search for: ecog

Hi, I'm sure this should be simple but I can't figure it out! I want to get the median survival calculated by the survfit function and use the value rather than just be able to print it. Something like this: library(survival) data(lung) lung.byPS = survfit(Surv (time, status) ~ ph.ecog, data=lung) # lung.byPS Call: survfit(formula = Surv(time, status) ~ ph.ecog, data = lung) 1 observation deleted due to missingness n events median 0.95LCL 0.95UCL ph.ecog=0 63 37 394 348 574 ph.ecog=1 113 82 306 268 429 ph.ecog=2 50 44...

...ve 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 the predictor, whereas ridge penalty term contains variables `age' and `ph.ecog'. Could someone explain what it means to regularize on parameters which are not part of the model? Based on definition of Cox ri...

...d anyone please provide some R codes to plot the below survival data to compare two groups (0 vs 1) after 2 simulations (TRL)? need 95% prediction interval on the plot from these 2 trials. I would like to simulate 1000 trials later. Thanks a lot for your great help and consideration! yan TRL ID ECOG BASE PTR8 GROUP POP ST ind 1 1 1 1 2.2636717 0.255634126 1 1 99.4 F 3 1 2 1 24.7719223 0.756458142 0 1 8.1 T 5 1 3 0 4.7685832 0.908777937 1 0 2.6 T 7 1 4 0 2.5934492 -0.231564522 1 1 7.4 T 9 1 5 0 4.7309141 -2.455991696 0 1 0.4 T 11 1 6 1 3.5904766 -0.459694803 1 1 13.1 T 13 1 7 1 4.8592495 0.983...

...model.frame in predict.coxph, which is why I went that route, but never noted the difference till now (preparing for my course in Nashville). Could someone shed light on the rationale for non-preservation? Terry T. Simple example > library(survival) > lfit <- lm(time ~ factor(ph.ecog) + ns(age, 3), data=lung) > ltemp <- model.frame(lfit, data=lung[1:2,]) > ltemp time factor(ph.ecog) ns(age, 3).1 ns(age, 3).2 ns(age, 3).3 1 306 1 -0.1428571 0.4285714 0.7142857 2 455 0 0.0000000 0.0000000 0.0000000 > lfit$model[1:2...

...variates 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, dist='weibull',scale=1) > summary(tmp) Call: survreg(formula = Surv(futime, fustat) ~ ecog.ps + rx, data = ovarian, dist = "weibull", scale = 1) Value Std. Error z p (Intercept) 6.962 1.322 5.267 1.39e-07 ecog.ps -0.43...

...m had a long question about drawing survival curves after fitting a Weibull model, using pweibull, which I have not reproduced. It is easier to get survival curves using the predict function. Here is a simple example: > library(survival) > tfit <- survreg(Surv(time, status) ~ factor(ph.ecog), data=lung) > table(lung$ph.ecog) 0 1 2 3 <NA> 63 113 50 1 1 > tdata <- data.frame(ph.ecog=factor(0:3)) > qpred <- predict(tfit, newdata= tdata, type='quantile', p=1:99/100) > matplot(t(qpred), 99:1/100, type='l') The result of p...

Hello, is there a R package that provides a log rank trend test for survival data in >=3 treatment groups? Or are there any comparable trend tests for survival data in R? Thanks a lot Markus -- Dipl. Inf. Markus Kreuz Universitaet Leipzig Institut fuer medizinische Informatik, Statistik und Epidemiologie (IMISE) Haertelstr. 16-18 D-04107 Leipzig Tel. +49 341 97 16 276 Fax. +49 341 97 16

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, 2L, 1L, 2L, 2L, 2L, : argument lengths...

...lem is detailed here. I fit a cox model to my data and validate the Somer's Dxy using the Design package. (Because of computation time problem, i only try 10 bootstrap samples for the time being) This is the model without stratification: > library(Design) > cox1a<-cph(surv.obj~factor(ecog)+factor(grade)+factor(tumor)+factor(extra),x=T,y=T) > coef1a<-coef(cox1a) > coef1a ecog=1 ecog=2 grade=2 grade=3 tumor=2 tumor=3 extra=1 0.3578954 0.8993140 0.4834090 0.5716166 0.7600330 1.5974558 0.8112942 > validate(cox1a,dxy=T,method=...

