similar to: formulas with non-syntactic names and an Error() term

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

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 + ... >

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 <-

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

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

I think there are two bugs in aov() that shows up when the right hand side of `formula' contains both `-1' and an Error() term, e.g., aov(y ~ a + b - 1 + Error(c), ...). Without `-1' or `Error()' there is no problem. I've included and example, and the source of aov() with suggested fixes below. The first bug (labeled BUG 1 below) creates an extra, empty stratum inside

I believe you are right, but can you please explain why anyone would want to fit this model? It differs only in the coding from aov(y ~ a + b + Error(c), data=test.df) and merely lumps together the top two strata. There is a much simpler fix: in the line if(intercept) nmstrata <- c("(Intercept)", nmstrata) remove the condition (and drop the empty stratum later if you

SImplify your call to lm using the "." argument instead of manipulating formulas. > strt <- lm(y1 ~ ., data = dat) and you do not need to explicitly specify the "1+" on the rhs for lm, so > frm2<-as.formula(paste(trg," ~ ", paste(xvars,collapse = "+"))) works fine, too. Anyway, doing this gives (but see end of output)" bst <-

> somenames <- c("Heigh", "Ho", "Silver", "Away") > as.formula(paste("~(",paste(somenames, collapse="+"),")^2")) ~(Heigh + Ho + Silver + Away)^2 > > On Mar 29, 2025, at 14:30, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > somenames <- c("Heigh", "Ho", "Silver",

Failed? What was the error message? 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 County" comic strip ) On Tue, Aug 22, 2017 at 8:17 AM, Stephen O'hagan <SOhagan at manchester.ac.uk> wrote: > I'm trying to use boot.stepAIC for

The error is "the model fit failed in 50 bootstrap samples Error: non-character argument" Cheers, SOH. On 22/08/2017 17:52, Bert Gunter wrote: > Failed? What was the error message? > > Cheers, > > Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka

# Your mailer is set to "none" (default on Windows), # hence we cannot send the bug report directly from R. # Please copy the bug report (after finishing it) to # your favorite email program and send it to # # r-bugs@r-project.org # ###################################################### Is this intended behavior for the read.fwf(na.strings="-999")? I anticipated that

Hello R Gurus, I am perplexed by the different results I obtained when I ran code like this: set.seed(100) test1<-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200) predict(test1, newdata=cbind(NewBinaryY, NewXs), type="response") and this code: set.seed(100) test2<-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200, xtest=NewXs, ytest=NewBinarY) The

This feels like inconsistent behavior. latticeParseFormula works the way I anticipated for factor, but not for ordered. I want the behavior I see with tmp2, but not with tmp. My next step is to use the right.name to isolate the tmp2[,c("a","b")] columns. tmp <- data.frame(y=(1:12)+.1, a=factor(rep(1:3,4)), b=ordered(rep(1:4,

OK, here's the problem. Continuing with your example: strt1 <- lm(y1 ~1, dat) strt2 <- lm(frm1,dat) > strt1 Call: lm(formula = y1 ~ 1, data = dat) Coefficients: (Intercept) 41.73 > strt2 Call: lm(formula = frm1, data = dat) Coefficients: (Intercept) 41.73 Note that the formula objects of the lm object are different: strt2 does not evaluate the formula. So

I'm trying to use boot.stepAIC for feature selection; I need to be able to specify the name of the dependent variable programmatically, but this appear to fail: In R-Studio with MS R Open 3.4: library(bootStepAIC) #Fake data n<-200 x1 <- runif(n, -3, 3) x2 <- runif(n, -3, 3) x3 <- runif(n, -3, 3) x4 <- runif(n, -3, 3) x5 <- runif(n, -3, 3) x6 <- runif(n, -3, 3) x7

Consider: x <- list(`a b` = 1) x$a<tab> (i.e., press the 'tab' key after typing 'x$a') The auto-complete mechanism will fill the buffer like so: x$a b This is not particularly helpful because this is now a syntax error. It seems to me there's a simple fix -- in utils:::specialCompletions(), we can wrap the result of utils:::specialOpCompletionsHelper() with

Great suggestion! I've started a patch: https://bugs.r-project.org/show_bug.cgi?id=18479 On Wed, Mar 1, 2023 at 1:56 AM Ivan Krylov <krylov.r00t at gmail.com> wrote: > > ? Wed, 1 Mar 2023 01:36:02 -0800 > Michael Chirico via R-devel <r-devel at r-project.org> ?????: > > > +comps[non_syntactic] <- paste0("`", comps[non_syntactic], "`") >

There turn out to be a few more things to fix. One problem is easy to solve: vapply() needs a third argument specifying the type of the return value. (Can we have unit tests for tab completion?) The other problem is harder: `comps` defaults to an empty string, and you can't have a symbol consisting of an empty string, because this value is internally reserved for missing function arguments.