similar to: bootstrap validation of LR error message

I have an R script with the following applicable lines: xshort <- window(s, start=st, end=ed) . . . xshort <- ts(xshort, frequency=1, start=1) . . . m1 <- m2 <- m3 <- m4 <- m5 <- m6 <- NULL m1 <- tslm(xshort ~ trend) I get an error: Error in get(dataname) : object 'xshort' not found When I do traceback() I get: 3: get(dataname) 2: tslm(xshort ~

Hey everybody, i am using the rugarch-package and its great! I have a pretty easy problem, but i just dont get it, so thanks if you can help me. Normally i use: / data(DATANAME) spec = ugarchspec() fit = ugarchfit(data = x[,1], spec = spec) fit slotNames(fit) names(fit at fit) coef(fit) infocriteria(fit) likelihood(fit) nyblom(fit) signbias(fit) head(as.data.frame(fit)) head(sigma(fit))

At 10:37 PM 5/1/00 -0400, E. S. Venkatraman wrote: >I have the following problem. I have survival data (time, status) along >with several covariates (X1, X2,..., Xn). I want to fit a Cox model for >each of the covariate (univariately) and obtain the fitted probability of >survival at a fixed time point t0 and covariate value Xi0. I tried to do >this in a for loop where the index

Can anyone tell me what I might be doing incorrectly for an ordinal logistic regression for lrm? I cannot get R(2.9.1)to run either summary nor will it let me bootstrp to validate. ### Y is a 5 value measure with a range from 1-5, the independent variables are the same. N=75 but when we knock out the NAs it comes down to 51#### > lrm(formula = Y ~ permemp + rev + gconec + scorpstat, data =

I have been using R the past couple of years to run models on data we are collecting. I recently got a new computer and updated to a new version of R (2.60 -> 2.90). Since the update, I cannot get my syntax to run. I have tried copying the file it is looking for into many different directories to try and run it. In the last version, I found that it was easiest if the file was copied into the R

I am writing a function where the arguments are names of objects or variable names in a data frame. To convert the strings to the objects I am using eval(parse(text=name)): f.graph.two.vbs<-function(dataname,v1){ val<-paste(dataname,v1,sep="$") val<-eval(parse(text=val)) val } However running this returns an error:

Hi, I was wondering whether this query was addressed on how to perform validation through boostrapping. I am currently trying to implement a boostrapping approach to validation but don't know where to start. Help please. Thank you and Regards, Vivienne Ozohili Risk Model Validation Manager Group Risk Independent Control Unit A5, Bank of Ireland Head Office Lower Baggot Street Dublin 2 Tel:

The generic rule is that R is not a macro language, so looping of names of things gets awkward. It is usually easier to use compound objects like lists and iterate over them. E.g. datanames <- paste0("aa_", 2000:2007) datalist <- lapply(datanames, get) names(datalist) <- datanames col1 <- lapply(datalist, "[[", 1) colnum <- lapply(col1, as.numeric) (The 2nd

Hi R-users! Being new to R, and a fairly advanced Stata-user, I guess part of my problem is that my mindset (and probably my language as well) is wrong. Anyway, I have what I guess is a rather simple problem, that I now without success spent days trying to solve. I have a bunch of datasets imported from Stata that is labelled aa_2000 aa_2001 aa_2002, etc. Each dataset is imported as a matrix, and

Hi! Thanks for the replies! I understand people more accustomed to R doesn't like looping much, and that thinking about loops is something I do since I worked with Stata a lot. The syntax from Peter Dalgaard was really clever, and I learned a lot from it, even though it didn't solve my problem (I guess it wasn't very well explained). My problem was basically that I have a data matrix

Um, if you insist on doing it that way, at least use assign(varname, as.vector(get(varname))) -pd > On 4 Dec 2017, at 22:46 , Love Bohman <love.bohman at sociology.su.se> wrote: > > Hi! > Thanks for the replies! > I understand people more accustomed to R doesn't like looping much, and that thinking about loops is something I do since I worked with Stata a lot. The

Dear list Is there anywhere I could find further information on how to interpret the output for a logistic regression for validate() from Design package?. I tried ?validate and google but I cannot find information on what the rows and the columns represent. Thanks David

Dear R-devel, I've made some potential extensions to writeForeignSAS in 'foreign' that I wanted to pass along if anyone is interested. I've attached the diff -u output against the version found in foreign_0.8-15 and an .R file with my changes. (In this .R file, the function is named writeForeignSAS7 to simplify testing/comparisons.) I've tried to alter the current

I have a dataframe with names in the first column and wait times between decisions in the second column. Since individuals make multiple decisions, I want the average for each individual. For example, the data might look like this name wtime jo 1 jo 2 jo 1 jo 3 tim 3 tim 2 tim 2 ro 1 ro 2 etc. I'm hoping there is something like mean(dataname$wtime[name]) which will just

I am trying to plot survival curves using the following code as an example: >rs1799964.coxph<-(coxph(Surv(sassurvmonths,status)~age+stage+rs1799964_TNFA,method="efron")) >plot(rs1799964.coxph,lyt=c(1,3),xlab="Survival in Months",ylab="Proportion Surviving") I am gettingthe following error message: >Error in xy.coords(x, y, xlabel, ylabel, log) :

I am trying to obtain the AICc after performing logistic regression using the Design package. For simplicity, I'll talk about the AIC. I tried building a model with lrm, and then calculating the AIC as follows: likelihood.ratio <- unname(lrm(succeeded~var1+var2,data=scenario,x=T,y=T)$stats["Model L.R."]) #Model likelihood ratio??? model.params <- 2 #Num params in my model AIC

Does anyone know how to get the C-index from a logistic model - not using the dataset that was used to train the model, but instead using a fresh dataset on the same model? I have a dataset of 400 points that I've split into two halves, one for training the logistic model, and the other for evaluating it. The structure is as follows: column headers are "got a loan" (dichotomous),

I am new to using R. Currently, I am using the logistf package to run logistic regression analysis. When I run the following line of code: attach(snpriskdata) logisticpaper<-logistf(sascasecon~saspackyrs+newsbmi+EDUCATION+sasagedx+sasflung+condobst+sasadultasprev) I get the following error message: Error in sum(y) : invalid 'type' (character) of argument What does this error

:-) I don't insist on anything, I'm just struggling to learn a new language and partly a new way of thinking, and I really appreciate the corrections. I hope I someday will be able to handle lists in R as easy as I handle loops in Stata... Thanks again! Love -----Ursprungligt meddelande----- Fr?n: peter dalgaard [mailto:pdalgd at gmail.com] Skickat: den 4 december 2017 23:09 Till:

Hi, I'm developing an experiment with logistic regression. I've come across the lrm function in the Design library. While I understand and can use the basic functionality, there are a ton of options that go beyond my knowledge. I've carefully read the help page for lrm, but don't understand many of the arguments and optional return values. (penalty, penalty.matrix,