similar to: bug in eval

Full_Name: Anantha Prasad Version: 1.1.1 OS: Linux Submission from: (NULL) (199.131.134.30) I am trying to convert some S-PLUS code to R (a tcl/tk application that uses R)... here is the error I got in R (but not in S-PLUS)...so I am wondering if it is a bug. Eg., the foll. extract from a function runs fine in S-PLUS but gives the error: Error in x[[j]] : subscript out of bounds in R code

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

I've got a data frame with two numeric variables, df$flow and df$flow1. > tl <- lm(flow~flow1,df,na.action=na.exclude) > tlo <- loess(flow~flow1,df,na.action=na.exclude) Both loess and a simple linear model fit the data well. summary(tl) and summary(tlo) seem reasonable. As do plots such as: plot(predict(tl),df$flow) plot(predict(tlo),df$flow) I want to replace missing values

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)

Hello, It seems that the parsing information attached to expressions parsed by the parse() function when keep.source=TRUE is not provided to the eval() function. Please consider this code: path <- tempfile() code <- '(function() print( str( sys.calls() ) ))()' writeLines(code, path) sys.source(path, envir=globalenv(), keep.source=TRUE) > OUTPUT: Dotted pair list of 4 $ :

> 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)

Here is a case where the current scheme fails: > with(datasets::mtcars, xyplot(mpg~wt|gear)$call) xyplot(substitute(expr), data, enclos = parent.frame()) Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, May 11, 2017 at 1:09 AM, Deepayan Sarkar <deepayan.sarkar at gmail.com> wrote: > On Wed, May 10, 2017 at 2:36 AM, William Dunlap via R-devel > <r-devel at

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))) }

I don't have time now to investigate myself, and I'm not feeling like deciding myself if the following is a bug: myplot <- function(dat, cex = 1.2, ...) { if(!is.data.frame(dat <- as.data.frame(dat))) stop("`dat' must be a data.frame") if(any(is.na(match(c("x","y"), names(dat))))) stop("`dat' must have a `x' and a

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 +

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)?

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

Hello. I have run the command imp<-mice(mydata, im=c("","pmm","logreg","logreg"),m=5)  for a variable with no missing data, a numeric one and two variables with binary data. I got the following message: There were 37 warnings (use warnings() to see them) > warnings() Warning messages: 1: In any(predictorMatrix[j, ]) ... : coercing argument of

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

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,

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

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(

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