similar to: Difficult debug

I haven't done any R memory debugging lately, but https://www.mail-archive.com/rcpp-devel at lists.r-forge.r-project.org/msg10289.html shows how I used to have gdb break where valgrind finds a problem so you could examine the details. Also, running your code after running gctorture(TRUE) can help track down memory problems. -Bill On Wed, Feb 7, 2024 at 12:03?PM Therneau, Terry M., Ph.D.

Hello all, thanks for your time and patience. I'm looking for a method in R to analyse the following data: Time to waking after anaesthetic for medical procedures repeated on the same individual. > str(mysurv) labelled [1:740, 1:2] 20 20 15 20 30+ 40+ 50 30 15 10 ... - attr(*, "dimnames")=List of 2 ..$ : NULL ..$ : chr [1:2] "time" "status" -

Hi, I am computing the Nelson-Aalen (NA) estimate of baseline cumulative hazard in two different ways using the "survival" package. I am expecting that they should be identical. However, they are not. Their difference is a monotonically increasing with time. This difference is probably not large to make any impact in the application, but is annoyingly non-trivial for me to just

First a statistical issue: The survfit routine will produce predicted survival curves for any requested combination of the covariates in the original model. This is not the same thing as an "adjusted" survival curve. Confusion on this is prevalent, however. True adjustment requires a population average over the confounding factors and is closely related to the standardized

I have some data which is censored and I want to determine the median. Its actually cost data for a cohort of patients, many of whom are still on treatment and so are censored. I can do the same sort of analysis for a survival curve and get the median survival... ...but can I just use the survival curve functions to plot an X axis that is $ rather than date? If not is there some other way to

I think there is a memory bug in `substr` that is triggered by a UTF-8 corner case: an incomplete UTF-8 byte sequence at the end of a string.? With a valgrind level 2 instrumented build of R-devel I get: > string <- "abc\xEE"??? # \xEE indicates the start of a 3 byte UTF-8 sequence > Encoding(string) <- "UTF-8" > substr(string, 1, 10) ==15375== Invalid read of

Hello, I want to know if two quadratic regressions are significantly different. I was advised to make the test using step 1 bootstrapping both quadratic regressions and get their slope coefficients. (Let's call the slope coefficient *â*^1 and *â*^2) step 2 use the slope difference *â*^1-*â*^2 and bootstrap the slope coefficent step 3 find out the sampling distribution above and

Hello, Can anyone suggest a simple way to generate a Kaplan-Meier plot with 2 survfit objects, just like this one:? https://drive.google.com/file/d/1fEcpdIdE2xYtA6LBQN9ck3JkL6-goabX/view?usp=sharing Suppose I have 2 survfit objects: fit1 is for the curve on the left (survtime has been truncated to the cutoff line: year 5), fit2 is for the curve on the right (minimum survival time is at the

You also need to reply-all so the mailing list stays in the loop. -- Sent from my phone. Please excuse my brevity. On December 19, 2017 4:00:29 PM PST, Timothy Axberg <axbergtimothy at gmail.com> wrote: >Sorry about that. Here is the code typed directly on the email. > >qe = (Qmax * Kl * ce) / (1 + Kl * ce) > >##The data >ce <- c(15.17, 42.15, 69.12, 237.7, 419.77)

Dear R users: Recently, I learn to use penalized logistic regression. Two packages (penalized and glmnet) have the function of lasso. So I write these code. However, I got different results of coef. Can someone kindly explain. # lasso using penalized library(penalized) pena.fit2<-penalized(HRLNM,penalized=~CN+NoSus,lambda1=1,model="logistic",standardize=TRUE) pena.fit2

Dear R users, Could somebody please help me to find a way of comparing nonlinear, non-nested models in R, where the number of parameters is not necessarily different? Here is a sample (growth rates, y, as a function of internal substrate concentration, x): x <- c(0.52, 1.21, 1.45, 1.64, 1.89, 2.14, 2.47, 3.20, 4.47, 5.31, 6.48) y <- c(0.00, 0.35, 0.41, 0.49, 0.58, 0.61, 0.71, 0.83, 0.98,

Hello, I am reading in a file called fit2.txt (Limma). fit2.txt has 38 columns but when I dim(fit2) I only get 6 columns. The first column that it does not read in is df.residual. fit2<-read.table(fit2, file="fit2.txt",sep="\t",quote="",comment.char="",as.is=TRUE) The first few lines of fit2.txt (does not include all 38 columns) looks like this:

Hello All, Does anyone know why length(fit1$time) < length(fit2$n) in survfit.coxph output? Why is the predicted time length is not the same as the number of samples (n)? I tried: example(survfit.coxph). Thanks, parmee > fit2$n [1] 241 > fit2$time [1] 0 31 32 60 61 152 153 174 273 277 362 365 499 517 518 547 [17] 566 638 700 760 791

Hello all, I wish to extract the terms from an rpart object. Specifically, I would like to be able to know what is the response variable (so I could do some manipulation on it). But in general, such a method for rpart will also need to handle a "." case (see fit2) Here are two simple examples: fit1 <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) fit1$call fit2 <-

Hey, I have stacked a couple of garchFit objects in a list with names $fit1, $fit2, ..., $fiti assigning objects names using a loop, i.e. after running the loop modelStack = list($fit1, $fit2,...,$fiti). Thus the following apply; a = modelStack$fit2, then a is the second garchFit object of formal class 'fGarch' with 11 slots, @call, @formula... etc. I then want to extract information in

I stumbled across a mild glitch when trying to compare the result of gam() fitting with the result of lm() fitting. The following code demonstrates the problem: library(gam) x <- rep(1:10,10) set.seed(42) y <- rnorm(100) fit1 <- lm(y~x) fit2 <- gam(y~lo(x)) fit3 <- lm(y~factor(x)) print(anova(fit1,fit2)) # No worries. print(anova(fit1,fit3)) # Likewise. print(anova(fit2,fit3)) #

Hi, I am trying to do logistic regression for data of 104 patients, which have one outcome (yes or no) and 15 variables (9 categorical factors [yes or no] and 6 continuous variables). Number of yes outcome is 25. Twenty-five events and 15 variables mean events per variable is much less than 10. Therefore, I tried to analyze the data with penalized regression method. I would like please some of the

Hi, I just encountered what I thought was strange behavior in MDS. However, it turned out that the mistake was mine. The lesson learned from my mistake is that one should plot on a square pane when plotting results of an MDS. Not doing so can be very misleading. Follow the example of an equilateral triangle below to see what I mean. I hope this helps others to avoid this kind of headache.

In R1.7 and above (including R 1.9 alpha), 'update.formula' forgets to copy any offset(...) term in the original '.' formula: test> df <- data.frame( x=1:4, y=sqrt( 1:4), z=c(2:4,1)) test> fit1 <- glm( y~offset(x)+z, data=df) test> fit1$call glm(formula = y ~ offset(x) + z, data = df) test> fit1u <- update( fit1, ~.) test> fit1u$call glm(formula = y ~ z,

Hey all, I seem to be having trouble fitting a spline to a large set of data using PSpline. It seems to work fine for a data set of size n=4476, but not for anything larger (say, n=4477). For example: THIS WORKS: ----------------------------- random = array(0,c(4476,2)) random[,1] = runif(4476,0,1) random[,2] = runif(4476,0,1) random = random[order(random[,1]),] plot(random[,1],random[,2])