similar to: request for additional test in R CMD check

This is still not repaired in R version 3.6.0 Patched (2019-05-17 r76528) > library(latticeExtra) > a <- xyplot(1 ~ 1) > c(a,a) Warning message: In formals(fun) : argument is not a function Can we have it in R-3.6.1 that Peter just announced? Rich On Mon, Apr 2, 2018 at 4:08 AM Deepayan Sarkar <deepayan.sarkar at gmail.com> wrote: > > On Fri, Mar 23, 2018 at 7:58 AM,

Thanks, Rich. I thought of that, too, but it violates the spirit of my restraints (to avoid character strings), which I unfortunately did not clearly articulate. So my apologies for that failure. My concern is that with more complex model formula, using as.formula, etc. to parse/convert character strings can get a bit hairy. But in most cases, as here maybe, it may be perfectly fine. So think of

Hi, I am trying to change a SAS macro to R. here is my code. I get an error at the last line. attach(fram) dset1<-cbind(AGE,BMI,DEATH) > BMIGRP<-cut(BMI,breaks=3,right=TRUE) > AGEGRP<-floor(AGE/10)-2 > dset<-cbind(AGEGRP,BMIGRP,DEATH) > maxage<-max(dset[,1]) > minage<-min(dset[,1]) > #maxcls<-dset[,2] > #mincls<-dset[,2] >

This problem is still present in > version _ platform x86_64-w64-mingw32 arch x86_64 os mingw32 system x86_64, mingw32 status Under development (unstable) major 4 minor 0.0 year 2019 month 12 day 03 svn rev 77513 language R version.string R Under development (unstable)

Hello, I am in Win-XP R:2.3.0 latest rtools and Perl - of today I got Rcmdr.HH source code and tried to compile it myself copy all directory to R/R-2.3.0/src/library/Rcmdr.HH from R/R-2.3.0/src/library I typed: ..\..\bin\R CMD build --force --binary --auto-zip Rcmdr.HH * checking for file 'Rcmdr.HH/DESCRIPTION' ... OK * preparing 'Rcmdr.HH': * checking DESCRIPTION meta-information

my take of the assignment was to avoid 'parse' specifically. we start with a character vector, so avoiding characters is not possible. i was dealing with the fortune "if parse is the answer, you have the wrong question" Sent from my iPhone On Mar 29, 2025, at 15:39, Bert Gunter <bgunter.4567 at gmail.com> wrote: ? Thanks, Rich. I thought of that, too, but it violates

Hello, I thought of answering "reformulate can solve the problem" but how do you create quadratic terms with reformulate? ~(Heigh + Ho + Silver + Away)^2 is still a problem with no solution that I know of but paste/as.formula. Or Bert's bquote or substitute. Rui Barradas ?s 23:18 de 29/03/2025, Ebert,Timothy Aaron escreveu: > The general formula is y ~ a + b + c + ... >

# R for Windows will not send your bug report automatically. # Please copy the bug report (after finishing it) to # your favorite email program and send it to # # r-bugs at r-project.org # ###################################################### The R documentation for "is.R" says "## 'which()' only exists in R:" This is no longer true. In S-Plus 8.0 >

Hi Everyone, I am stuck with a simple problem. Suppose I have a vector x, and I want to calculate y[i]=3Dx[i+1]-x[i], it is very easy. I just need to write y<-x[2:length(x)]-x[1:length(x)-1]. Now if I know y, and want to know the vector x defined by x[i]=3Dx[i-1]+y[i-1] for all i, how can I do this without a loop? Thanks, Geoffrey PS. Sorry if you see a duplicate message. The previous

## Can someone suggest a simpler expression than either of these, with the goal ## of taking a long matrix into a wide one with exactly one of the factors converted to ## columns and all the rest retained as factors. I want something that generalizes beyond ## the three factors illustrated here. ## Rich meltTest <- data.frame(A=rep(c("B","C"), each=12),

The general formula is y ~ a + b + c + ... There is this approach: formula <- reformulate(independent_vars, response = "y") model <- lm(formula, data = mydata) summary(model) It does not generate a string object, but the formula is still a string even if it is of class formula. Also, in this approach you only get + and if you want interactions or such you will need to code them

I am confused. Richard's answer that Bert did not like did not use parse explicitly. Richard pasted together a string that a function like lm() will have to parse to run the analysis. However, the answers so far do not use parse(). In the reply to Richard, Bert indicated we cannot use strings. Even if I pass a vector where R can assume that the first variable is the dependent variable and all

## I would like help in using variable values in plotmath expressions ## in lattice x <- 1:10 y <- 1:10 pval <- .95 plot(y ~ x, ## works as I want in base graphics main=substitute(list(alpha * " = " * group("",list(p),"")), list(p=pval))) plot(y ~ x, ## doesn't work as intended: "pval" is displayed main=substitute(list(alpha *

The warning message in the last line of this email is incorrect. This is behavior which Duncan Murdoch labeled a bug in https://stat.ethz.ch/pipermail/r-help/2017-December/450494.html This is a fresh install of R-devel (2018-03-21 r74436) R Under development (unstable) (2018-03-21 r74436) -- "Unsuffered Consequences" Copyright (C) 2018 The R Foundation for Statistical Computing

I have Windows XP machines with different software configurations. My laptop has Adobe Reader 8 and Adobe Acrobat 6. My desktop has Adobe Reader 8 and does not have Adobe Acrobat. On both machines, PDF extensions are registered to Adobe Reader 8.0, and double-clicking a pdf file in Windows Explorer opens Adobe Reader 8.0. In R-2.5.0, with all Adobe windows closed, the command

R : Copyright 2001, The R Development Core Team Version 1.3.1 (2001-08-31) > tmp <- factor(1:3) > get("print.factor")(tmp) [1] 1 2 3 Levels: 1 2 3 > print.a <- function(x,...) { + print("this is a") + NextMethod("print", x, quote = FALSE, right = TRUE, ...) + } > > get("print.a") function(x,...) { print("this is a")

... and here is a non-dplyr rsolution: > z <-gsub("[^[:digit:]]"," ",dd$Lower) > sapply(strsplit(z," +"),function(x)sum(as.numeric(x),na.rm=TRUE)) [1] 105 67 60 100 80 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom

Dear R-Group, I have a function that sorts a data frame and oneo of the lines in the function is: vars <- unlist(strsplit(formc, "[\\+\\-]")) The function works fine and the above line is always reached. However, when I include the function in a package and run "R CMD check pkgname" it gives this error message: '\+' is an unrecognized escape in character

Hi all, One of the things I find most problematic in R is the partial matching of names in lists. Robert and I have discussed this and we believe that having a mechanism that does not do partial matching would be of significant benefit to R programmers. To that end, I have written a patch that modifies the behavior of "[[" as follows: 1. [[ gains an 'exact' argument with

As always, I would like to thank all who responded for their insights and suggestions. I have learned from them. Thus far, my own aesthetic preference -- and therefore not to be considered in any sense as a "best" approach -- is to use Duncan's suggestion to produce the call directly with call() rather than substitute in my simple for() loop; i.e. somenames <-