similar to: R-3.3.3/R-3.4.0 change in sys.call(sys.parent())

Here is a case where the current scheme fails: > with(datasets::mtcars, xyplot(mpg~wt|gear)$call) xyplot(substitute(expr), data, enclos = parent.frame()) Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, May 11, 2017 at 1:09 AM, Deepayan Sarkar <deepayan.sarkar at gmail.com> wrote: > On Wed, May 10, 2017 at 2:36 AM, William Dunlap via R-devel > <r-devel at

On Wed, May 10, 2017 at 2:36 AM, William Dunlap via R-devel <r-devel at r-project.org> wrote: > Some formula methods for S3 generic functions use the idiom > returnValue$call <- sys.call(sys.parent()) > to show how to recreate the returned object or to use as a label on a > plot. It is often followed by > returnValue$call[[1]] <- quote(myName) > E.g., I see

> lm2 <- function(...) lm(...) > lm2(mpg ~ wt, data=mtcars) Call: lm(formula = ..1, data = ..2) Coefficients: (Intercept) wt 37.285 -5.344 > lm2(mpg ~ wt, weights=cyl, data=mtcars) Error in eval(expr, envir, enclos) : ..2 used in an incorrect context, no ... to look in Can anyone explain why this is happening? (Obviously this is a manufactured example, but it

Hi all, It seems that there is a use case for obtaining the environment for the "top" promise. By "top", I mean following the promise chain up the call stack until hitting a non-promise. S4 data containers often mimic the API of base R data structures. This means writing S4 methods for functions that quote their arguments, like with() and subset(). The methods package

Hello All, A simple question. I get some return from the R in my C++ program (via Rcpp package). The result come, as SEXP and it should be a simple numeric variable. How to convert it to double? The code, what i use: stringstream ss; ss << "p <- predict(fit_ar11, n.ahead = 2, doplot=FALSE);" << "p$pred[1]"; SEXP ans; int iRet =

This is an expanded version of the question I tried to ask last night - I thought I had it this morning, but it's still not working and I just do not understand what is going wrong. What I am trying to do is write a wrapper for lattice xyplot() that passes a whole bunch of its secondary arguments, so that I can produce similarly formatted graphs for several different data sets. This is what

Hi, I am using RInside for one of the applications that I'm using. The error that occured is as follows. When I use R.parseEval("xyz") if the parse fails then the mb_m is not cleared and hence all subsequent calls to R.parseEval fail. example: when R.parseEval("stringConcatfactoryrandom") is executed the R instance looks like this (gdb) p R $7 = (RInside &)

Hi everyone, This is about documentation for the model.frame() page. The get_all_vars() function (added in R 2.5.0) is a great addition, but the behavior of its '...' argument is different from that of model.frame() with which it is documented and this creates ambiguity. The current docs read: \item{\dots}{further arguments such as \code{data}, \code{na.action}, \code{subset}. Any

I am trying to plot a graph but the points on the graph should be different symbols and colors. It should represent what is in the legend. I tried using the points command but this does not work. Is there another command in R that would allow me to use different symbols and colors for the points? Thank you kindly. data(mtcars) plot(mtcars$wt,mtcars$mpg,xlab= "Weight(lbs/1000)",

The lattice package uses special logic to allow for multiple left-hand-side variables in a formula, e.g. y1 + y2 ~ x. Is there an elegant way to do this outside of lattice? I'm trying to implement a data summarization function that logically takes multiple dependent variables. The usual invocation of model.frame( ) causes R to try to do arithmetic addition to create a single dependent

Hi, I would like to make a suggestion for a small syntactic modification of FUN argument in the family of functions [lsv]apply(). The idea is to allow one-liner expressions without typing "function(item) {...}" to surround them. The argument to the anonymous function is simply referred as ".". Let take an example. With this new feature, the following call

Dear list, I want the 1st, 2nd, 5th, and 6th columns of mtcars. After copying them, the columns become numeric class rather than data frame. But, when I copy rows, they data frame retains its class. Why is this? I don't see why copying rows vs columns is so different. > class(mtcars) [1] "data.frame" > head(mtcars) mpg cyl disp hp drat wt qsec vs

I'm sure this exists elsewhere, but, as a trade-off, could you achieve what you want with a separate helper function F(expr) that constructs the function you want to pass to [lsv]apply()? Something that would allow you to write: sapply(split(mtcars, mtcars$cyl), F(summary(lm(mpg ~ wt,.))$r.squared)) Such an F() function would apply elsewhere too. /Henrik On Thu, Apr 16, 2020 at 9:30 AM

Before I go and do this another way - can I check if anyone has a way of looping through data in odfWeave (or possibly sweave) to do a repeating analysis on subsets of data? For simplicity lets use mtcars dataset in R to explain. Dataset looks like this: > mtcars mpg cyl disp hp drat wt ... Mazda RX4 21.0 6 160 110 3.90 2.62 ... Mazda RX4 Wag 21.0 6 160 110 3.90

I have applied the same linear model to several different subsets of a dataset. I recently read that in R, code should never be repeated. I feel my code as it currently stands has a lot of repetition, which could be condensed into fewer lines. I will use the mtcars dataset to replicate what I have done. My question is: how can I use fewer lines of code (for example using a for loop, a function or

Hi everyone, I've run into a bit of strange problem with implicit vs explicit printing and the call stack. I've included an example at the bottom of this email. The basic problem is that I have an S3 object with a print method. When the object is implicitly printed (ie. typed directly into the console) the function arguments in the call stack are exploded out to their actual values,

I want to use a strip.custom as with useOuterStrips for three conditioning variables. useOuterStrips restricts this to only two conditioning variables, and I cannot figure out how to write strip.custom properly to do this. library(lattice) mtcars$HP <- equal.count(mtcars$hp) #with two factors x2<-xyplot(mpg ~ disp | HP + factor(cyl), mtcars) useOuterStrips(x2) #with three factors

Dear List I have math test scores for male and female students where gender is a dummy code (female =1). I also have a variety of other demographic variables. However to begin, I want to create a very simple stripchart where female math scores are a blue circle and male scores are a red triangle. I am having difficulty using conditional statements to accomplish this. Thank you. ------

Dear R users, Does the results below make any sense? Can the the interval of the correlation coefficient be between *-1.0185* and -0.8265 at 95% confidence level? Liviu > library(boot) > data(mtcars) > with(mtcars, cor.test(mpg, wt, met="spearman")) Spearman's rank correlation rho data: mpg and wt S = 10292, p-value = 1.488e-11 alternative hypothesis: true rho is not

Dear Sirs, I am Professor at Indira Gandhi Krishi Vishwavidyalaya, Raipur, Chhattisgarh, India. While taking classes, I found the *by() *function producing following error when I use FUN=mean or median and some other functions, however, FUN=summary works. Given below is the output of the example I used on a built-in dataset "mtcars", along with error message reproduced herewith: >