similar to: Unexpected behaviour when comparing (==) long quoted expressions

>>>>> Daniel Chen >>>>> on Fri, 12 Jul 2019 13:53:21 -0500 writes: > 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 <-

The deparse options used by default by 'deparse' and 'dput' are c("keepNA", "keepInteger", "niceNames", "showAttributes") but Defn.h still has #define DEFAULTDEPARSE 1089 /* KEEPINTEGER | KEEPNA | NICE_NAMES, used for calls */ i.e., with the SHOWATTRIBUTES bit turned off. Is that on purpose? Note that this leads to weird

When large calls cause errors R may stall for extended periods.? This is particularly likely to happen with `do.call`, as in this example with a 24 second stall: ??? x <- runif(1e7) ??? system.time(do.call(paste0, list(abs, x)))? # intentional error ??? ## Error in (function (..., collapse = NULL)? : ??? ##?? cannot coerce type 'builtin' to vector of type 'character' ??? ##

Luke, thanks for considering the issue.? I would like to try to separate the problem into two parts, as I _think_ your comments address primarily part 2 below: 1. How can we avoid significant and possibly crippling ?? stalls on error with these non-standard calls. 2. What is the best way to view these non-standard calls. I agree that issue 2. requires further thought and discussion under a

We also have a few other suggestions and wishes about backtrace storage and display on the one hand, and display of constructed calls on the other hand. Perhaps it would be better to open a different wishlist item for traceback() to keep the discussions focused? FWIW I think deparsing backtraces lazily is a great idea. Displaying 1 line per call by default in interactive sessions, while being

Hello everybody, I'm running R 1.8.1 on both Linux and OS X compiled with gcc 3.2.2 and 3.3, respectively. The following call seems to freeze the interpreter on both systems: > chisq.test(matrix(c(233, 580104, 3776, 5786104), 2, 2), simulate.p.value=TRUE) By freeze, I mean, the function call never returns (running > 10 hours so far), the process is unresponsive to SIGINT (but I

1) Definitely yes. They are on CRAN. Just type install.packages( c("dplyr", "tidyr" ) ) at the R console. 2) Don't know, but most likely the answer is yes. Since all R packages on Linux are compiled by R when installed, you either need to activate your virtual machine, compile the packages, and save the state for future use, or you have to build the package library on

Hi Team, Please suggest me on below :- 1).are There any linux based library for dplyr and tidyr? I could find these r-libraries for windows and Mac but not for linux. 2). is it possible to add external library to portableR zip(http://nafiux.github.io/portableR/) file if yes then how to add external library to poratbleR zip file? In our application AWS lambda invoking the R-script by using

Hi Jeff, 1). I have tried install.packages( c("dplyr", "tidyr" ) ) into the r-script file got below error:- > install.packages( c("dplyr", "tidyr" ) ) Installing packages into ?/tmp/1776492876238396447/lib/portableR-master/site-library? (as ?lib? is unspecified) Error in install.packages(c("dplyr", "tidyr")) : This version of R is

Hi all, Today I had an R CMD build that failed while building a vignette because the vignette needs tidyr (>= 1.0, declared in DESCRIPTION Suggests) but my system had a previous version installed. It did not take me too long to figure out the issue (solved by upgrading tidyr) but it would have been even faster / easier if R CMD build failed early, with an error message that says something

UseRs, When doing some data manipulations using the tidyverse, I am repeatedly getting the same error message in now three separate situations. I can write up a reproducible example, but want to lay out the high-level issues in case someone recognizes exactly what is happening here. The error is: Error in mut_env_parent(overscope$.top_env, lexical_env) : object 'rlang_mut_env_parent'

Buenas tardes, estoy haciendo una gráfica de múltiples lineas pero no he podido generar las legendas. Alguno de ustedes me podría colaborar. library(ggplot2) #### Con b=-2 t=seq (-4, 4, by=0.01) l=exp(t+2)/(1+(exp(t+2))) ##con b igual a -1 t=seq (-4, 4, by=0.01) o=exp(t+1)/(1+(exp(t+1))) ### Con b igual a 0.7 t=seq (-4, 4, by=0.01) i=exp(t-0.7)/(1+(exp(t-0.7))) ### Con b igual a 2 t=seq

Thanks for the feedback Jeff. Before I pursue a bug report, let me give a full example: ###### begin console output R version 3.4.1 (2017-06-30) -- "Single Candle" Copyright (C) 2017 The R Foundation for Statistical Computing Platform: i386-w64-mingw32/i386 (32-bit) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions.

This sounds an awful lot like a bug. Read the Posting Guide to know what to do about bugs. And delaying making the reprex is _always_ a bad idea. -- Sent from my phone. Please excuse my brevity. On August 14, 2017 7:26:32 AM PDT, "Szumiloski, John" <John.Szumiloski at bms.com> wrote: >UseRs, > >When doing some data manipulations using the tidyverse, I am repeatedly

¡Gracias! Espectacular. Me ha servido mucho. El mié., 18 de jul. de 2018 1:15 PM, Víctor Granda García < victorgrandagarcia en gmail.com> escribió: > Hola Sebastián. > > Entiendo que tratas de que aparezca una leyenda con el tipo de curva > (l,o,u,i). Si quieres aprovechar las ventajas de ggplot (como las leyendas > automáticas) normalmente tienes que asignar linetype a una

Thank you very much for the support. I have just used the reshape library and my problem was solved. Kind regards, Peter -----Original Message----- From: Jeff Newmiller [mailto:jdnewmil at dcn.davis.ca.us] Sent: 10 July 2017 03:52 PM To: r-help at r-project.org; Mangalani Peter Makananisa; r-help at r-project.org Cc: Vito Ricci (vito_ricci at yahoo.com) Subject: Re: [R] dplyr help I am

> On Aug 14, 2017, at 8:37 AM, Szumiloski, John <John.Szumiloski at bms.com> wrote: > > Thanks for the feedback Jeff. Before I pursue a bug report, let me give a full example: > > ###### begin console output > > R version 3.4.1 (2017-06-30) -- "Single Candle" > Copyright (C) 2017 The R Foundation for Statistical Computing > Platform:

Buenos días , por favor alguien me puede ayudar necesito separar en tres columnas una columna fecha que viene en el siguiente formato (01/01/1983) Muchas gracias Saludos [[alternative HTML version deleted]]

Hello, Inline. ?s 19:03 de 12/06/2024, Yuan Chun Ding via R-help escreveu: > I am sorry that I know I should provide a dataset that allows to replicate my problem. > > It is a research dataset and quite large, so I can not share. > > Both Bert and Tim guessed my problem correctly. I also thought about the conflicting issue between different packages and function masking. > I

I am pretty sure that this is not a question about dplyr... it is a question about tidyr. Look at the help file ?tidyr::spread. If I understand your question (I may not, because you gave no example of input/output data), the answer is no, the column names come from the column named by the key parameter and the values that fill those columns come from the column named by the value parameter.