similar to: Passing in arguments into function

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

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

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)",

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(),

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

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: >

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 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, 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 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 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

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. ------

Suppose you have a vector of data in x and response values in y. How do you plot together both the points (x,y) and the curve that results from the fitted model, if the model is not y ~ x, but a higher order polynomial, e.g. y~poly(x,2)? (In other words, abline doesn't work for this case.) Thanks, --Paul -- Paul Lynch Aquilent, Inc. National Library of Medicine (Contractor)

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

Hi folks, I'm trying to estimate bias-corrected percentile (BCP) confidence intervals on a vector from a simple for loop used for resampling. I am attempting to follow steps in Manly, B. 1998. Randomization, bootstrap and monte carlo methods in biology. 2nd edition., p. 48. PDF of the approach/steps should be available here: https://wyocoopunit.box.com/s/9vm4vgmbx5h7um809bvg6u7wr392v6i9 If

Le 19/04/2020 ? 20:46, Gabor Grothendieck a ?crit?: > You can get pretty close to that already using fn$ in the gsubfn package: >> library(gsubfn) fn$sapply(split(mtcars, mtcars$cyl), x ~ >> summary(lm(mpg ~ wt, x))$r.squared) > 4 6 8 0.5086326 0.4645102 0.4229655 Right, I thought about similar syntax but this implementation has similar flaws pointed by Simon, i.e. it reduces

Hello list, I am running a regression using lm(Y~A+B+log(C)+log(D)) Now, I would like to test if glm can produce similar results. So the code was revised as glm(Y~A+B+C+D, family=poisson) (code 1) However, I found some example using glm for lm. It suggests that the code should be revised like glm(Y~A+B+log(C)+log(D), family=poisson) (code 2) Please kindly advise which code is correct.

Thanks everyone for their responses. My data is organized in a data.table.? My goal is to perform analyses according to some groups.? The results of analysis are objects.? If these objects could be stored as elements of a data.table, this would help downstream summarizing of results. Let me try another example. carsdt <- setDT(copy(mtcars)) carsdt[, unique(cyl) |> length()] #[1] 3