similar to: how to plot a graph with different pch

Hello: I am a bit confused by this problem. Can anyone give me some advice on this I would greatly appreciate it. Thank you for all your help. Need to create a for loop that saves the estimate of pi from each 0f 100 separate iterations and store it in a numeric vector (of length 100). The for loop should be placed in a function that allows the user to vary the sample size, the simulation size,

Also, it is possible to change "ylim" also? 2012/7/1 li li <hannah.hlx@gmail.com> > Dear all, > I have a few questions regarding the boxplot output from the > "geom_boxplot" function. > Attached is the output I get. Below are my questions: > > 1. How can I define the xlab and ylab myself? > Also I would like to remove

Hello, When plotting a barchart with ggplot it drops the levels of the factor for which no counts are available. For example: library(ggplot) mtcars$cyl<-factor(mtcars$cyl) ggplot(mtcars[!mtcars$cyl==4,], aes(cyl))+geom_bar() levels(mtcars[!mtcars$cyl==4,]) This shows my problem. Because no counts are available for factorlevel '4', the label 4 dissapears from the plot. However, I

Dear all Is there an easier way to retrieve the name of an object? For example, > tmp <- 1:10 > as.character(quote(tmp)) [1] "tmp" > as.character(quote(mtcars$cyl)) [1] "$" "mtcars" "cyl" > as.character(quote(mtcars$cyl))[3] [1] "cyl" The last call more than anything seems a hack. Is there a better way? Thank you Liviu

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

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 all, Consider the model below: > x <- lm(mpg ~ cyl * disp * hp * drat, mtcars) > summary(x) Call: lm(formula = mpg ~ cyl * disp * hp * drat, data = mtcars) Residuals: Min 1Q Median 3Q Max -3.5725 -0.6603 0.0108 1.1017 2.6956 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.070e+03 3.856e+02 2.776 0.01350 * cyl

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 all I'm unable to find an example of extracting the rotated scores of a principal components analysis. I can do this easily for the un-rotated version. data(mtcars) .PC <- princomp(~am+carb+cyl+disp+drat+gear+hp+mpg, cor=TRUE, data=mtcars) unclass(loadings(.PC)) # component loadings summary(.PC) # proportions of variance mtcars$PC1 <- .PC$scores[,1] # extract un-rotated scores of

Hello Given the following data for a data set called airquality. To identify the nature of the objects from the data set airquality example "Ozone" would it be best to use the command is. like is.character(airquality$Ozone) ....... I tried attributes(airquality$Ozone) but it came up null. Would there be a better way to identify these objects. Thanking you in advance for your

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

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 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

> 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 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, I am using levelplot, and would like remove from each panel (condition) its unused x levels. e.g. Remove from panel vs=1 the cyl level=8. data(mtcars) levelplot(mpg~factor(cyl)*factor(gear)|factor(vs)) Thanks for your help, Ronny -- View this message in context: http://r.789695.n4.nabble.com/Lattice-levelplot-remove-unused-levels-per-panel-tp4656087.html Sent from the R help mailing

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 R-Devel, I seem to no longer be able to access the bug-reporting system, so am doing this by e-mail. My report concerns models where variables are explicitly referenced (or is it "dereferenced"?), such as: cars.lm <- lm(mtcars[[1]] ~ factor(mtcars$cyl) + mtcars[["disp"]]) I have found that it is not possible to predict such models with new data. For example:

How can I export an ftable object in the same format that appears in R command window? For testing that i was using this example that is in help of this function. ## Start with a contingency table. ftable(Titanic, row.vars = 1:3) ftable(Titanic, row.vars = 1:2, col.vars = "Survived") ftable(Titanic, row.vars = 2:1, col.vars = "Survived") ## Start with a data frame. x <-