similar to: Function call error in cph/survest (package Design)

Given a cox model: library(Hmisc); 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

Hi, I am trying to run a conditional logistic model on a nested case-control study using cph() and then estimate survival based on the model. The data came from Prof Bryan Langholz website where he also has the SAS code to this, so I am trying to replicate the SAS results. The data attached. Basically, the variables are: rstime: risk set age rsentry: fake entry time, just before rstime setno:

Dear Prof 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

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 example in "help cph". The 95% confidence intervals

Dear R-helpers, I am working with 'cph' models from 'rms' library. When I build simple survival models, based on 'Surv(time, event)', everything is fine and I can make nice plots using plot(Predict(f, time=3)). However, recently I tried to be more specific and used 'Surv(start, stop, event)' type model. Using this model 'plot(Predict(f))' works OK, but

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, 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

Hi, I've lost my mind on it... I have to scatterplot two vectors, grouped by a third variable, with two different dimensions according to whether each cell line in the plot is sensitive or resistant to a given drug, and with a different color for each of 9 tissues of origin. Here's what I've done:

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

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,

Dear all, I am using R 3.4.3 on Windows 10. I am preparing some teaching materials and I'm having trouble matching the by-hand version with the R code. I have fitted a Cox model - let's use the ovarian data as an example: library(survival) data(ovarian) ova_mod <- coxph(Surv(futime,fustat)~age+rx,data=ovarian) If I want to make predict survival for a new set of individuals at 100

Hi all, Using: R version 2.8.1 Patched (2009-03-07 r48068) on OSX (10.5.6) with survival version: Version: 2.35-3 Date: 2009-02-10 I get the following using the first example in ?summary.survfit: > summary( survfit( Surv(futime, fustat)~1, data=ovarian)) Call: survfit(formula = Surv(futime, fustat) ~ 1, data = ovarian) time n.risk n.event survival

Dear R users, I am trying to use "muhaz" and "plot.muhaz" functions in "muhaz" package to estimate and plot hazard funciton. However function "muhaz" always gives error message "Error in Surv(times, delta) : object "times" not found". I could not even run their sample codes in the user's manual as follows: data(ovarian)

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 off, as it

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

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

Hi, Is it normal to get intercept in the list of covariates in the output of survreg function with standard error, z, p.value etc? Does it mean that intercept was fitted with the covariates? Does Value column represent coefficients or some thing else? Regards, ------------------------------------------------- tmp = survreg(Surv(futime, fustat) ~ ecog.ps + rx, ovarian,

Dear all, I am doing library(survival) fit <- coxph(Surv(futime,fustat) ~ rx, ovarian) plot(survfit(fit,newdata=ovarian),col=c(1,2)) legend("bottomleft", legend=c("rx = 0", "rx = 1"), lty=c(1,2),col=c(1,2)) Is this correct to compare these two groups? Is the 0.31 the p-value that the median f two groups are equal Why lty does not work here? Many thanks

Dear All: I have some questions about the output of coxph. Below is the input and output: ---------------------------------------- > coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = + ovarian, x = TRUE) Call: coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = ovarian, x = TRUE) coef exp(coef) se(coef) z p age 0.147 1.158

I am trying to fit a lognormal distribution on interval-censored data. Some of my intervals have a lower bound of zero. Unfortunately, it seems like survreg() cannot deal with lower bounds of zero, despite the fact that plnorm(0)==0 and pnorm(-Inf)==0 are well defined. Below is a short example to reproduce the problem. Does anyone know why survreg() must behave that way? Is there an alternate