similar to: on.exit() & sys.on.exit(): Calling them via eval() does not work as hoped

(Apologies if this is better suited for R-help.) On my system (macOS Sierra, late 2014 MacBook Pro; R 3.4.1, Homebrew build), I found that it is faster to construct a function using eval(call("function", ...)) than using as.function(list(...)). Example: make_fn_1 <- function(a, b) eval(call("function", a, b), env = parent.frame()) make_fn_2 <- function(a, b)

I made the following example to see what are the difference between expression and quote. But I don't see any difference when they are used with eval()? Could somebody let me know what the difference is between expression and quote? expr=expression(2*3) quo=quote(2*3) eval(expr) str(expr) class(expr) typeof(expr) mode(expr) attributes(expr) eval(quo) str(quo) class(quo) typeof(quo)

On Sat, Oct 15, 2016 at 2:00 AM, Martin Maechler <maechler at stat.math.ethz.ch> wrote: >>>>>> Michael Lawrence <lawrence.michael at gene.com> >>>>>> on Wed, 12 Oct 2016 15:21:13 -0700 writes: > > > Thanks, this was what I expected. There is a desire to > > eliminate the usage of pairlist from user code, which > >

In ?eval, it says ? ? ... ?If ?envir? is not ? ? specified, then the default is ?parent.frame()? (the environment ? ? where the call to ?eval? was made). I tried the following example with "eval(expr)" ?and "eval(expr, parent.frame())" in f(). The results are different, which are not consistent with the help. Could somebody let know whether I misunderstand the help? Or there

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, I have a script which I source, which evaluates a changing expression call hundreds of times. It works, but it prints to screen each time, which is annoying. There must be simple way to suppress this, or to use a slightly different set of commands, which will be obvious to those wiser than I... Here is a simpler mockup which shows the issue: x = data.frame(rbind(c(1,2,3),c(1,2,3)))

Dear all, I have the following problem. Given an expression object 'expr' containing a certain set of symbols (say 'a', 'b', 'c'), I would like to translate the expression object in an R function of, say, 'a', programmatically. Here an example of what I mean. Given: > expr <- expression(a+b+c) a call like: > asFunctionOf(expr, 'a',

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,

Hi, is there a way to retrieve the line number of where en error occurred when sourcing a file in a tryCatch statement? Is it stored somewhere accessible? It is not found in the error object. Consider the following code/output and note the difference in the traceback between source (has line number) and sys.source (has no line number). Thank you, Renaud ######## # code ######## codefile <-

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)

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)?

> 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

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 $ :

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 <-

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) + } > >

In the following example, the inner evaluation pulls in the global value of subset (a function) rather than the one I thought I was passing in (a vector). Can anyone help me understand what's going on, and what I need to do to fix the problem? f0 <- function(formula, data, subset, na.action ) { f1(formula, data, subset,

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 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

Full_Name: Wacek Kusnierczyk Version: 2.10.0 r48269 OS: Ubuntu 8.04 Linux 32 bit Submission from: (NULL) (129.241.199.164) In the following example (and many other cases): quote(a=1) # 1 the argument matching is apparently incorrect wrt. the documentation (The R Language Definition, v 2.8.1, sec. 4.3.2, p. 23), which specifies the following algorithm for argument matching: 1. Attempt to