similar to: some question about partial prediction in survival

Hi! I came across R just a few days ago since I was looking for a toolbox for cox-regression. I?ve read "Cox Proportional-Hazards Regression for Survival Data Appendix to An R and S-PLUS Companion to Applied Regression" from John Fox. As described therein plotting survival-functions works well (plot(survfit(model))). But I?d like to do some manipulation with the survival-functions

Hello: Using the built-in dataset aml as an example: data(aml) If I use instead dummy variables: aml$x1 = (aml$x=="maintained")aml$x2 = (aml$x=="unmaintained") and I want to plot the survival curve using x1, x2, and I just want the 2 levels, rather than 4 curves from: fit <- survfit(Surv(time, status) ~ x1+x2, data=aml) plot(fit) I guess because there are 2 levels

Hello friends and fellow R users, I have successfully tabulated and entered my survival data into R and have generated survival curves. But I would like to be able to determine what the survival rates are now at one month, three months, six months and one year. I have a data set, via.wall, which I have entered into R, and which generates the following Surv object: Surv(Days,Status==1) [1]

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Hi, I am new to time-dependent Cox model to estimate time dependent hazard ratios. Let me use aml dataset from survival package: > aml3<-survSplit(aml2,cut=c(5,10,20),end="time",start="start", event="status",episode="i") If I want to esimate hazard ratio for each of the time intervals 0-5, 5-10, 10-20 and >=20, would the following calculate

Hello, When I try to to obtain the expected risk for a new dataset using coxph in the survival package I get an error. Using the example from ?coxph: > test1 <- list(time= c(4, 3,1,1,2,2,3),+ status=c(1,NA,1,0,1,1,0),+ x= c(0, 2,1,1,1,0,0),+ sex= c(0, 0,0,0,1,1,1))> cox<-coxph( Surv(time, status) ~ x + strata(sex), test1)

Hi there, I have a question about one command sentence when I follow the example in the book of "Survival analysis in S": > aml1<-aml[aml$group==1] but I got the error warning: NULL data frame with 23 rows Thus, I couldn't keep going on the next command: esf.fit<-survfit(Surv(aml1,status)~1). and also when I try > aml1<-aml[aml$group==1,]

Hi there: recently i try to use LogicReg package for a tree model(logistics fit ) . i list my code and error below: > dim(model.dat) [1] 48000 745 > fit1 <- logreg(resp = model.dat[,745], bin=model.dat[, 9:700], type = 3, select = 3, ntrees = c(1,2), nleaves=c(1,7), ) Insufficient declaration LGCn1MAX in logreg() is 20000 LGCn1MAX should be at least 48000 Please fix and

Hello all. I've been attempting to utilize the "survival" pkg ( http://cran.r-project.org/web/packages/survival/index.html), while reading through this guide (http://www.ms.uky.edu/~mai/Rsurv.pdf). I figured working through the guide would be the best way to go, before attempting my own data. I tried to utilize the Kaplain-Meier estimator as shown in the guide:

Hi, I am having trouble with the survival library, particualrily the coxph function. the following works coxph(jtree9$cph.call,z,rep(1,dim(z)[1])) Call: coxph(formula = jtree9$cph.call, data = z, weights = rep(1, dim(z)[1])) coef exp(coef) se(coef) z p SM 0.2574 1.294 0.0786 3.274 1.1e-03 Sex -0.1283 0.880 0.1809 -0.709

Hi List, A friend of mine recently asked the same question as Heinz T?chler. Since I've already written the code I'd like to share with the list. # x is an object returned by "survfit"; # "smed" returns a matrix of 5 columns of # n, events, median, 0.95LCL, 0.95UCL. # The matrix returned has rownames as the # group labels (eg., treatment arms) if any. smed <-

I am using plot(survfit(Surv(time,status) ~...) and would like to add error bars rather than the confidence intervals. Am I able to do this at specified times? e.g. when time = 20 & 40. leukemia.surv <- survfit(Surv(time, status) ~ x, data = aml) plot(leukemia.surv, lty = 2:3,xlim = c(0,50)) #can i add error bars at times 20 & 40? legend(100, .9, c("Maintenance", "No

Hello, I'm using plot.survfit to plot cumulative incidence of an event. Essentially, my code boils down to: cox <-coxph(Surv(EVINF,STATUS) ~ strata(TREAT) + covariates, data=dat) surv <- survfit(cox) plot(surv,mark.time=F,fun="event") Follow-up time extends to 54 weeks, but the last event occurs at week 30, and no more people are censored in between. Is there a

I would like to produce a complimentary log-log survival plot with only the points appearing on the graph. I am using the code below, taken from the plot.survfit page of help for the the survival package (version 2.35-7). I am running in R 2.10.0 on Windows XP, and the list of packages following the error is loaded. Is there some specific 'type= ' syntax, or an additional parameter that

Hello R community, I have two questions about using R. The first is about dividing each element of a list with another similar sized list. So, if the first list has two elements and so does the second, then the result should also be a list with two elements. For example, the inputs are: list(matrix(1:6,ncol=2),matrix(1:6,ncol=2))->l1 l2<-list(1:3,2) I want to get a list, l3 with the

Dear all, I used cph() function from Frank harrell's Design package to create a survival model, then used functions 'Function' and 'sascode' to generate prediction equation based on the saved survival model. But it failed. I included a stratified variable in the model. If I removed the stratification, they were working well. Does that mean that function 'Function'

> Dear All, > > I have built a survival cox-model, which includes a covariate * time interaction. (non-proportionality detected) > I am now wondering how could I most easily get survival predictions from my model. > > My model was specified: > coxph(formula = Surv(event_time_mod, event_indicator_mod) ~ Sex + > ageC + HHcat_alt + Main_Branch + Acute_seizure +

Hi, I fit a Cox PH model to estimate the cause-specific hazards (in a competing risks setting). Then , I compute the survival estimates for all the individuals in my data set using the `survfit' function. I am currently playing with a data set that has about 6000 observations and 12 covariates. I am finding that the survfit function is very slow. Here is a simple simulation example

Though tapply(x, factor, fun, simplify =TRUE) should be equivalent to sapply(split(x, factor), fun, simplify=TRUE), note simplify=TRUE, it is not so if fun() returns a vector rather than a scalar, e.g. > tapply(1:6, c(0,0,0,1,1,1), function(x) c(min=min(x), max=max(x)), simplify=TRUE) $"0" min max 1 3 $"1" min max 4 6 > sapply(split(1:6, c(0,0,0,1,1,1)),

I am unable to get the basehaz function to apply a proportional hazards model to a new data frame. I replicated my specific situation with the example for coxph in the help, where I changed the x value of the first record from 0 to 1. Is there something incorrect in the syntax that I am using? Thanks in advance! test1 <- list(time= c(4, 3,1,1,2,2,3), status=c(1,NA,1,0,1,1,0),