similar to: Bug in comparison of language objects?

Hi everyone: I?m one of the interns at RStudio this summer working on a project that helps teachers grade student code. I found an unexpected behaviour with the |==| operator when comparing |quote|d expressions. Example 1: |u <- quote(tidyr::gather(key = key, value = value, new_sp_m014:newrel_f65, na.rm = TRUE)) s <- quote(tidyr::gather(key = key, value = value,

My points: - The 'withCallingHandlers' construct that is used in current 'stopifnot' code has no effect. Without it, the warning message is the same. The overridden warning is not raised. The original warning stays. - Overriding call in error and warning to 'cl.i' doesn't always give better outcome. The original call may be "narrower" than 'cl.i'. I

A private reply by Martin made me realize that I was wrong about stopifnot(exprs=TRUE) . It actually works fine. I apologize. What I tried and was failed was stopifnot(exprs=T) . Error in exprs[[1]] : object of type 'symbol' is not subsettable The shortcut assert <- function(exprs) stopifnot(exprs = exprs) mentioned in "Warning" section of the documentation similarly fails

Dear R List, I encountered a serious problem regarding the registration of ".C" when following the documentation "Writing R Extensions" that leads to a segmentation fault (tested on windows and mac os x). The registration mechanism for ".C" routines via R_registerRoutines and the R_CMethodDef structure has been enhanced recently with the addition of two fields, one

Below is code that introduces a list comprehension syntax into R, allowing expressions like: > .[ sin(x) ~ x <- (0:11)/11 ] [1] 0.00000000 0.09078392 0.18081808 0.26935891 0.35567516 0.43905397 [7] 0.51880673 0.59427479 0.66483486 0.72990422 0.78894546 0.84147098 > .[ .[x*y ~ x <- 0:3] ~ y <- 0:4] [,1] [,2] [,3] [,4] [,5] [1,] 0 0 0 0 0 [2,] 0 1 2

Another possible shortcut definition: assert <- function(exprs) do.call("stopifnot", list(exprs = substitute(exprs), local = parent.frame())) After thinking again, I propose to use ??? ? ? stop(simpleError(msg, call = if(p <- sys.parent()) sys.call(p))) - It seems that the call is the call of the frame where stopifnot(...) is evaluated. Because that is the correct context, I

I ran into this and found the result very surprising: identical( quote({ a }), quote({ a }) ) # FALSE It seems related to curly braces. For example, parens work fine: identical( quote(( a )), quote(( a )) ) # TRUE Is this expected behavior? I can't seem to find anything in the help for identical that relates to this. -Winston

Full_Name: Juan Gea Version: R version 2.6.2 OS: Fedora Core 6 Submission from: (NULL) (79.153.48.49) Abort: objeS <- matrix("AAA",1000000) class(objeS) outTxt <- textConnection("vaClob", open = "w", local = FALSE) dput(objeS,outTxt) close(outTxt) R version 2.6.2 (2008-02-08) Copyright (C) 2008 The R Foundation for Statistical Computing ISBN

I put the idea below into a function that gives nicer looking results. Here's the new code: dupnames <- function(path = ".") { Rfiles <- pkgload:::find_code(path) allnames <- data.frame(names=character(), filename=character(), line = numeric()) result <- NULL for (f in Rfiles) { exprs <- parse(f, keep.source = TRUE) locs <-

Dear R developers, while experimenting with repeated measures ANOVA, I found out that it is possible to construct a model specification that is syntactically valid, but causes aov() to abort with an error. A minimal reproducer and its output are attached to this mail. I was able to reproduce this problem with the latest SVN revision. The root cause is similar to that of bug 15377: aov()

>From https://github.com/HenrikBengtsson/Wishlist-for-R/issues/70 : ... and follow up note from 2018-03-15: Ouch... in R-devel, stopifnot() has become yet 4-5 times slower; ... which is due to a complete rewrite using tryCatch() and withCallingHandlers(). >From https://stat.ethz.ch/pipermail/r-devel/2017-May/074256.html , it seems that 'tryCatch' was used to avoid the following

I am more than a little puzzled by your question. In the construct {expr1; expr2; expr3} all of the expressions expr1, expr2, and expr3 are evaluated, in that order. That's what curly braces are FOR. When you want some expressions evaluated in a specific order, that's why and when you use curly braces. If that's not what you want, don't use them. Complaining about it is like

Hi, Vince and I have noticed a problem with non-syntactic names in data frames and some modeling code (but not all modeling code). The following, while almost surely as documented could be a bit more helpful: m = matrix(rnorm(100), nc=10) colnames(m) = paste(1:10, letters[1:10], sep="_") d = data.frame(m, check.names=FALSE) f = formula(`1_a` ~ ., data=d) tm =

Dear List members, I really, like the feature that one can call R functions with mathematical expressions, e.g. curve(x^2, 0, 1) I wonder, how I can construct in a simple way a function like mycurve = function (expr) {...} such that that a call mycurve(x^2) has the same effect as the call curve(x^2, -100,100) Below is some code that works, but it seems much to complicated: it first

Dear members, I have the following code: > TB <- {x <- 3;y <- 5} > TB [1] 5 It is consistent with the documentation: For {, the result of the last expression evaluated. This has the visibility of the last evaluation. But both x AND y are created, but the "return value" is y. How can this be advantageous for solving practical problems?

Unless you do something special within a function, only the value(s) returned are available to the caller. That is the essence of functional-type programming languages. You need to read up on (function) environments in R . You can search on this. ?function and its links also contain useful information, but it may too terse to be explicable to you. There are of course many available references on

I was initially pretty shocked by the result in this question: https://stackoverflow.com/questions/57527434/when-do-i-need-parentheses-around-an-if-statement-to-control-the-sequence-of-a-f Briefly, the following returns 0, not 3 as might be expected: if (TRUE) { 0 } else { 2 } + 3 At first I thought it the question was simply one of syntax precedence, but I believe the result is too

I suspect akshay is (or was? Not sure) unclear about what braces do. They are not closures... they create an expression that wraps multiple expressions into one expression... they are a little more like parentheses than closures. They are not intrinsically associated with creation of environments for holding variables. Functions are closures... they run in a call-specific environment that

>>>>> Henrik Bengtsson <henrik.bengtsson at gmail.com> >>>>> on Sat, 24 Sep 2016 11:31:49 -0700 writes: > Martin, did you post your code for withAutoprint() anywhere? > Building withAutoprint() on top of source() definitely makes sense, > unless, as Bill says, source() itself could provide the same feature. I was really mainly asking

Hello, I am running a program in r that calls a function, which calls another function, which calls another etc. These functions are of the form: example<- function(x,y,z) {x, y, and z are defined within curly braces like this} Here's my question. To start the main function, I input as an initial parameter a matrix of regressors of the form: MyMatrix<-cbind(this.one,that.one)