similar to: rpart

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

Hello R users,  I am trying to obtain a direct adjusted survival curve. I am sending my whole code (see below). It's basically the larynx cancer data with Stage 1-4. I am using the cox model using coxph option, see the fit3 coxph. When I use the avg.surv option on fit3, I get the following error: "fits<-avg.surv(fit3, var.name="stage.fac", var.values=c(1,2,3,4), data=larynx)

Well, I would suggest using the code already in place in the survival package. Here is my code for your problem. I'm using a copy of the larynx data as found from the web resources for the Klein and Moeschberger book. larynx <- read.table("larynx.dat", skip=12, col.names=c("stage", "time", "age", "year",

I'm having difficulties with plot.locfit.3d, at least I think that is the problem. I have a large dataframe (about 4 MM cases) and was hoping to see a non-parametric estimate of the hazard plotted against two variables: > fit <- locfit(~surv.yr+ ur_protein + ur_creatinine, data=TRdta, cens = 1-death, family = "hazard", xlim=c(0,10)) # it took somewhere between 1 and 2

Hello All, I am stuck in the following problem. Cexpt=c(0,25,50,100,150,300,250,125,40) t=c(0,0.2,0.4,0.6,1,4,8,12,24) theta0= vector of 6 parms (My initial parameter) A=Constraint matrix (hopefully 6*6) B= Constraint vector of length 6) Cfit=function(t,theta){ J(t)=function(theta,t) Cfit=function(J(t),constant) return(Cfit) } loss=function(theta,t,Cexpt) {

Hi is it possible to fit a trend line (or some other panel function) through each of multiple data series plotted on the same graph? Specifically, while one can do something like xyplot(a+b+c~x) which plots three series, a,b & c, but can one automatically fit lines through each of them? I suppose one could generate three more variables afit, bfit, and cfit with a model & predict and

Dear All, Sorry to bother you again. I have a model: coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma) and I'm trying to do a plot of Schoenfeld residuals using the code: plot(cox.zph(coxfita)) abline(h=0,lty=3) The error message I get is: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In sqrt(x$var[i, i] * seval) : NaNs

i want to make survival plots for a coxph object using survfit function. mod.phm is an object of coxph class which calculated results using columns X and Y from the DataFrame. Both X and Y are categorical. I want survival plots which shows a single line for each of the categories of X i.e. '4' and 'C'. I am getting the following error: > attach(DataFrame) >

Hi All, I have to say upfront that I am a complete neophyte when it comes to programming. Nevertheless I enjoy the challenge of using R because of its incredible statistical resources. My problem is this .........I am running a regression tree analysis using "rpart" and I need to run the calculation repeatedly (say n=50 times) to obtain a distribution of results from which I will pick

Hello, and thank you in advance. I would like to capture the expected survival from a coxph model for subjects in an observational study with recurrent events, but the survexp statement is failing due to memory. I am using R version 2.13.1 (2011-07-08) on Windows XP. My objective is to plot the fitted survival with the Kaplan-Meier plot. Below is the code with output and [unfortunately]

I am not sure if this is the right place for that kind of questions, but I wondered that the recommended package survival did not pass R's check procedure without warnings: 1) unbalanced braces: * Rd files with unbalanced braces: * man/Surv.Rd * man/cluster.Rd * man/cox.zph.Rd * man/coxph.Rd * man/coxph.detail.Rd * man/date.ddmmmyy.Rd * man/lines.survfit.Rd *

Hello, I'm using mvpart option xv="1se" to compute a regression tree of good size with the 1-SE rule. To better understand 1-SE rule, I took a look on its coding in mvpart, which is : Let z be a rpart object , xerror <- z$cptable[, 4] xstd <- z$cptable[, 5] splt <- min(seq(along = xerror)[xerror <= min(xerror) + xvse * xstd]) I interprete this as following: the

I am running some models (for the first time) using rpart and am getting results I don't know how to interpret. I'm using cross-validation to prune the tree and the results look like: Root node error: 172.71/292 = 0.59148 n= 292 CP nsplit rel error xerror xstd 1 0.124662 0 1.00000 1.00731 0.093701 2 0.064634 1 0.87534 1.08076 0.092337 3 0.057300 2

Hi, I've searched online, in a few books, and in the archives, but haven't seen this. I believe that xerror is scaled to rel error on the first split. After fitting an rpart object, is it possible with a little math to determine the percentage of true classifications represented by a xerror value? -seth -- View this message in context:

Full_Name: Gregory R. Warnes Version: R 1.3.1 OS: Solaris 2.8 Submission from: (NULL) (192.77.198.200) rpart library version 3.1-2 Error message: > plotcp(fit.thirds.1,ylim=c(0.7,1.5)); Error in plot.default(ns, xerror, axes = FALSE, xlab = "cp", ylab = "X-val Relative Error", : formal argument "ylim" matched by multiple actual arguments > This can be

Hi Listers I need to calculate and then plot a frequency histogram of the best tree calculated using the 1-se rule. I have included some code that has worked well for me in the past but it was only for selecting the minimum cross-validation error. I include the code for my model, some relevant output and the code for selecting and plotting the frequency histogram of minimum xerror. Here is the

Hello list, I'm trying to generate classifiers for a certain task using several methods, one of them being decision trees. The doubts come when I want to estimate the cross-validation error of the generated tree: tree <- rpart(y~., data=data.frame(xsel, y), cp=0.00001) ptree <- prune(tree, cp=tree$cptable[which.min(tree$cptable[,"xerror"]),"CP"]) ptree$cptable

Hi All, I have some questions on using library rpart. Given my data below, the plotcp gives me increasing 'xerrors' across different cp's with huge xstd (plot attached). What causes the problem or it's not a problem at all? I am thinking 'xerror's should be decreasing when 'cp' gets smaller. Also what the 'xstd' really tells us? If the error bars for

Dear R-list members, I am doing a CART analysis in R using the rpart function in the rpart package: Phrag.rpart=rpart(PhragDiff~., data = Phrag, method="anova", xval=10). I used the xerror values in the CP table to prune the tree to 4 nsplits: CP nsplit rel error xerror xstd 1 0.098172 0 1.00000 1.02867 0.12768 2 0.055991 3 0.70548 1.00823 0.12911 3

> 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 +