similar to: Schoenfeld Residuals

Hi All, This sounds a relatively simple query, and I hope it is! I am looking at a continuous variable, age. I am looking at time to 12-month remission and can calculate the HR and 95% confidence interval are follows: coxfita = coxph(Surv(rem.Remtime,rem.Rcens)~nearma$all.age,data=nearma) exp(coxfita$coefficients) exp(confint(coxfita)) However, because I am looking at age as a continuous

Dear All, I am attempting to use forward and/or backward selection to determine the best model for the variables I have. Unfortunately, because I am dealing with patients and every patient is receiving treatment I need to force the variable for treatment into the model. Is there a way to do this using R? (Additionally, the model is stratified by randomisation period). I know that SAS can be

Dear R-Help, I am using the 'mfp' package. It produces three plots (as I am using the Cox model) simultaneously which can be viewed together using the following code: fit <- mfp(Surv(rem.Remtime,rem.Rcens)~fp(age)+strata(rpa),family=cox,data=nearma,select=0.05,verbose=TRUE) par(mfrow=c(2,2)) plot(fit) They can be viewed separately but the return key must be pressed after each graph

-- begin included message ----- I am looking at a continuous variable, age. I am looking at time to 12-month remission and can calculate the HR and 95% confidence interval are follows: coxfita = coxph(Surv(rem.Remtime,rem.Rcens)~nearma$all.age,data=nearma) exp(coxfita$coefficients) exp(confint(coxfita)) However, because I am looking at age as a continuous variable I cannot draw a Kaplan-Meier

hi sorry if this has been discussed before, but I'm wondering why the scaled Schoenfeld residuals do not follow the defining formula for obtaining them from the ordinary Schoenfeld residuals, but are instead offset by the estimated parameter values. e.g. library(survival) attach(ovarian) sv<-Surv(futime,fustat) f1<-coxph(sv~age+ecog.ps) f1

Dear all, I use the Andersen plot to check for proportional hazards assumption for a factor (say x) in the Cox regression model and obtained a straight line that pass through the origin. However, the formal test done by the R-function cox.zph, which is based on the plot of Schonefeld residuals against time, indicates that proportional hazards assumption is violated. Further, a plot of the score

Hi all, I'm very confused! I've been using the same code for many weeks without any bother for various covariates. I'm now looking at another covaraite and whenever I run the code you can see below I get an error message: "Error in rep(0, nrow(data)) : invalid 'times' argument" This code works: # remove 'missing' cases from data # snearma <-

Hello, I got error message when applying the plot function to the cox.zph object to create the Schoenfeld residual plots. > plot(zph.revasFit[1]) Error in plot.window(xlim, ylim, log, asp, ...) : need finite 'ylim' values In addition: Warning messages: 1: NaNs produced in: sqrt(x$var[i, i] * seval) 2: no non-missing arguments to min; returning Inf in: min(x) 3: no

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

Dear R-help, I am using R 2.14.1 on Windows 7. I would like to produce a plot like the attached - although simplified to actual vs. Predicted survival time with distinguishing marks for censored and observed points. I have a dataset and have fitted a Cox model to it. In an attempt to visualise how accurate the model is it would be ideal if I could plot the actual survival times against the

To enhance my understanding, and that of my students, I have a question about cox.zph in the survival package. If I have correctly gleaned the high-level point from the 1994 Biometrika paper of Grambsch and Therneau, it looks to me like cox.zph provides a mechanism to test for a simple trend in plots of a function of time, g(t) versus the scaled schoenfeld residuals and it also provides some

Dear all, I am struggling with calculation of Schoenfeld residuals of my Cox Ph models. Based on the formula as attached, I calculated the Schoenfeld residuals for both non tied and tied data, respectively. And then I validated my results with R using the same data sets. However, I found that my results for non-tied data was ok but the results for tied data were different from R's. How

Hi, I have a coxph model like coxph(Surv(start, stop, censor) ~ x + y, mydata) I would like to calculate the Schoenfeld residuals for the null, i.e the same model where the beta hat vector (in practical terms, the coeff vector spat out by summary()) is constrained to be all 0s --all lese stays the same. I could calculate it by hand, but I was wondering if there is a way of doing it with

Dear all, I am using Windows and R 2.9.2 for my analyses. I have a large dataset and I am particularly interested in looking at time to an event for a continuous variable. I would like to produce a plot of log(relative risk) or relative risk (also known as hazard ratio) against the continuous variable. I have spent a long time looking for advice on how to do this but my search has proved

Dear All, I have analysed time to event data for continuous variables by considering the multivariable fractional polynomial (MFP) model and comparing this to the untransformed and log transformed model to determine which transformation, if any, is best. This was possible as the Cox model was the underlying model. However, I am now at the situation where the assumption that the competing risks

Hi everybody, this is a real dummy thing. I sorted a matrix based on a given column, and what I get is right, until it comes to columns of negative and positive values; than, "order" orders everything from max to min in the negative values, and then AGAIN from max to min in the positive values!!! Why isn't everything order from max to min, and that's it? Thank you!!! Attached

Hi I am trying to understand how to get the validate() function in Design to work with the subset option. I tried this: ovarian.cph=cph(Surv(futime, fustat) ~ age+factor(ecog.ps)+strat(rx), time.inc=1000, x=T, y=T, data=ovarian) validate(ovarian.cph) #fine when no subset is used, but the following two don't work: > validate(ovarian.cph, subset=ovarian$ecog.ps==2) Error in

Hi, I''ve used old ZPH patch under squid 2.4 Stable4 and it works great ! Now I want to patch squid 2.4 stable 5, with new patch, on http://www.it-academy.bg/zph/ I''ve patched and installed squid 2.5 stable 5 succefully, but I can''t get ZPH works. I''m trying with ... $TC class add dev $LANDEV parent 1: classid 1:7 htb rate 1Mbit $TC filter add dev $LANDEV

Dear r-help-list: If I use the rpart method like cfit<-rpart(y~.,data=data,...), what kind of tree is stored in cfit? Is it right that this tree is not pruned at all, that it is the full tree? If so, it's up to me to choose a subtree by using the printcp method. In the technical report from Atkinson and Therneau "An Introduction to recursive partitioning using the rpart

You report an error message: > plot(zph.revasFit[1]) Error in plot.window(xlim, ylim, log, asp, ...) : need finite 'ylim' values I have never seen this error before, and I cannot guess what causes it. You need to provide more information, and likely a small data set that produces the problem. Perhaps you have an x variable that is a constant? Terry Therneau