...question is: *What common measures exists for ranking/measuring variable importance of participating variables in a CART model? And how can this be computed using R (for example, when using the rpart package)* ---end ---- Consider the following printout from rpart summary(rpart(time ~ age + ph.ecog + pat.karno, data=lung)) Node number 1: 228 observations, complexity param=0.03665178 mean=305.2325, MSE=44176.93 left son=2 (81 obs) right son=3 (147 obs) Primary splits: pat.karno < 75 to the left, improve=0.03661157, (3 missing) ph.ecog < 1.5 to the right, impr...

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 0.0463 3.17 0.0015 rx -0.815 0.443 0.6342 -1.28 0.2000 ecog.ps 0.103 1.109...

...og-likelihood for the final estimates of coefficients. I would expect to get the penalized log-likelihood. I would like to check if this is as expected. If not, I am prepared to give more details. Cheers,DK. For example, if we have a model > fit <- coxph(Surv(futime, fustat) ~ ridge(rx, age, ecog.ps, theta=1),data=ovarian) > fit$loglik [1] -34.98494 -27.17558 > fit Call: coxph(formula = Surv(futime, fustat) ~ ridge(rx, age, ecog.ps, theta = 1), data = ovarian) coef se(coef) se2 Chisq DF p ridge(rx) -0.780 0.5862 0.5589 1.77 1 0.1800 ridge(age)...

...put below: 106, 2, 3). The variable names represented in my function (ee) are shown below, but none of those variables correspond to the column of events as shown in the output. ----------------------------- The easiest is to use the summary function: fit <- survfit(Surv(time, status) ~ ph.ecog, data=lung) temp <- summmary(fit) temp$table records n.max n.start events median 0.95LCL 0.95UCL ph.ecog=0 63 63 63 37 394 348 574 ph.ecog=1 113 113 113 82 306 268 429 ph.ecog=2 50 50 50 44 199 156...

...olution to this problem? Sincerely, Jerome Asselin library(survival) data(ovarian) newovarian <- ovarian newovarian$lower59 <- newovarian$futime-59 newovarian$time59 <- Surv(newovarian$lower59,newovarian$futime, event=rep(3,nrow(newovarian)),type="interval") survreg(time59~ecog.ps+rx,data=newovarian,dist="lognormal") #THIS DOES NOT WORK BECAUSE ONE OF THE LOWER BOUNDS IS ZERO #Error in survreg(time59 ~ ecog.ps + rx, data = newovarian, # dist = "lognormal") : # Invalid survival times for this distribution -- Jerome Asselin (J?r?me) Statistica...

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 all..!! [[alternative HTML version deleted]]

...core test is not calculated for ridge coxph(), i.e score nor rscore components are not included in the coxph object when ridge is specified. Please find the code below. I use 2.9.2 R with 2.35-4 version of the survival package. Thanks DK. > fit <- coxph(Surv(futime, fustat) ~ ridge(rx, age, ecog.ps, theta=1),data=ovarian) > a<-summary(fit) Call: coxph(formula = Surv(futime, fustat) ~ ridge(rx, age, ecog.ps, theta = 1), data = ovarian) n= 26 coef se(coef) se2 Chisq DF p ridge(rx) -0.780 0.5862 0.5589 1.77 1 0.1800 ridge(age) 0.123 0.038...

...#39;s fine with the cox model, or if the degrees of freedom are specified directly by the user), and I was wondering if there is some reason for this? Simple example: library(survival) attach(cancer) # cox model with specified degs of freedom - works fine fit1 <- coxph(Surv(time, status) ~ ph.ecog + pspline(age,3), cancer) # cox model with aic - works fine fit1 <- coxph(Surv(time, status) ~ ph.ecog + pspline(age,method="aic"), cancer) # weibull model with specified degs of freedom - works fine fit1 <- survreg(Surv(time, status) ~ ph.ecog + pspline(age,3), cancer, dist="w...

...iven 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="lp")) Thank you. Regards, Axel. [[alternative HTML version deleted]]

...' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. > library(coxme) Loading required package: survival Loading required package: splines Loading required package: bdsmatrix > fit <- coxme(Surv(time, status) ~ age + (1|ph.ecog), lung) > ranef(fit) $ph.ecog Intercept 0.1592346 > library(lme4) Loading required package: Matrix Loading required package: lattice Attaching package: 'lme4' The following object(s) are masked from package:coxme : fixef, ranef > ranef(fit) Error...

...at 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) plot(survfit(fit,type="breslow")) summary(survfit(fit,type="breslow")) #### for the first two failure points, we have s(59|x1)=0.971, s(115|x2)=0.942 how can we guarantee that s(59|x1) is always greater than s(115|x2)? since s(59|x1)=s_0(59)^exp(\beta'x1) a...