similar to: survConcordance with 'counting' type Surv()

--- included message ---- Thus, my 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,

Hi, I am new to model selection by coefficient shrinkage method such as lasso. And I became particularly interested in variable selection in Cox regression by lasso. I became aware of the coxpath() in R package glmpath does lasso on Cox model. I have tried the sample script on the help page of coxpath(), but I have difficult time understanding the output. Therefore, I would greatly appreciate if

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

R Folk: Please forgive what I'm sure is a fairly na?ve question; I hope it's clear. A colleague and I have been doing a really simple one-off survival analysis, but this is an area with which we are not very familiar, we just happen to have gathered some data that needs this type of analysis. We've done quite a bit of reading, but answers escape us, even though the question below

I was looking at how the model.frame method for lm works and comparing it to my own for coxph. The big difference is that I try to retain xlevels and predvars information for a new model frame, and lm does not. I use a call to 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

Calum 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

Hello R helpers: *My first message didn't pass trough filter so here it's again* I would like to obtain probability of an event for one single patient as a function of time (from survfit.coxph) object, as I want to find what is the probability of an event say at 1 month and what is the probability of an event at 80 months and compare. So I tried the following but it fails miserably. I

A user reported a problem with the survdiff function and the use of variables that contain a space.? Here is a simple example.? The same issue occurs in survfit for the same reason. lung2 <- lung names(lung2)[1] <- "in st"?? # old name is inst survdiff(Surv(time, status) ~ `in st`, data=lung2) Error in `[.data.frame`(m, ll) : undefined columns selected In the body of the code

Dear R users, My apologies for the simple question, as I'm starting to learn the concepts behind the Cox PH model. I was just experimenting with the survival and rms packages for this. I'm simply trying to obtain the expected survival time (as opposed to the probability of survival at a given time t). I can't seem to find an option from the "type" argument in the predict

I believe this has to do terms() making "term.labels" (hence the dimnames of "factors") with deparse(), so that the backquotes are included for non-syntactic names. The backquotes are not in the column names of the input data.frame (nor model frame) so you get a mismatch when subscripting the data.frame or model.frame with elements of terms()$term.labels. I think you can

Hello: I would like to obtain probability of an event for one single patient as a function of time (from survfit.coxph) object, as I want to find what is the probability of an event say at 1 month and what is the probability of an event at 80 months and compare. So I tried the following but it fails miserably. I looked at some old posts but could not figure out the solution. Here's what I did

On examining non-linearity of Cox coefficients with penalized splines - I have not been able to dig up a completely clear description of the test performed in R or S-plus. >From the Therneau and Grambsch book (2000 - page 126) I gather that the test reported for "linear" has as its null hypothesis that the spline coefficient is the same at the center of basis. Thus, in the example

Thanks to Bill Dunlap for the clarification. On follow-up it turns out that this will be an issue for many if not most of the routines in the survival package: a lot of them look at the terms structure and make use of the dimnames of attr(terms, 'factors'), which also keeps the unneeded backquotes. Others use the term.labels attribute. To dodge this I will need to create a

Hi, I'm trying to using pspline in bic.surv{BMA}. ############################# library(BMA) library(survival) data(veteran) test.bic.surv<- bic.surv(Surv(time,status) ~ karno+pspline(age,df=3)+diagtime+prior, data = veteran, factor.type = TRUE) summary(test.bic.surv, conditional=FALSE, digits=2) ############################# The results are:

--begin inclusion --- I am trying to extract output information from the survfit function in order to generate a matrix of select output for multiple factors. Specifically, I am interested in extracting the number of events (in the output 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

> I'm using rpart function for creating regression trees. > now how to measure the fitness of regression tree??? > > thanks n Regards, > Vibha I read R-help as a digest so often come late to a discussion. Let me start by being the first to directly answer the question: > fit <- rpart(time ~ age +ph.ecog,lung) > summary(fit) Call: rpart(formula = time ~ age +

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)

Meant to respond to this but forgot. I didn't write a new terms() function -- I added an attribute to the terms() (a vector of the names of the constructed model matrix), thus preserving the information at the point when it was available. I do agree that it would be preferable to have an upstream fix ... On Thu, Mar 8, 2018 at 9:39 AM, Therneau, Terry M., Ph.D. via R-devel <r-devel

Hi, I am trying to use the rpart library with my own set of functions on a survival object. I get an immeadiate segmentation fault when i try calling rpart with my list of functions. I get the same problem with the logrank example from Therneau,s S-rpart library though their anova example works. Should I report this as a bug, as even if my functions are structured improperly, that should lead to

>>>>> Ben Bolker <bbolker at gmail.com> >>>>> on Thu, 8 Mar 2018 09:42:40 -0500 writes: > Meant to respond to this but forgot. > I didn't write a new terms() function -- I added an attribute to the > terms() (a vector of the names > of the constructed model matrix), thus preserving the information at > the point when