search for: cph

Hi all, I am trying to get bootstrap resampled estimates of covariates in a Cox model using cph (Design library). Using the following I get the error: > ddist2.abr <- datadist(data2.abr) > options(datadist='ddist2.abr') > cph1.abr <- cph(Surv(strt3.abr,loc3.abr)~cov.a.abr+cov.b.abr, data=data2.abr, x=T, y=T) > boot.cph1 <- bootcov(cph1.abr, B=100, coef.reps=TR...

Hello, I am modeling some survival data wih cph (Design). I have modeled a predictor which showed non linear effect with restricted cubic splines. I would like to retrieve the se(coef) for other, linear, predictors. This is just to make nice LateX tables automatically. I have the coefficients with coef(). How do I do that? Thanks, David B...

Dear UseR, I do not know if this a problem with me, my data or cph/survest in package design. The example below works with a standard data set, but not with my data, but I cannot locate the problem. Note that I am using an older package of survival to avoid a problem with the newly renamed function in survival meeting Design. Dieter # First, check standard exa...

...3,55,33,36,53,48,25,54,58,60,34, 47,23,34,60,39,34,22,30,41,55,64,48,34,54)) frame.bc # preparing for a simple univariate Cox regression: dd.bc <- datadist(frame.bc[, c('bc1','age')], adjto.cat='first') options(datadist = 'dd.bc') # a univariate Cox regression: cph.bc <- cph(formula = Surv(time1,status1)~bc1, data = frame.bc, x=TRUE, y=TRUE, surv=TRUE) anova(cph.bc) cph.bc summary(cph.bc) # the validate command for the Cox model: val.cph.bc <- validate(cph.bc, B=200, dxy=TRUE , pr=TRUE) ---------------------- Output from the validate command:...

I'd appreciate some tips, I'm making a simple mistake trying to extract elements from a list of fit objects. The following command works, coef( cph.list[[1]] ) ,.... coef( cph.list[[3]] ) and so forth. These extract regression coefficients from the appropriate list element. When I try to run this inside a for loop, or an lapply (lapply(cph.list,coef)) I get this error message > for (ii in (1:22)){ + cat(ii,fill=T) + print(...

Hi, I'm trying to print the p-values from the output of a CPH test onto a Kaplan Meier plot. Can this be done? I only really want the p-values from the CPH test to appear but if this can't be done I am willing to have the entire CPH output. This is what I am currently trying: (it doesn't print the CPH output) plot_KM <- function(field) { l...

Can someone help me. I am very new to R. I am fitting a Cox model using Frank Harrell's cph as I want to produce a Nomogram. This is what I have done: Srv<- Surv(time,cens) f.cox<- cph(Srv~ v1+v2+v3+v4, x=T, y=T, surv=T) As soon as I press enter, Windows XP crashes. If I remove surv=T, then it works. I have R version 2.9.0. Is there a way of displaying the parameter estimates (ie...

Hi there, I encountered a weird problem using cph() with Design package: I have 2 datasets, say "dat1" and "dat2", both data frames with 3 columns "time","status" and "scores", all numeric If I run the following: dd<-datadist(dat1) options(datadist='dd') dd time statu...

Hello, I am using Dr. Harrell's rms package to make a nomogram. I was able to make a beautiful one. However, I want to change 5-year survival probability to 5-year failure probability. I couldn?t get hazard rate from Hazard(f1) because I used cph for the model. Here is my code: library(rms) f1 <- cph(Surv(retime,dfs) ~ age+her2+t_stage+n_stage+er+grade+cytcyt+Cyt_PCDK2 , data=data11, surv=T, x=T, y=T, time.inc=5) surv<- Survival(f1) haz<- Hazard(f1) Here is the Error in UseMethod("Hazard") : no applicable method fo...

Hi, in Design package, a plot of survival probability vs. a covariate can be generated by survplot() on a cph object using the folliowing code: n <- 1000 set.seed(731) age <- 50 + 12*rnorm(n) label(age) <- "Age" sex <- factor(sample(c('male','female'), n, TRUE)) cens <- 15*runif(n) h <- .02*exp(.04*(age-50)+.8*(sex=='Female')) dt <- -log(runif(n))/h l...

