similar to: unevaluated expression

Hi, How to turn a character string into an unevaluated R object? I want to load some files in a directory into data matrix R objects. I could do this with read.table and assign (see below). Then, I want to turn the character string representing a file name (the evaluated expression of i) into an unevaluated R object. Basically, I want to create matrices whose names are the same as the related file

Hi R-users, I would like to create an expression without evaluating it. Then paste that expression to an object. Example: Result <- paste('Result', 1, sep="") paste(Result, substitute(apply(exp.des[1:10,], 1, one.row, parms=parameters)), sep="<-") However this pastes EACH element of the unevaluated expression. Instead I just would like the expression

Hi, I'm trying to get ...as a list of unevaluated arguments, ie. substitute(list(...)) gives me an unevaluated list of the arguments, but I want a list of the unevaluated arguments. My attempts so far: (function(...) substitute(...))(a=1, b=a) # Only returns first (function(...) substitute(list(...)))(a=1, b=a) # Unevaluated list, not list of unevaluated (function(...)

I would like R CMD build to run some R code which does some stuff and save the result as a file in /inst/docs folder. Is there any way of doing this. Thank you. Johan ____________________________________________________________________________________ We won't tell. Get more on shows you hate to love [[alternative HTML version deleted]]

HI, As I'm trying to compute Taylor series, I'm having problems in adding and multiplying unevaluated expressions. I searched for a solution but found none. my Taylor function works fine for evaluating functions as you can see here: rTaylorVal=function(exp,x0,dx,n) { ls=list(x=x0) newexp=eval(exp,ls) exp0=exp for (i in 1:n){ exp0=D(exp0,"x")

I have a question about "deparse" function in R What is the reason that "deparse" use an argument like "width.cutoff" ? Why the maximum cutoff is 500? I was manipulating an R formula and used "deparse". Since the length of user's formula was greater then 500, my code didnt work. thanks Johan johan Faux <johanfaux@yahoo.com> wrote: I have a

Hi, I have a S3 package with namespace called "myS3Pkg". Inside my package I would like to create a S4 class which extends (adds 2 slots) another S4 class from some other package. The class should be created in "myPkg" environment (and not global environment). Using: setClass("myS4class", representation("otherS4class", mydata = "numeric"),

I would like to turn off all the messages during library(aPackage) or require(aPackage) I tried different commands: invisible, capture.output, sink but none of them is working. For example, loading VGAM, gives a lot of unnecessary messages: > library(VGAM) Attaching package: 'VGAM' The following object(s) are masked from package:splines : bs The

Hi, I am looking for a way to find out the last expression that was entered by the user, similar to ".Last.value", but for the unevaluated expression instead of the evaluated one. Example: > x <- 4 > x*x + 3 [1] 19 > .Last.value # that's the evaluated last expression [1] 19 # but I am looking for the unevaluated expression (".Last.expression", so to

Hello , a<-c(1) for(i in 2:length(a)) do.something with a[[i]] I get : Error in a[[i]] : subscript out of bounds Am I missing something here? Doesnt R check the value of i inside "for" and if the condition is not tru, dont do anything ???? thanks, johan --------------------------------- Got holiday prints? See all the ways to get quality

This may have been asked before, but is there an elegant way to check whether an variable/argument passed to a function is a "parse tree" for an (unevaluated) expression or not, *without* evaluating it if not? Currently, I do various rather ad hoc eval()+substitute() tricks for this that most likely only work under certain circumstances. Ideally, I'm looking for a isParseTree()

-------- Original-Nachricht -------- Betreff: Re: [R] Adding and Multiplying two Unevaluated Expressions Datum: Tue, 01 Dec 2009 23:49:39 +0100 Von: Benjamin M?ller <ben_mueller.bm at web.de> An: Rolf Turner <r.turner at auckland.ac.nz> Referenzen: <20091201144125.316310 at gmx.net> <8E40E49F-E8FC-4FBD-8CC5-93789FFB0E53 at auckland.ac.nz> This works fine for your

I finally upgraded to Firefox 3.05 from 2.x, and now I can reproduce a bug a colleague has been complaining about but which I hadn't been able to reproduce before. In Windows, set Firefox as the default browser. Then in Rgui (seems to affect all versions up to R-devel), use the menu to open HTML Help, and choose "Search Engine and Keywords". Enter a keyword (e.g. plot), and

Interestingly, as.list(substitute(...())) also works. On Sun, Aug 12, 2018 at 1:16 PM, Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > On 12/08/2018 4:00 PM, Henrik Bengtsson wrote: >> >> Hi. For any number of *known* arguments, we can do: >> >> one <- function(a) list(a = substitute(a)) >> two <- function(a, b) list(a = substitute(a), b =

Duncan, Thank you for the clarification on how delayedAssign works. Should R-level interfaces to promise objects ever become available, I expect they would at time come in handy. On the subject of substitute and delayedAssign, I do have a follow-up question for the list. I'm trying to convert a named list of expression objects into an environment of promise objects. After conversion, each

Hi, Consider the following (much simplified) Sweave example: -------------- First, we set the value of $x$: <<chunk1,eval=FALSE>>= x <- 1 @ Then we set the value of $y$: <<chunk2,eval=FALSE>>= y <- 2 @ Thus, the overall algorithm has this structure: <<combined,eval=FALSE>>= <<chunk1>> <<chunk2>> @

Folks: Herein is a suggestion for a little R convenience function mainly to obtain unevaluated ... function arguments. It arose from a query on R-help on how to get these arguments. The standard (I think) idiom to do this is via match.call(expand.dots=FALSE)$... However, Bill Dunlap pointed out that this repeats the argument matching of the function call and suggested a couple of alternatives

I am sure it is just me not understanding how R works, but could somebody explain why curve(cos(x)) works and curve(expression(cos(x)) does not? I have done some investigating and here is what I found. If I comment out the line of curve indicated below, both calls work fine. function (expr, from, to, n = 101, add = FALSE, type = "l", ylab = NULL, log = NULL, xlim =

Hi all, ##################################### integ=function(f,n){ # computes vector y = f(x) x=runif(10000) y=f hatI=mean(y) hatI } # example of use integ(x^2+x+100,100) ##################################### it returns an error says no obj 'x' how do I 'tell' R to treat 'f' as an input expression? Thanks. casper -- View this message in context:

R-devel, I used the 'substitute' function to create labels for objects inside an environment, without actually evaluating the objects, as the objects might be promises. However, I was surprised to see that 'substitute' returns the expression slot of the original promise even after the promise has been forcibly evaluated. (Doesn't the promise go away after evaluation?) This