search for: survest

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 example to...

I wonder if anyone know why survest (a function in Design package) and standard survfit.coxph (survival) returned different confidence intervals on survival probability estimation (say 5 year). I am trying to estimate the 5-year survival probability on a continuous predictor (e.g. Age in this case). Here is what I did based on an...

...Harrell and everyone, My PC: Window XP service pack 3 and service pack 2 R version 2.8.1 and 2.9 alpha For the last 3 days, after updating R, my two computers have been facing problems when running existing and runable R commands that involves with Design package I attempt to use 'survest', but I failed all the times with R (both 2.8.1 and 2.9 alpha) being shut down immediately with following error report messages. AppName: rgui.exe AppVer: 2.90.48212.0 ModName: survival.dll ModVer: 0.0.0.0 Offset: 00007749 However, if I run these commands on other computers which have...

...coef se(coef) z p cr500 1.491 0.184 8.08 6.66e-16 smoke25 0.461 0.182 2.53 1.16e-02 These results are the same as what SAS reported! Next, I want to estimate the survival for stratum age 50-69 (rstime2=2), so I used the following, this is where I got the error message: > survest(fit, expand.grid(cr500=c(0,1),smoke25=c(0,1),rstime2=2, logw=0), >times=unique(uminers$rstime[uminers$rstime2=='2']), conf.int=.95) Error in factor(rep(1:nstrat, scount), labels = names(fit$strata)) : invalid labels; length 3 should be 1 or 2 Anyone has any suggestions what went wr...

Hello, I apologize for the post as I am certainly overlooking a simple solution to my difficulties with getting a legend to print on a survplot from the rms package. I am plotting the following: survplot(survest(fita), n.risk=T, conf='none', cex.n.risk=.85, dots=T, col='gray10', lty=2) survplot(survest(fit), n.risk=F, conf='none', add=T) survplot(survest(fitb), n.risk=F, conf='none', col='gray20', add=T, lty=4) I have tried to place 'label.curves=list(method=&qu...

Dear R users, I have a long function that among other things uses the "survest" function from the Design package. This function generates the warning: In survest.cph (...) S.E. and confidence intervals are approximate except at predictor means. Use cph(...,x=T,y=T) (and don't use linear.predictors=) for better estimates. I would like to turn this specific warning...

Dear R users, My question is more methodology related rather than specific to R usage. Using time on study as time in a cox model, eg: library(Design) stanf.cph1=cph(Surv(time, status) ~ t5+id+age, data=stanford2, surv=T) #In this case the 1000-day survival probability would be: stanf.surv1=survest(stanf.cph1, times=1000) #Age in this case is a covariate. #I now want to compare the above estimate to the 1000-day survival probability I get using age at entry and exit as my time-scale: stanf.cph2=cph(Surv(age,age+time, status) ~ t5+id, data=stanford2, surv=T) stanf.surv2=survest(stanf.cph2,...

...need results='asis' * latex methods - set file default to '' * GiniMd: moved to Hmisc package * plot.nomogram: fixed bug where abbreviations were being ignored. Thanks: Zongheng Zhang * nomogram: improved examples in help file * survplot.npsurv: fixed n.risk when competing risks * survest.cph, survfit.cph, others: fixed large problems due to incompatibility with survival 2.39-5 for survival predictions; changed object Strata to strata in cph to be compatible with survival package * new test survest.r to more comprehensively check survfit.cph and survest.cph * ggplot.Predict: quit ig...

...need results='asis' * latex methods - set file default to '' * GiniMd: moved to Hmisc package * plot.nomogram: fixed bug where abbreviations were being ignored. Thanks: Zongheng Zhang * nomogram: improved examples in help file * survplot.npsurv: fixed n.risk when competing risks * survest.cph, survfit.cph, others: fixed large problems due to incompatibility with survival 2.39-5 for survival predictions; changed object Strata to strata in cph to be compatible with survival package * new test survest.r to more comprehensively check survfit.cph and survest.cph * ggplot.Predict: quit ig...

...sult of the modeling of the baseline cumulative hazard function by natural cubic splines (in contrast to using the crude hazard function itself)." ? Does R provide a package/function to do this particular method for estimating survival probability as a function of a continuous variable? Is the survest.cph() in rms package doing estimation with just the crude hazard function? ? Thanks very much! ? John

....bj, validate.cph, validate.ols * print.validate: added B argument to limit how many resamples are printed summarizing variables selected if BW=TRUE * print.calibrate, print.calibrate.default: added B argument * Added latex method for results produced by validate functions * Fixed survest.cph to convert summary.survfit std.err to log S(t) scale * Fixed val.surv by pulling surv object from survest result * Clarified in predict.lrm help file that doesn't always use the first intercept * lrm.fit, lrm: linear predictor stored in fit object now uses first intercept an...

