similar to: Bug? plot.formula does need support plot.first / plot.last param in plot.default

? Fri, 05 Jul 2024 14:35:40 +0300 "Erez Shomron" <r-mails at erezsh.org> ?????: > This works as expected: > with(mtcars, plot(wt, mpg, plot.first = { > plot.window(range(wt), range(mpg)) > arrows(3, 15, 4, 30) > })) I think you meant panel.first, not plot.first. At least I cannot find any mention of plot.first in the R source code. In this example,

That definitely looks like a bug, but not one that anyone will be eager to fix. It's very old code that tried to be clever, and that's the hardest kind of code to fix. Remember Kernighan's Law: "Everyone knows that debugging is twice as hard as writing a program in the first place. So if you?re as clever as you can be when you write it, how will you ever debug it?"

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)",

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

Erez, I think the API is very explicit about this, NULL is not an accepted input for any function taking SEXP by design. The special case of try*Eval() return values can be taken as a case where the resulting object is not actually SEXP but rather a special type which can be NULL (=failure) or SEXP. It may be even perhaps useful to declare it as a separate type to make this clearer, but I

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

Thank you Ivan, At this point, without it being documented explicitly, I tend to lean on the safe side. If the non-null assumption is ever incorrect, on debug and safe builds unwrapping is an assert that will guarantee to crash R. While the source code has plenty of NULL checks, also for some SEXP, it's hard to tell just from grepping if any are related to the public API or not. Secondly

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

> 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

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 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. ------

Simon, Thanks for replying. In what follows I won't try to argue (I understood that you find this a bad idea) but I would like to make clearer some of your point for me (and may be for others). Le 16/04/2020 ? 16:48, Simon Urbanek a ?crit?: > Serguei, >> On 17/04/2020, at 2:24 AM, Sokol Serguei <sokol at insa-toulouse.fr> >> wrote: Hi, I would like to make a

Hi, I am wondering if anyone knows of an easy way to fit a saturated model using the sem package on raw data? Say the data were: mtcars[, c("mpg", "hp", "wt")] The model would estimate the three means (intercepts) of c("mpg", "hp", "wt"). The variances of c("mpg", "hp", "wt"). The covariance of mpg with

Thanks Simon, Now, I see better your argument. Le 16/04/2020 ? 22:48, Simon Urbanek a ?crit?: > ... I'm not arguing against the principle, I'm arguing about your > particular proposal as it is inconsistent and not general. This sounds promising for me. May be in a (new?) future, R core will come with a correct proposal for this principle? Meanwhile, to avoid substitute(),

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: >

Dear All, I am learning R so it's a very simple problem but I do not understand while I am not able to generate a graph from two vectors. when I type this code, it generates a very nice graph. pdf("mygraph.pdf") > attach(mtcars) > plot(wt,mpg) > abline(lm(mpg~wt)) > title("Regreesion of mpg") > detach(mtcars) > dev.off() But I am trying to create a

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 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

I think you are not using the best function for what your intentions are. Try: > by(data=mtcars, INDICES=list(as.factor(mtcars$am)), FUN=colMeans) : 0 mpg cyl disp hp drat wt qsec vs 17.1473684 6.9473684 290.3789474 160.2631579 3.2863158 3.7688947 18.1831579 0.3684211 am gear carb 0.0000000

Dear All, Thanks for your help. However, I would like to draw your attention to the following: Actually, I was replicating the Example 2.3, using the dataset "brainsize.txt" given in Section 2.3.3 ("Summarize by group") at page 55, of a famous book "R by Example" written by "Jim Albert and Maria Rizzo" published in Springers (2012) in a Use R! Series. The