similar to: Lattice: shifting strips to left of axes

I am using R 2.4.0 and lattice to produce some xyplots conditioned on a factor and a shingle. The shingle merely chops up the data along the x-axis, so it is easy to identify which part of the shingle a panel is in by looking at the x-axis markings. I only want to have a strip at the top for the factor. Is this possible? I looked into calculateGridLayout() and it seems to me that there

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

I would like to be able to do something like this: <<echo=F,fig=T,width=mywidth>>= ... @ with mywidth set in a previous code chunk. Is there a way to do this in Sweave? (Sorry for two questions in a row, I have been saving these up.) -- Michael

I would like to be able to do something like this: <<echo=F,fig=T,width=mywidth>>= ... @ with mywidth set in a previous code chunk. Is there a way to do this in Sweave? (Sorry for two questions in a row, I have been saving these up.) -- Michael

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)",

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

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

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

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

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,

Hi, I'm working on some lattice wireframe figures that have two conditioning factors, and I want the strips labelled on the top and left of the entire plot, rather than above each individual panel. useOuterStrips() does this, but it draws internal axis lines, even after I explicitly set axis.line to 0. Is there a way to use useOuterStrips but without axis boxes? I've included a short

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

Hi All, I am writing a function to predict values based on a model. It works fine as long as the formula just uses regular variable names. I am having a problem when the variables are wrapped with a function call. For example: m <- lm(mpg ~ factor(cyl), data = mtcars) ## I get the column names using as.character(attr(terms(m), "variables"))[-1L] ## which gives the same column

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 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

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