similar to: `eval' and environment question

I initially thought, this should better be posted to r-devel but alas! no response. so I try it here. sory for the lengthy explanation but it seems unavoidable. to quickly see the problem simply copy the litte example below and execute f(n=5) which crashes. called with n != 5 (and of course n>3 since there are 3 parameters in the model...) everything is as it should be. in detail:

I've noticed the following oddity where capture.output() prevents eval() from evaluating an expression in the specified environment. I'm not sure if it is an undocumented feature or a bug. It caused me many hours of troubleshooting. By posting it here, it might save someone else from doing the same exercise. Start by defining foo() which evaluates an expression locally in a given

Hi R users! I was looking at some of the example code for the "environment" function. Here it is: e1 <- new.env(parent = baseenv()) # this one has enclosure package:base. e2 <- new.env(parent = e1) assign("a", 3, envir=e1) ls(e1) ls(e2) exists("a", envir=e2) # this succeeds by inheritance exists("a", envir=e2, inherits = FALSE)

Dear List, I'm using glmulti for model averaging in R. There are ~10 variables in my model, making exhaustive screening impractical - I therefore need to use the genetic algorithm (GA) (call: method = "g"). I need to include random effects so I'm using glmulti as a wrapper for lme4. Methods for doing this are available here

Hi, this relates to the question "How to set a former environment?" asked yesterday. What is the best way to to return a function with a minimal environment from a function? Here is a dummy example: foo <- function(huge) { scale <- mean(huge) function(x) { scale * x } } fcn <- foo(1:10e5) The problem with this approach is that the environment of 'fcn' does not

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

Hello, Suppose I have an expression, E, which accesses some variables present in an environment V. I do this via eval(E,envir=V) however all assignments end up in V. I would like the results of assignments in E to end up in the .GlobalEnv ? Or at least the calling environment. Is there a quick way to this instead of iterating over all objects E and assigning into .GlobalEnv? Thank you Saptarshi

I've been looking at the help page for eval for a while, but I can't make sense of why this example does not work. show.a <- function() { a } init.env <- function() { a <- 200 environment() } my.env <- init.env() ls(envir=my.env) # returns this: # > ls(envir=my.env) # [1] "a" # but this does not work: eval(expression(show.a()),envir=my.env) # >

I?m wondering which is the most efficient (time, than memory usage) way to obtain a multivariate time series object from a data frame (the easiest data structure to get data from a database trough RODBC). I have a starting point using timeSeries or xts library (these libraries can handle time zones), below you can find code to test. Merging parallelization (cbind) is something I?m thinking at

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 <-

x<-1 f<-function(){ x<-3 eval(substitute(x+y,list(y=10))) } f() #13 x<-1 f<-function(){ x<-3 eval(substitute(x+y,list(y=10)), envir = sys.frame(sys.parent())) } f() #11 x<-1 f<-function(){ x<-3 eval.parent(substitute(x+y,list(y=10))) } f()#11 the help page says: "If 'envir' is not specified, then 'sys.frame(sys.parent())', the

Would it be possible to have the value of eval() preserve the "visibility" of the value of the expression? "PROBLEM": # Invisible > x <- 1 # Visible > eval(x <- 2) [1] 2 "TROUBLESHOOTING": > withVisible(x <- 1) $value [1] 1 $visible [1] FALSE > withVisible(eval(x <- 2)) $value [1] 2 $visible [1] TRUE WORKAROUND: eval2 <-

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 mailing list, I have found some strange behaviour which I think relates to parent frames and eval. Can anyone explain what's going on here? First example: > test.parent.funcs_ function() { outer.var_ 5 subfunc1_ function() substitute( outer.var, envir=parent.frame()) print( subfunc1()) subfunc2b_ function() eval( quote( outer.var), envir=parent.frame()) print(

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

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 +

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

I am trying, without success, to find out how to formulate correctly the parameters of "eval". My code snippet looks like: proutside <- function(txt) { cat("\n",txt,"\n"); print(eval(parse(text = txt))) } vari <- function(Ob) { prininside <- function(txt) { cat("\n",txt,"\n"); print(eval(parse(text = txt))) }

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) + } > >