model contains a cluster variable? The output looks strange with the cluster variable. My intervals are based on calendar time and the clustering variable is related to the season the event occurs in. model1<-coxph(Surv(Start,Stop,Event)~LagAOO+I(LagAOO^2)+ FirstSeen+TSLE+strata(CumPOO.rc)+cluster(quarter),data=data8, x=T)

Hi I'm trying to interpret the output from the coxph command. One of the outputs is: Rsquare= 0.01 (max possible= 0.069 ) Given that the methods for calculating R squared and its validity with proportional hazards regression are debated in the literature - can anyone tell me exactly how this part of the coxph output is calculated and how I derive my R-squared from it? Thanks Helen

SImplify your call to lm using the "." argument instead of manipulating formulas. > strt <- lm(y1 ~ ., data = dat) and you do not need to explicitly specify the "1+" on the rhs for lm, so > frm2<-as.formula(paste(trg," ~ ", paste(xvars,collapse = "+"))) works fine, too. Anyway, doing this gives (but see end of output)" bst <-

The error is "the model fit failed in 50 bootstrap samples Error: non-character argument" Cheers, SOH. On 22/08/2017 17:52, Bert Gunter wrote: > Failed? What was the error message? > > Cheers, > > Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka