similar to: Fitting problem for Cox model with Strata as interaction term

Hi all, The last e-mails about beginners gave me the courage to post a question; from a beginner's perspective, there are a lot of questions that I'm tempted to ask. But I'm trying to find the answers either in the documentation, either in the about 15 free books I have, either in the help archives (I often found many similar questions posted in the past). Being an (still actual)

dear community, I am a beginer in R , and can't predict with logistic model in package logistf, could anyone help me ? thanks ! the following is my command and result : >library(logistf) >data(sex2) >fit<-logistf(case ~ age+oc+vic+vicl+vis+dia, data=sex2) >predict(fit,newdata=sex2) Error in predict(fit, newdata = sex2) : no applicable method for "predict"

To Danardonos concern of splitting time for records with delayed entry: This can fairly easily be accomodated, by simply splitting time in small intervals of time since entry into the study, and then compute the value of the other timescales for each of these e.g.: current.age <- time.from.entry + age.at.entry but the cut on the other timescales will not be exactly where you may want them

Greetings, I have stumbled across some unexpected behavior (potential a bug) in, what I suspect to be R's (2.6.2 on Ubuntu Linux) t.test function; then again the problem may exist in my code. I have shutdown R and started it back up, re-run the code and re-experienced the error. I have searched on Google for the abnormal termination error message "(stderr < 10 * .Machine$double.eps *

Dear All, since years I am struggling with Surv objects in data.frames. The following seems to have to do with it. See below the modified example from the help page of survSplit. The original works, as expected. If, however, a Surv object is added to the data.frame, each record gets doubled. Is there some solution other than avoiding Surv objects in data.frames? Thanks, Heinz

Dear r-devel list members, I'm posting this message here because it concerns the nlme package, which is maintained by R-core. The problem I'm about to describe is somewhere between a bug and a feature request, and so I thought it a good idea to ask here rather posting a bug report to the R bugzilla. I was made aware (by Ben Bolker) that the car::Anova() method for "lme"

in wine/dlls/qcap/v4l.c: ReadThread() --> OutputPin_GetDeliveryBuffer((OutputPin *)capBox->pOut, &pSample, NULL, NULL, 0); --> IMediaSample_SetTime(*ppSample, tStart, tStop); in wine/dlls/quartz/memallocator.c: StdMediaSample2_GetTime(IMediaSample2 * iface, REFERENCE_TIME * pStart, REFERENCE_TIME * pEnd) --> if tstart/stop is NULL, leave pStart/pEnd not setted. in

Hello! I'm looking for an easy way to filter data frames. Some rows in my data frame needs to be left out, for instance, rows with invalid data. Right now, if I want to pair two columns of this data frame, I have to do this: > pairs(as.data.frame(list(tstart=df$tstart[valid],tend=df$tend.sdr[valid]))) or this: >

Dear list members, After further testing, I found that the following simplified version of model.matrix.lme(), which omits passing xlev to the default method, is more robust. The previous version generated spurious warnings in some circumstances. model.matrix.lme <- function(object, ...){ data <- object$data if (is.null(data)){ NextMethod(formula(object),

Hi everyone, I was trying to mode the following exercise using R. The question: Set up a one pool model using numericintegration. The model will run from time 1 to time 30 using a time step of 1.The pool (A) will be fed by flux "inA" at a rate of 5 units per hour anddrained by flux "outA" at a rate of 20% per hour. At time 0, A has 5units. At time 30, what is the pool size of

Hi All, I am currently conducting some survival analyses. I would like to extract coefficients at each level of the IVs. I read on a previous posting that dummy regression using coxph was not possible. Therefore I though, hey why not categorize the variables (I realize some folks object to categorization but the paper I am replicating appears to have done so ...) and turn the variables

Hi im trying to add error bars to my barplots, there very basic, i have a few grapghs where the y variable is different but on all the X variable is Age (Adult and Juvenile) however this is split into two levels so i have males and females, so my graph basically has four bars on it. I know how to add eror bars for instance when there is only one level eg lookng at the diffrence between male and

tftp-hpa-0.35 and tftp-hpa-0.36 show wrong transmission time. >tftp -V tftp-hpa 0.36, with readline >tftp -m octet -v localhost -c get aaa /tmp/aaa Connected to localhost.localdomain (127.0.0.1), port 69 getting from localhost.localdomain:aaa to /tmp/aaa [octet] Received 33554432 bytes in -3.3 seconds [-81407477 bit/s] ^^^^ ^^^^^^^^^ By the way, why

Dear R-project! How do i create 1 dummy from 2 already existing dummys. To be more precise, I want to create a dummy from a dummy called "sex" and another called "sex1" when both thoose dummys are 1 I want my created dummy "samesex" to take 1. Thanks for the help! Paulie [[alternative HTML version deleted]]

Dear R-users, I use the survsplit function in the survival package to change my data into counting-process format and the transformed format is as follow: (a) start stop event DP age .... 0 5 0 1 20 5 10 0 1 20 10 25 1 1 20 looking at the above three entries that belong to the same person, if an event happen at

I'm trying to convert a dataset from the time-independent analysis form > head(addicts) id clinic status survtime prison meth clinic01 1 1 1 1 428 0 50 1 2 2 1 1 275 1 55 1 3 3 1 1 262 0 55 1 into the "counting data format" necessary to perform extended Cox regression. I got

Dear R users, I am doing Cox regression using coxph(Survival) or cph(Design). I have time varying effects (diagnosed with schoenfeld residuals Chi2 test and graph) so I first want to split time into 2 separate intervals : t<6months and t>=6months, to estimate one hazard ratio (hr) for each interval. I am analysing Overall survival according to 3 prognostic factors (age,deep,ldh).

  Hi there,   I am beginner in R and I have some basic question. Suppose I run a common procedure such as a t test or cox model like below:   out<-coxph( Surv(tstart,tstop, death1) ~ x1+x1:log(tstop+1) , test1,method=c("breslow"))    Which yields the following result:   Call: coxph(formula = Surv(tstart, tstop, death1) ~ x1 + x1:log(tstop +     1), data = test1, method =

I have data that looks like this time_stamp (seconds) user_id The data is (partial) ordered by time - in that sometimes transactions occur at the same timestamp. The output I want is collated by transaction time on a per user basis, normalized by the maximum number of transactions per user, and aggregated over each day. So, if the users have 50 transactions in the first day and 20 transactions

Greetings to you all, I am performing a semi parametric bootstrap in R on a Gamma Distributed data and a Binomial distributed data. The main challenge am facing is the fact that the residual variance depends on the mean (if I am correct). I strongly feel that the script below may be wrong due to mean-variance relationship #####R code####### fit1s