similar to: The default argument 'envir' of eval()

After many hours of debugging code, I came to the conclusion that I have a fundamental misunderstanding regarding eval, and hope that someone here can explain to me, why the following code acts as it does: foo <- function(expr) { eval(substitute(expr), envir=list(a=5), enclos=parent.frame()) } bar <- function(er) { foo(er) } > foo(a) [1] 5 > bar(a) Error in eval(expr,

R-users & helpers: I am using Amelia, mitools and cmprsk to fit cumulative incidence curves to multiply imputed datasets. The error message that I get "Error in eval(expr, envir, enclos) : invalid 'envir' argument" occurs when I try to fit models to the 50 imputed datasets using the "with.imputationList" function of mitools. The problem seems to occur

my understanding of `eval' behaviour is obviously incomplete. my question: usually `eval(expr)' and `eval(expr, envir=parent.frame())' should be identical. why does the last `eval' (yielding `r3') in this code _not_ evaluate in the local environment of function `f' but rather in the global environment (if `f' is called from there)?

I am absolutely new to R and I am aware of only a few basic command lines. I was running a robust regression in R, using the following command line library (MASS) rfdmodel1 <- rlm (TotalEmployment_2004 ~ MISSISSIPPI + LOUISIANA + TotalEmployment_2000 + PCWhitePop_2004 + UnemploymentRate_2004 + PCUrbanPop2000 + PCPeopleWithACollegeDegree_2000 + PCPopulation.of.or.over.65.years.of.age_2004)

> datafilename="E:/my documents/r/sex/bysex1.csv" > data.sex=read.table(datafilename,header=T) > data.sex y.sex.age.region.c.n 1 1980,F,A,N,-18.15,13.61 2 1980,F,A,N,-18.61,13.04 3 1980,F,A,N,-18.81,12.32 4 1990,F,A,N,-21.12,11.7 5 1990,F,A,N,-20.77,11.58 6 1990,F,A,N,-21.6,13.34 7 1990,F,A,N,-21.78,12.6 > model.anova<-aov(c~age*sex,data=data.sex)

Dear friends, I am testing glm as at page 514/515 of THE R BOOK by M.Crawley, that is on proportion data. I use glm(y~x1+,family=binomial) y is a proportion in (0,1), and x is a real number. I get the error: In eval(expr, envir, enclos) : non-integer #successes in a binomial glm! But that is exactly what was suggested in the book, where there is no mention of a similar warning. Where am I

Sorry to intervene. Argument passed to 'eval' is evaluated first. So, eval(x <- 2) is effectively like { x <- 2; eval(2) } , which is effectively { x <- 2; 2 } . The result is visible. eval(expression(x <- 2)) or eval(quote(x <- 2)) or evalq(x <- 2) gives the same effect as x <- 2 . The result is invisible. In function 'eval2', res <-

Hi R-listers, I am trying to make a trellis boxplot with the HSuccess (y-axis) in each Rayos (beach sections) (x-axis), for each Aeventexhumed (A, B, C) - nesting event. I am not able to do so and keep receiving: Error in eval(expr, envir, enclos) : object 'Rayos' not found Please advise, Jean require(plyr) resp <- read.csv("ABC Arribada R File Dec 12 Jean

Hi R-listers, I am trying to better undertand what we would call "functional paradigm" use of S/R to better map my programming activities in other languages. This little function is aimed to create an object (at the end end, it would have it's own class): -- myObject =function(){ list( a=1, foo=function(b) { cat("b:",b)

I am working with a largish dataset of 25k lines and I am now tying to use predict. pred = predict(cuDataGlmModel, length + meanPitch + minimumPitch + maximumPitch + meanF1 + meanF2 + meanF3 + meanF4 + meanF5 + ratioF1ToF2 + rationF3ToF1 + jitter + shimmer + percentUnvoicedFrames + numberOfVoiceBreaks + percentOfVoiceBreaks + meanIntensity + minimumIntensity + maximumIntensity +

Is this a bug ? > foo <- function(a, ...) + { + tl <- substitute(list(a=a)) + + ## the foll two should give the same results + ## according to defaults for eval + + print(eval(tl )) + print(eval(tl, parent.frame())) + } > > > foo1 <- function(...) + { + foo(...) + } > > test <- function(a = 6) + { + foo1(a = a) + } > >

I'm running an LSODA to generate some graphs, but I need to stop at a certain point and use those values to generate another LSODA output. This is working fine, but when I try to run the second LSODA, I get the "Error in eval(expr, envir, enclos) : object 'N' not found". Any ideas what can be causing this? I have no object 'N' anywhere in the script. I made an

I have a for-loop in my code that calls another .R file: source("estimation.R") This file runs through without any problems, so the program completes the loop one time. However, when the loop starts a second time and it comes time to call the file "estimation.R" again, program stops and prints the following error message: "Error in eval.with.vis(expr, envir, enclos) :

Dear all I tried to use regression to predicted mu data, but it has error like this: > IWJR.complete x y [1,] 33.17635 2.4705021 [2,] 81.61225 3.3815620 [3,] 65.47392 1.6518975 [4,] 57.97806 1.6110785 [5,] 76.05528 2.1601246 [6,] 41.36090 1.5498132 [7,] 68.77844 2.8078691 [8,] 55.57040 2.1183063 [9,] 41.29287 1.8015709 [10,] 65.43935 2.3483183 [11,] 22.44821

Dear all, I have a problem when passing parms from one function to another when the argument list is just '...'. Consider this example: foo<-function(){ xx <- 111222 bar(x=xx) } bar <- function(...){ cl <- match.call(expand.dots=TRUE) print(cl) x <- eval(cl$x) print(x) } foo() > bar(x = xx) > Error in eval(expr, envir, enclos) : Object "xx" not

Hi there I'm trying to fit a logistic regression model to data that looks very similar to the data in the sample below. I don't understand why I'm getting this error; none of the data are proportional and the weights are numeric values. Should I be concerned about the warning about non-integer successes in my binomial glm? If I should be, how do I go about addressing it? I'm

Hi, I am intending to save a path-describing character object in a slot of a class I'm working on. In order to have the option to use "system.file" etc in such string-saved path definitions, I wrote this ExpressionEvaluator <- function(x){ x <- tryCatch( expr=base::parse(text=x), error = function(e){return(as.expression(x))}, finally=TRUE) return(x) } This

Why does the following eval() call on sys.on.exit() not return what I expect/evaluate in the proper environment? foo <- function() { cat("foo()...\n"); on.exit( message("exiting") ) cat("sys.on.exit():\n") res <- sys.on.exit() print(res) cat("eval(sys.on.exit()):\n") expr <- quote(sys.on.exit()) print(expr) res <- eval(expr)

Hi everybody: I need to get the names of the arguments in an object of class "expression". I've got the following expression: > x <- expression(rho * cos(omega)) Is there any function or procedure that returns the names of the arguments (in the example: "rho" and "omega")? I tried a rough approach implemented in the function expr.args() shown below. As

Dear list, A while ago, I posted a question asking how to use data or subset arguments in a user-defined function. Duncan Murdoch suggested the following solution in the context of a data argument: data <- data.frame(a=c(1:10),b=c(1:10)) eg.fn <- function(expr, data) { x <- eval(substitute(expr), envir=data) return(mean(x)) } eg.fn(a,data) I've