similar to: R CMD check -- non S3 method warning

Is putting a variable into a list a deep copy (and is tracemem the correct way to confirm)? warmstrong at krypton:~/dvl/R.packages$ R > x <- rnorm(1000) > tracemem(x) [1] "<0x3214c90>" > x.list <- list(x.in.list=x) tracemem[0x3214c90 -> 0x2af0a20]: > Is it possible to put a variable into a list without causing a deep copy (i.e. if you _really_ want the

I've been working on a package that requires a shared library to be loaded. I have used the NAMESPACE file to load the library according to: http://cran.r-project.org/doc/manuals/R-exts.html#Load%20hooks <http://cran.r-project.org/doc/manuals/R-exts.html#Load%20hooks> my shared library is "excelpoi.so" hence I have added "useDynLib(excelpoi)" to my NAMESPACE file.

Since, gcc was using upwards of 2gb of ram to compile my package, I just split all the functions into individual files. I guess I'm too clever for myself, because now I get hit with the "Argument list too long" error. Is there a way to deal with this aside from writing my own configure script (which could possibly feed the gcc commands one by one). -Whit RHEL 5 [whit at

Dear R-help, I am trying to estimate a Cox model with nested effects basing on the minimization of the overall AIC; I have two frailties terms, both gamma distributed. There is a error message (theta2 argument misses) and I don?t understand why. I would like to know what I have wrong. Thank you very much for your time. fitM7 <- coxph(Surv(lifespan,censured) ~ south + frailty(id,

Dear r-help subscribers, A couple of weeks ago I sent the following message to the r-help mail list. It hasn't generated any response, and I could really use some help on this. Anyone able to help? Thanks again, Roger Dungan >> I am working on some survival analysis of some interval censored failure time data in R. I have done similar analysis before using PROC LIFEREG in SAS. In

Dear R users: I'm generating the following survival data: set.seed(123) n=200 #sample size x=rbinom(n,size=1,prob=.5) #binomial treatment v=rgamma(n,shape=1,scale=1) #gamma frailty w=rweibull(n,shape=1,scale=1) #Weibull deviates b=-log(2) #treatment's slope t=exp( -x*b -log(v) + log(w) ) #failure times c=rep(1,n) #uncensored indicator id=seq(1:n) #individual frailty indicator

Dear R-developers, I am currently developing an R package called RLadyBug. When developing under Linux "R CMD check ." works fine without a warning. However, when I do "Rcmd check ." under Windows (version 2.4.0 and earlier) I get a ?syntax error" when checking the examples. This puzzles me somewhat, because a manual source("RLadyBug-Ex.R") on Windows works

Hi, I want to create upper and lower 95% confidence intervals for a p-p plot of an empirical distribution with a theoretical gamma distribution. This is my code: x<-rgamma(100,shape=2, rate=1) # empirical data fitdistr(x,"gamma") # fit a gamma distribution dist<-pgamma(x,shape=1.9884256 ,rate=0.8765314 ) # fitted distribution, using the loglikelihood estimated parameters

Dear R help list, I am modeling some survival data with coxph and survreg (dist='weibull') using package survival. I have 2 problems: 1) I do not understand how to interpret the regression coefficients in the survreg output and it is not clear, for me, from ?survreg.objects how to. Here is an example of the codes that points out my problem: - data is stc1 - the factor is dichotomous

Hi, I am trying to replicate Lambert (1992)'s simulation with zero-inflated Poisson models. The citation is here: @article{lambert1992zero, Author = {Lambert, D.}, Journal = {Technometrics}, Pages = {1--14}, Publisher = {JSTOR}, Title = {Zero-inflated {P}oisson regression, with an application to defects in manufacturing}, Year = {1992}} Specifically I am trying to recreate Table 2. But my

If I have an R script that I am executing from a command line in linux, do you know how I can return the value of the variable in my R script to the linux environment without writing it to a file in my R script and then reading the file through cat? For example, if I had a simple one line R script that just did string <- 'TEST', when I call /usr/local/bin/R

1. The weibull is the only distribution that can be written in both a proportional hazazrds for and an accelerated failure time form. Survreg uses the latter. In an ACF model, we model the time to failure. Positive coefficients are good (longer time to death). In a PH model, we model the death rate. Positive coefficients are bad (higher death rate). You are not the first to be confused

Hi All, could someone shed some light on what the difference between the estimated dispersion parameter that is supplied with the GLM function and the one that the 'gamma.dispersion( )' function in the MASS library gives? And is there consensus for which estimated value to use? It seems that the dispersion parameter that comes with the summary command for a GLM with a Gamma dist. is

I have the following questions about the variance of the random effects in the survreg() function in the survival package: 1) How can I extract the variance of the random effects after fitting a model? For example: set.seed(1007) x <- runif(100) m <- rnorm(10, mean = 1, sd =2) mu <- rep(m, rep(10,10)) test1 <- data.frame(Time = qsurvreg(x, mean = mu, scale= 0.5, distribution =

Thanks for sharing the questions and responses! Is it possible to appreciate how much the coefficients matter in one or the other model? Say, using Biau's example, using coxph, as.factor(grade2 == "high")TRUE gives hazard ratio 1.27 (rounded). As clinician I can grasp this HR as 27% relative increase. I can relate with other published results. With survreg the Weibull model gives a

Hi, I was wondering if someone in the mailing list has any insight into this segfault error that I consistently find when running a script containing heatmap() in R 2.8.1 and 2.8.0 on a Linux 64-bit machine. Some points: 1. This occurs when running heatmap(). 2. Interestingly, if I source() the script or copy and paste the script in its entirety, this error occurs. However, if I run the

Please can someone explain why I seem to get these contradictory results? choose(5,2) [1] 10 gamma(6)/(gamma(3)*gamma(4)) [1] 10 gamma(6)/(gamma(3)*gamma(4)) == choose(5,2) [1] TRUE # all's well so far. # now look what happens: gamma(21)/(gamma(6)*gamma(16)) == choose(20,5) [1] FALSE # check individual terms: gamma(21)/(gamma(6)*gamma(16)) [1] 15504 choose(20,5) [1] 15504 # so they are the

I am trying to generate predictions from a weibull survival curve but it seems that the predictions assume that the shape(scale for survfit) parameter is one(Exponential but with a strange rate estimate?). Here is an examle of the problem, the smaller the shape is the worse the discrepancy. ### Set Parameters scale<-10 shape<-.85 ### Find Mean scale*gamma(1 + 1/shape) ### Simulate Data

Dear R experts, How should lambda and gamma (with std.errors) be calculated for a weibull model with age as an independent predictor? I have assumed that this can be done with survreg with e. g. (summary(survreg(Surv(time, status) ~ age, dist = 'weibull')) ) and predict.survreg with e.g. (predict(model, se.fit = T, newdata = data.frame(age = seq(50, 80, 5)) but unfortunately I'm

Hi! I would like to perform an F-Test over more than one variable within a generalized mixed model with Gamma-distribution and log-link function. For this purpose, I use the package mgcv. Similar tests may be done using the function "anova", as for example in the case of a normal distributed response. However, if I do so, the error message "error in eval(expr, envir, enclos) :