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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 appears (Click or hit ENTER for next page). I'd like to isolate the second plot produced (the estimated function...

Dear All, Sorry to bother you again. I have a model: coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma) and I'm trying to do a plot of Schoenfeld residuals using the code: plot(cox.zph(coxfita)) abline(h=0,lty=3) The error message I get is: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In sqrt(x$var[i, i] * seval) : NaNs pr...

...ph and cph that may enable some plots to be produced but none that specifically look at the relative risk one. In addition to the survival analysis, I have incorporated the mfp function (from package mfp). I currently use code such as, library(mfp) library(Design) coxfit1 <- coxph(Surv(rtime,rcens)~cts,data=data1) or coxfit2 <- mfp(Surv(rtime,rcens)~fp(cts),family=cox,data=data1,select=0.05,verbose=TRUE) plot(coxfit1) nor plot(coxfit2) produce the relevant relative risk vs. continuous variable that I need. Can anyone help? Thank you, Laura [[alternative HTML version deleted]]

...- 3 if(data$all.type[i]=="G" && data$all.aa[i]>=1 && data$all.tc[i]>=1) stype[i] <- 3 if(data$all.otc[i]>=1) stype[i] <- 4 if(data$all.o[i]>=1) stype[i] <- 4 } return(stype) } fita <- survdiff(Surv(rem.Remtime,rem.Rcens)~stypea(nmarma)+strata(nmrpa),data=nmarma) fita lrpvalue=1-pchisq(fita$chisq,3) xx <- cuminc(nmarma$rem.Remtime/365,nmarma$rem.Rcens,stypea(nmarma),strata=nmrpa) plot(xx,curvlab=c("Simple/Complex","SC+2gentc or 2gentc","TC or My/Ab or My/Ab+gentc","Other"...

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 variable I cannot draw a Kaplan-Meier curve. Instead I need to draw a plot of hazard against age. How do I do this? I don't think plot(nearma$all.age,coxfita$lin...

-- 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 curve. Instead I need to draw a plot of hazard against age. How do I do this? I don't think plot(nearma$all.age,coxfita$lin...

...iables, 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 Cox model, is there anyway I can centre the variables? Do I have to do it before I fit them into the model and if so, how? fit2=coxph(Surv(rem.Remtime,rem.Rcens)~x1(partial)+x2(partial),data=partial,method="breslow") Thank you, Laura

...ed by randomisation period). I know that SAS can be used to do this but my SAS coding is poor and consequently it would be easier for me to use R, especially given the fractional polynomial transformations! Currently the model is as follows (without treatment). coxfita=coxph(Surv(rem.Remtime,rem.Rcens)~sind(nearma)+fsh(nearma)+fdr(nearma)+th1(nearma)+th2(nearma)+fp(cage)+fp(fint)+fp(tsb)+strata(rpa),data=nearma) Thank you for your help, Laura

Dear R-help, I am using R 2.14.1 on Windows 7 with the 'gfcure' package (cure rate model). I have included the treatment variable in the cure part of the model as shown below: Ø ref_treat <- gfcure(Surv(rem.Remtime,rem.Rcens)~1,~1+strata(drpa)+factor(treat(delcure)),data=delcure,dist="loglogistic") >From that I can obtain the coefficients, standard errors etc as per alternative models (with covariates only fitted to the survival part of the model say). > summary(ref_treat) However, only one standard...

...l.gmail.com> Content-Type: text/plain 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 variable I cannot draw a Kaplan-Meier curve. Instead I need to draw a plot of hazard against age. How do I do this? I don't think plot(nearma$all.age,coxfita$lin...