similar to: Predictions from a Cox model - understanding centering of binary/categorical variables

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

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 make sure library(Design)

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

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)

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

Hello, I am having some trouble with the 'censoring-adjusted C-index' by Uno et al, in the package survAUC. The relevant function is UnoC. The question has to do with what happens when I specify a time point t for the upper limit of the time range under consideration (we want to avoid using the right-end tail of the KM curve). Copying from the example in the help file: TR <-

I am wondering how to estimate the survival curve for a particular case(s) given a coxph model using this example code: #fit a cox proportional hazards model and plot the #predicted survival curve fit <- coxph( Surv(futime,fustat)~resid.ds+strata(rx)+ecog.ps+age,data=ovarian[1:23,]) z <- survfit(fit,newdata=ovarian[24:26,],individual=F) zs <- z$surv zt <-

Hello, I have questions regarding penalized Cox regression using survival package (functions coxph() and ridge()). I am using R 2.8.0 on Ubuntu Linux and survival package version 2.35-4. Question 1. Consider the following example from help(ridge): > fit1 <- coxph(Surv(futime, fustat) ~ rx + ridge(age, ecog.ps, theta=1), ovarian) As I understand, this builds a model in which `rx' is

I am confused when trying the function survfit. my question is: what does the survival curve given by plot.survfit mean? is it the survival curve with different covariates at different points? or just the baseline survival curve? for example, I run the following code and get the survival curve #### library(survival) fit<-coxph(Surv(futime,fustat)~resid.ds+rx+ecog.ps,data=ovarian)

I would like to be able to construct hazard rates (or unconditional death prob) for many subjects from a given survfit. This will involve adjusting the ( n.event/n.risk) with (coxph object )$linear.predictors I must be having another silly day as I cannot reproduce the linear predictor: fit <- coxph(Surv(futime, fustat) ~ age, data = ovarian) fit$linear.predictors[1] [1] 2.612756

Hello everyone, I tried to use coxph.detail() to get the hazard function. But a warning messge always returns to me, even in the example provided by its help document: > ?coxph.detail > fit <- coxph(Surv(futime,fustat) ~ age + rx + ecog.ps, ovarian, x=TRUE) > fitd <- coxph.detail(fit) Warning message: data length [37] is not a sub-multiple or multiple of the number of rows

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

Hello, i have this dataset http://www.umass.edu/statdata/statdata/data/pharynx.txt. the variables GRADE, T_STAGE anda N_STAGE are qualitative or quantitative variables??? i only have this simple doubt...! another example: why in the dataset ovarian (library survival) the variable ecog.ps: ECOG performance status (1 is better, see reference) it is consider quantitative? Thank's for

Hello: In the example below (or for a censored data) using survfit.coxph, can anyone point me to a link or a pdf as to how the probabilities appearing in bold under "summary(pred$surv)" are calculated? Do these represent acumulative probability distribution in time (not including censored time)? Thanks very much, parmee *fit <- coxph(Surv(futime, fustat) ~ age, data = ovarian)*

[code]>require("survival") > coxph(Surv(futime,fustat)~age + strata(rx),ovarian) Call: coxph(formula = Surv(futime, fustat) ~ age + strata(rx), data = ovarian) coef exp(coef) se(coef) z p age 0.137 1.15 0.0474 2.9 0.0038 Likelihood ratio test=12.7 on 1 df, p=0.000368 n= 26, number of events= 12 > coxph(Surv(futime,fustat)~age, ovarian, subset=rx==1)