Hello, I am constructing a nomogram using cph and nomogram commands in Dr. Harrell's Design/RMS package. The HR that I obtain for dichotomous and categorical variables are identical to those that I obtain using STATA stcox. However, the inter-quartile HR I obtain for continuous variables is obviously different, since STATA gives me HR for...

Hello, I am using Dr. Harrell's design package to make a nomogram. I was able to make a beautiful one. However, I want to change 5-year survival probability to 5-year failure probability. I couldn?t get hazard rate from Hazard(f1) because I used cph for the model. Here is my code: f1 <- cph(Surv(retime,dfs) ~ age+her2+t_stage+n_stage+er+grade+cytcyt+Cyt_PCDK2 , data=data11, surv=T, x=T, y=T, time.inc=5) surv<- Survival(f1) haz<- Hazard(f1) Here is the Error in UseMethod("Hazard") : no applicable method for 'Haza...

...",survname="fail tm12", censname="failcens") Warning message: Loglik converged before variable 2 ; beta may be infinite. in: fitter(X, Y, strats, offset, init, control, weights = weights, > ## warnings are just warnings when options()$warn=0: > lapply(qq,class) $cph.x1x2 [1] "coxph" $cph.x1x2.int [1] "coxph" > qq <- FitModels(vnames=c("cd3ncd56p","adrpcd8p"),dframe="ALLmth12",survname="failt m12", censname="failcens") Warning message: Loglik converged before variable 1 ; beta may be...

Hi. I am using Windows version of R 1.8.1. Being somewhat new to survival analysis, I am trying to compare cph (Design) with coxph (survival) for use with a survival data set. I was wondering why cph and coxph provide me with different confidence intervals for the hazard ratios for one of the variables. I was wondering if I am doing something wrong? Or if the two functions are calculating hazard ratios and...

...e a model based on Surv(time,response) object, so there is a single row per subject and no start,stop and no switching of strata. The newdata has many subjects and each subject has a strata and the survival based on the subject risk and the subject strata is needed. If I do newpred <- survfit(cph.approve,new=newapp,se=F) I get all strata for every subject. Attempting > newpred <- survfit(cph.approve,new=newapp,id=CertId,se=F) Error in survfit.coxph(cph.approve, new = newapp, id = CertId, se = F) : The individual option is only valid for start-stop data > newpred <- survfi...

Could someone explain the summary(cph.object)? The example is in the help file of cph. n <- 1000 set.seed(731) age <- 50 + 12*rnorm(n) label(age) <- "Age" sex <- factor(sample(c('Male','Female'), n, rep=TRUE, prob=c(.6, .4))) cens <- 15*runif(n) h <- .02*exp(.04*(age-50)+.8*(s...

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 order(c(1L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L,...

...amp;lpg=PP1&hl=it&pg=PA147#v=onepage&q&f=false http://finzi.psych.upenn.edu/R/Rhelp02/archive/52828.html It seems that SAS is more well-suited to this purpose, as explained in the links above, but i do not have enough time to learn SAS. I think i should use something like this: cph(Surv(start,end,status)~a.factor+a.factor:start) but i'm not sure, specially after Thernau's reply to my previous post. Assuming i'm doing well, i'd prefer to use cph because of the rms bootstrap validation facilities. Unfortunately cph doesn't wanto to model an interaction te...

Hello,   does anyone have an example on how to use restricted cubic splines function rcs within survfit.cph, if cph (Cox Proportional Hazard Regression) was done with restricted cubic splines (which I made to work)? Thank you. > [[alternative HTML version deleted]]

...experimenting with the survival and rms packages for this. I'm simply trying to obtain the expected survival time (as opposed to the probability of survival at a given time t). I can't seem to find an option from the "type" argument in the predict methods from coxph{survival} or cph{rms} that will give me expected survival times. library(rms) options(na.action=na.exclude) # retain NA in predictions fit <- coxph(Surv(time, status) ~ age + ph.ecog, lung) fit2 <- cph(Surv(time, status) ~ age + ph.ecog, lung) head(predict(fit,type="lp")) head(predict(fit2,type=&q...