similar to: Survival analysis extrapolation

Hi I am trying to determine the mean of a Weibull function that has been fit to a data set, adjusted for a categorical covariate , gender (0=male,1=female). Here is my code: library(survival) survdata<-read.csv("data.csv") ##Fit Weibull model to data WeiModel<-survreg(Surv(survdata$Time,survdata$Status)~survdata$gender) summary(WeiModel) P<-pweibull(n,

Hello, I am working with the nls() function and inserting a formula into it that uses the pweibull function. However the pweibull function is annoyingly producing NaNs, which nls() refuses to handle. I have put a sample of the code below. Is there a way to prevent these NaNs from interfering, for example a method to catch them? I get the following error when I try to run the code: res.nls <-

Hello everyone, I have a quick question but I am stuck with it and I do not know how to solve it. Imagine I need the distribution function of a Weibull(1,1) at t=3, then I will write pweibull(3,1,1). I want to keep the shape and scale parameters in a list (or a vector or whatever). Then I have parameters<-list(shape=1,scale=1) but when I write pweibull(3,parameters) I get the following

In rw1050, I found that > pweibull(3:10, 2) [1] 0.9998766 0.9999999 1.0000000 1.0000000 NaN NaN [7] NaN NaN Warning message: NaNs produced in: pweibull(q, shape, scale, lower.tail, log.p) more surprisingly, > pweibull(3:10, 2.1) [1] 0.9999566 1.0000000 1.0000000 -Inf NaN NaN [7] NaN NaN Warning message: NaNs produced in: pweibull(q,

Please allow me to add just a little more about this: nothing wrong with pweibull(), namely, the two cases I reported: pweibull(3:10, 2) and pweibull(3:10, 2.1), in rw1041 and earlier version. I wonder this might just due to the change from rw1041 to rw1050, however, I can't find anything relevant (seems to me) in the News or Readme. Thanks Sundar for the suggestion of using 1 -

Dear all - I have followed the thread the reply to which was lead by Thomas Lumley about using pweibull to generate fitted survival curves for survreg models. http://tolstoy.newcastle.edu.au/R/help/04/11/7766.html Using the lung data set, data(lung) lung.wbs <- survreg( Surv(time, status)~ 1, data=lung, dist='weibull') curve(pweibull(x, scale=exp(coef(lung.wbs)),

Dear R gurus, I have the following code, but I still not know how to estimate and extract confidence intervals (95%CI) from resampling. Thanks! ~Adriana #data penta<-c(770,729,640,486,450,410,400,340,306,283,278,260,253,242,240,229,201,198,190,186,180,170,168,151,150,148,147,125,117,110,107,104,85,83,80,74,70,66,54,46,45,43,40,38,10) x<-log(penta+1) plot(ecdf(x),

Dear R users, Please can you help me with a relatively straightforward problem that I am struggling with? I am simply trying to plot a baseline survivor and hazard function for a simple data set of lung cancer survival where `futime' is follow up time in months and status is 1=dead and 0=alive. Using the survival package: lung.wbs <- survreg( Surv(futime, status)~ 1, data=lung,

I have an R function (about 1000 lines long) that takes more than 20 times as long to run under R Windows 1.9.0 and 1.8.1 than it does under 1.7.1. Profile results indicate that the $<-.data.frame operation is the culprit, but I don't understand exactly what that is (assignment of data frame elements to another variable?), or why it's only a problem under 1.8.1 and 1.9.0. Any advice?

I should hope that there is trouble, since "type III" is an undefined concept for a Cox model. Since SAS Inc fostered the cult of type III they have recently added it as an option for phreg, but I am not able to find any hints in the phreg documentation of what exactly they are doing when you invoke it. If you can unearth this information, then I will be happy to tell you whether

Hello: So i've fit a hazard function to a set of data using kmfit<-survfit(Surv(int, event)~factor(cohort)) this factor variable, "cohort" has four levels so naturally the strata variable has 4 values. I can use this data to estimate the hazard rate haz<-n.event/n.risk and calculate the cumulative hazard function by H<--log(haz) Now, I would like to plot this

OK this is bound to be something silly as I'm completely new to R - having started using it yesterday. However I am already warming to its lack of 'proper' GUI... I like being able to rerun a command by editing one parameter easily... try and do that in a Excel Chart Wizzard! I eventually want to use it to analyse some chemotherapy response / survival data. That data will not be

Full_Name: M Welinder Version: 1.4 OS: (src) Submission from: (NULL) (192.5.35.38) It seems to me that pweibull can be improved in the lower_tail=TRUE and log_p=FALSE case by using expm1. Something like -expm1(-pow(x / scale, shape)), I think. -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list -- Read

(1) ...need to speed up a monte-carlo sampling...any suggestions about how I can get R to use all 8 cores of a mac pro would be most useful and very appreciated... (2) spent the last few hours trying to get pnmath to compile under os- x 10.5.4... using gcc version 4.2.1 (Apple Inc. build 5553) as downloaded from CRAN, xcode 3.0... ...xcode 3.1 installed over top of above after

Dear useRs, I am trying to fit a Cox PH model on survival data from a lung cancer dataset. I would like to include the patient staging (I-IV) as a covariate. For this I use the following function: coxph(Surv(time,status) ~ stage) The staging information is a categorical variable, and it is important to take the ordering into account (I<II<III<IV). Does the coxph function handle

Dear all, I have a question regarding performing test if the data fits chi-squared distribution. For example, using ks.test() I found in the examples how to fit it to gamma or weibull x<-rnorm(100) ks.test(x, "pweibull", shape=2,scale=1) for the gamma, pgamma can be used But I cannot find the value of this second parameter for the chi-squared distribution. Maybe someone

Dear R helpers, I need to use KS and AD test for Generalized Pareto and Generalized extreme value. E.g. if I need to use KS for Weibull, I have teh syntax ks.test(x.wei,"pweibull", shape=2,scale=1) Similarly, for AD I use ad.test(x, distr.fun, ...) My problem is fir given data, I have estimated the parameters of GPD and GEV using lmom. But I am not able to find out the distribution

R-users E-mail: r-help@r-project.org I have a quenstion on "gam()" in "gam" package. The help of gam() says: 'gam' uses the _backfitting algorithm_ to combine different smoothing or fitting methods. On the other hand, lm.wfit(), which is a routine of gam.fit() contains: z <- .Fortran("dqrls", qr = x * wts, n = n, p = p, y = y *

Hi: Could anyone help me to clarify this: are the weights normalized inside lm function (package:stats) before applied to the error term? For example: >lm (cost ~ material, weights=quatity, data=receipt) will lm normalize quatity such that sum(quatity) = 1? I traced to lm.wfit and then the weights get transferred into a precompiled FORTRAN module so I can't figure out. Thanks!

Hi, I am facing a problem in extrapolation of data series. It is a series of Bond yields, I am having the yield for 1 year to 30 years. I want to find the yield for 0.5 year and 30.5 years. I used the Langrange's Extrapolation but the extrapolation deviates from the normal trend ( as we can see in theoritical yield curves) very sharply, as go on increasing my years from 30 years to 35 years