similar to: ID parameter in model

I am unsure what is being returned, and what is supposed to be returned, when using 'predict' with "type='expected'" for an aftreg survival model. The code below first generates a weibull model, then uses predict to create a vector of the linear predictors, then attempts to create the 'expected' vector, which is empty. The final two steps in the code generate a

Dear all, I have some trouble using the "id"-argument with aftreg (accelerated failure time regression analysis from the eha library). As far as I understand it, the id argument is used to group individuals together if there are time-varying covariates and the data is arranged in counting process style. Unfortunately, i cannot figure out how to use the "id"-argument. The

Dear Prof. Brostr?m, Dear R-mailinglist, first of all thanks a lot for your great effort to incorporate time-varying covariates into aftreg. It works like a charm so far and I'll update you with detailled benchmarks as soon as I have them. I have one more questions regarding Accelerated Failure Time models (with aftreg): You mention that left truncation in combination with time-varying

Hi, I'm estimating a loglogistic aft (accelerated failure time) model, just a simple plain vanilla one (without time dependent covariates), I'm comparing the results that I obtain between aftreg (eha package) and survreg(surv package). If I don't use any covariate the results are identical , if I add covariates all the coefficients are the same until a precision of 10^4 or 10^-5 except

G?ran, thanks for the update, I'm just about to install it! Just wanted to drop you a short line about performance (as you once requested): aftreg takes ages on my windows machine to calculate a small set of 7 observations which are not even grouped together by "id". To be a bit more precise, it takes 2:40 mins on my Intel T9300 Core2 Duo @ 2.5 GHz. Bigger samples with about 700

Hi list! I'm getting an error message when trying to fit an accelerated failure time parametric model using the aftreg() function from package eha: > Error in optim(beta, Fmin, method = "BFGS", control = list(trace = > as.integer(printlevel)), : > non-finite finite-difference value [2] This only happens when adding four specific covariates at the same time in the

Dear R-users, I need to use the aftreg function in package 'eha' to estimate failure times for left truncated survival data. Apparently, survreg still cannot fit such models. Both functions should be fitting the accelerated failure time (Weibull) model. However, as G?ran Brostr?m points out in the help file for aftreg, the parameterisation is different giving rise to different

Dear all, does anyone know how I can extract specific p-values for covariates from an aftreg object? After fitting a model with aftreg I can find all different variables by using str(), but there's no place where p-values are kept. The odd thing is that print() displays them correctly. EXAMPLE: > testdata start stop censor groupvar var1 var2 1 0 1 0

Dear R-community, I have a short question: How do I interpret the result of a likelihood ratio test correctly? I am fitting a parametric survival model (with aftreg {eha}) and the output tells me the overall p-value of my model is < 0.001. My simple question is: Does the result mean my model fits the data well OR does it mean my model DOES NOT fit the data well? Some side information how the

Dear R-users, I have two samples with individuals that are in more than one of the samples and individuals that are only in one sample. I have been trying to do a pooled hazard model, stacking one sample below the other, with aftreg and time-dependent covariates. The idea behind is to see aggregate effects of covariates, but need to control for ther effects of same individuals in both samples

Dear R colleagues, I am trying to create a new column in a data frame, which converts values and NA's from another column into binary format. Essentially I need the NA's to become 1 and the rest to be 0. The code I wrote is returning the following error message: Error in if (mort[i, 4] != NA) mort[i, 8] <- 0 else if (mort[i, 4] == : missing value where TRUE/FALSE needed

Hello Folks, I have 5 columns for thousands of tree records that record whether that tree was alive or dead. I want to recode the columns such that the cell reads "found" when a live tree is first observed, "alive" for when a tree is found alive and is not just found, and "mort" when it was previously alive but is now dead. Given the following: > tree_live

Hi, I'm trying to plot a graph with error bars using xYplot in the Hmisc package. My data looks like this. mort stand site type 0.042512776 0.017854525 Plot A ST 0.010459803 0.005573305 PF ST 0.005188321 0.006842107 MSF ST 0.004276068 0.011592129 YSF ST 0.044586495 0.035225266 Plot A LD 0.038810662 0.037355408 PF

Hola. Como te dijo Carlos, el problema está en los nombres de las columnas y en los nombres de las filas. Cuando hice la importación (con dd<-read.csv('mortality.csv'), tuve problemas con las filas de nombre: - Malignant tumour of the larynx trachea bronchus and lungs - Malignant tumour of the lip pharynx and mouth - Other endocrinological metabolic and nutritional conditions

It was the return type which was i64. I changed it also to my abi_int_size and it works now. I have to take care of a few other type translations, but it looks like MCJIT is working now. Thank you. On 08/11/13 18:12, Yaron Keren wrote: > Something must be wrong with the Function Type. Try to debug into > runFunction to see which if condition fails. > Just a guess, if this is on 64

Something must be wrong with the Function Type. Try to debug into runFunction to see which if condition fails. Just a guess, if this is on 64 bit system the first argument type may be int64 but needs to be int32. Yaron 2013/11/8 edA-qa mort-ora-y <eda-qa at disemia.com> > That makes it more mysterious then since I am indeed only calling a main > function. Perhaps I have to invoke

> How can I draw a Hypercube sample for the variable mortality_probability so > that this variable exhibits the same pattern as the observed distribution? One simple way is to use the uniform random output of randomLHS as input to the quantile function for your desired distribution(s). For example: q <- randomLHS(1000, 3) colnames(q) <- c("A", "B",

I believe the memcpy is there just as a consequence of Clang's design - different parts of the compiler own different pieces of this, so in some sense one hand doesn't see what the other is doing. Part of it is "create an argument" (memcpying the local variable into an unnamed value) and then the next part is "oh, but that argument gets passed in registers, so decompose it

That makes it more mysterious then since I am indeed only calling a main function. Perhaps I have to invoke it a different way. Here's my call I have now: auto main = linker->getModule()->getFunction( "main" ); std::vector<llvm::GenericValue> args(2); args[0].IntVal = llvm::APInt( platform::abi_int_size, 0 ); args[1].PointerVal = nullptr; llvm::GenericValue gv =

Dear R-help Can anyone tell me why my curve appears the wrong way round on a rug plot? I am using the same code as on pg 596 of the Crawley R-book. mod<-glm(mort~logBd,binomial) par(mfrow=c(2,2)) xv<-seq(0,8,0.01) yv<-predict(mod,list(logBd=xv),type="response") plot(logBd,mort) lines(xv,yv) I've tried swapping xv and yv around but no luck. Thanks, Pete