similar to: error message - unexpected input

Hi, I appreciate you're busy so I just wanted to follow up on the email I sent you the other day, copy included below for reference. On Sat, Nov 12, 2016 at 12:40 PM, Danielle James <danielle.james at whoishostingthismail.com> wrote: Hi, I just wanted to let you know about a link that seems to be broken on this page http://www.hdt-project.org/. Here's the link

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

Hello, I am new to R. I often collect data at multiple sites and need to create separate graphs (such as scatterplots or histograms) of specific variables for each site. I have tried to do this by splitting the data frame and then using lapply, but it seems that the graphing commands cannot be called as functions. Here is a sample of my data, called "seeddist2": site

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:

Running dovecot 1.2.12 on ubuntu 10.10. Here is the problem: I have mysql backend for expire. I am using the dovecot-postfix package which creates /etc/dovecot/conf.d/01-mail-stack-delivery.conf which looks like this: # Some general options protocols = imap imaps managesieve disable_plaintext_auth = no ssl = yes ssl_cert_file = /etc/ssl/certs/ssl-mail.pem ssl_key_file =

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

Hi, I'm trying to do some basic social network analysis with igraph in R, but I'm new to R and haven't been able to find documentation on a couple basic things: I want to run igraph's community detection algorithms on a couple thousand small graphs but don't know how to automate igraph looking at multiple graphs described in a single csv file. My data look like something in

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

Hi, I would like to use the multinomRob function to test election results. However, depending on which independent variables I include and how many categories I have in the dependent variable, the model cannot be estimated. My data look like this (there are 68 observations): > head(database) RESTE09 GAUCHE09 PDC09 PLR09 UDC09 MCG09 RESTE05 GAUCHE05 PDC05 D1 1455

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

Hi,? I'm trying to obtain a table of coefficients and confidence intervals from a logistic regression analysis in R. My code is as follows: # read in csv file datafile<-read.csv("file.csv", row.names=1) # read in the variable list varlist<-names(datafile)[66:180] models<-lapply(varlist, function(x) {glm(substitute(outcome ~ i, list(i=as.name(x))), data=datafile, family =

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:-) 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:

Hello, I used the glm function in R to fit a dose-response relationship and then have been using dose.p to calculate the LC50, however I would like to calculate the NOEL (no observed effect level), ie the lowest dose above which responses start occurring. Does anyone know how to do this? [[alternative HTML version deleted]]

library(survival) library(boot) data=NULL lambda=NULL result=NULL pat=rep(1:102,each=1) trt=rep(c(1,0),51) status=rep(1,102) site=rep(1:51, each=2) nr.datasets=100 seed=2006 beta=log(1/2) for (i in 1:51) { lambda[i]=1+((3-1)/50)*(i-1)} lambda1=rep(lambda, each=2) dummy=rep(c(exp(beta),1),51) elf=lambda1*dummy r=70 #the number of bootstrap replicates