Hi, I have fitted a cox model with time-varying covariates (counting process style) using the cph function of the Design package. Now I want to know the survival probabilities at each time point given the history of a single individual. I know the survest function, but I am not sure how to interpretet its output when using time-varying covariates. Does it just give the probabilities as if it are independent individuals or can/does it take in consideration that it is the history of a single individual? Is this even possible? An example: Individual...

....bj, validate.cph, validate.ols * print.validate: added B argument to limit how many resamples are printed summarizing variables selected if BW=TRUE * print.calibrate, print.calibrate.default: added B argument * Added latex method for results produced by validate functions * Fixed survest.cph to convert summary.survfit std.err to log S(t) scale * Fixed val.surv by pulling surv object from survest result * Clarified in predict.lrm help file that doesn't always use the first intercept * lrm.fit, lrm: linear predictor stored in fit object now uses first intercept an...

...; library(survival); (library(Design); cox.model=cph(Surv(futime, fustat) ~ age, data=ovarian, surv=T) str(cox.model) What I need is the total estimated time until failure (death), not the probability of failing at a given time (survival probability), or hazard etc, which is what I get from survest and predict for example. I suspect the answer is embarrassing simple... Eleni Rapsomaniki Research Associate Strangeways Research Laboratory Department of Public Health and Primary Care [[alternative HTML version deleted]]

Hi, Sorry to trouble the list. I have a problem which I'm not sure how to resolve. I have a Cox model with 1 independent variable with 2 categories (and thus 2 survival curves on plotting survfit) How can I get an estimate of survival for each category at a particular time point, with standard error? Looking through ?cph and ?coxph, I'm not quite sure how to go about that. I would

...text html latex example missing link(s): validate.Design latex.ols pentrace text html latex example missing link(s): nlminb plot.Design text html latex example missing link(s): key matxv survest key mgp.axis.labels win.slide gs.slide pphsm text html latex example missing link(s): print.pphsm print.cph text html latex missing link(s): print.cphfit print.coxreg print.cph.fit text h...

...used 'Surv(start, stop, event)' type model. Using this model 'plot(Predict(f))' works OK, but 'plot(Predict(f, time=3))' throws an error: "Error in summary.survfit(g, print.it = FALSE, times = times) : object 'n.enter' not found" Actually, using 'survest(f, time=3)' has the same result. Has anybody encountered this kind of error? Is there a workaround? Best regards, -- Michal J. Figurski, PhD HUP, Pathology & Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413

I am fitting a coxph model on a large dataset (approx 100,000 patients), and then trying to estimate the survival curves for several new patients based on the coxph object using survfit. When I run coxph I get the coxph object back fairly quickly however when I try to run survfit it does not come back. I am wondering if their is a more efficient way to get predicted survival curves from a coxph

...um.intercepts,ols,ols.influence,oos.loglik,pantext,Penalty.matrix,Penalty.setup,pentrace,perimeter,perlcode,plot.xmean.ordinaly,pol,pphsm,predab.resample,Predict,psm,rcs,related.predictors,reVector,robcov,Rq,sascode,scored,sensuc,setPb,show.influence,specs,strat,Surv,"[.Surv",survdiffplot,survest,Survival,survplot,univarLR,validate,val.prob,val.probg,val.surv,vif,which.influence) importFrom(Hmisc) S3method(anova, rms) S3method(latex, anova.rms, bj, cph, Glm, Gls, lrm, naprint.delete, ols, pphsm, psm, rms, Rq, summary.rms, validate) When doing R CMD INSTALL I get: [using R 2.15.3] ** prep...

...calibrate (not shown) #....move on to validation cohort of n=800 Data2 #Validation data with same predictor variables A, B, cens, time # do I need to put data2 into datadist?? s2 <- Surv(data2$time, data2$cens) #able to derive 60 month estimates of survival using data2.est5 <- survest(cph.nomo, expand.grid(A=data2$A, B=data2$B), times=c(60), conf.int=0) rcorr.cens(data2.est5$surv, s2) # tests discrimination of the model #against the validation data observed censored data # I cant find a way to use calibrate in this setting though?? # Also, if I have the 5 year estim...