similar to: retrieve name of an object?

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

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

Dear all Is there an easy way to display only one half (top-right or bottom-left) of a correlation matrix? > require(Hmisc) > rcorr(as.matrix(mtcars[ , 1:4])) mpg cyl disp hp mpg 1.00 -0.85 -0.85 -0.78 cyl -0.85 1.00 0.90 0.83 disp -0.85 0.90 1.00 0.79 hp -0.78 0.83 0.79 1.00 n= 32 P mpg cyl disp hp mpg 0 0 0 cyl 0 0 0 disp 0 0

Dear all ?Hmisc::rcorr states that it takes as main argument "a numeric matrix". But is it normal that it fails in such an ugly way on a data frame? (See below.) If the function didn't attempt any conversion to a matrix, I would have expected it to state that in the error message that it didn't accept 'data.frame' objects in its input. Also, I vaguely remember having used

Dear all, What is the quickest and most efficient way to diff two data frames, so as to obtain a vector of indices (or logical) for rows/columns that differ in the two data frames? For example, > Xe <- head(mtcars) > Xf <- head(mtcars) > Xf[2:4,3:5] <- 55 > all.equal(Xe, Xf) [1] "Component 3: Mean relative difference: 0.6863118" [2] "Component 4: Mean relative

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)",

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 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

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

Dear R users I would like to do some spreadsheet style expansion of dates. For example, I would need to obtain a vector of months. I approached in an obviously wrong way: > paste(01:12) [1] "1" "2" "3" "4" "5" "6" "7" "8" "9" "10" "11" "12" > as.Date(paste(01:12),

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, Given the following vector: > (z <- c('R project', 'hello world', 'something Else')) [1] "R project" "hello world" "something Else" I know how to obtain all capitals or all lower case letters: > tolower(z) [1] "r project" "hello world" "something else" > toupper(z) [1] "R

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

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

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(),

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

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

A querent on StackOverflow asked about the with() function http://stackoverflow.com/questions/42283479/why-when-to-use-with-function#42283479 and asked about the example in ?with library(MASS) with(anorexia, { anorex.1 <- glm(Postwt ~ Prewt + Treat + offset(Prewt), family = gaussian) summary(anorex.1) }) which saves little or no typing

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 <-

hi, i am running R with JGR and Deducer under Ubuntu Mint Lucid. since i updated to R 2.12 i can no longer make graphs - R just crashes. the full error report was posted to launchpad (https://bugs.launchpad.net/ubuntu/+source/openjdk-6/+bug/675905) without any success so far. anyone with a similar problem or helpful suggestions? thanks! kat