similar to: extract variables from model formula

Displaying 20 results from an estimated 10000 matches similar to: "extract variables from model formula"

2007 Oct 01
4
how to plot a graph with different pch
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)",
2010 Nov 30
3
pca analysis: extract rotated scores?
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
2013 Apr 17
2
remove higher order interaction terms
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
2020 Apr 16
2
suggestion: "." in [lsv]apply()
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
2020 Apr 16
6
suggestion: "." in [lsv]apply()
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
2011 Aug 17
9
too many var in lm
Hello, It might be an easy question but if you have many variables to fit in the lm function, how do you take all without specifying var1+var2+...+var2100 in the terms parameter in response ~ terms? Cheers, Carol
2013 Jan 20
2
Lattice levelplot- remove unused levels per panel
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
2017 Mar 26
1
Documentation of model.frame() and get_all_vars()
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
2007 May 12
2
Implicit vs explicit printing and the call stack
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,
2020 Apr 16
2
suggestion: "." in [lsv]apply()
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
2020 Apr 17
2
suggestion: "." in [lsv]apply()
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(),
2011 Aug 19
2
display only the top-right half of a correlation matrix?
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
2018 Jul 20
2
Model formulas with explicit references
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$cyl) + mtcars[["disp"]]) I have found that it is not possible to predict such models with new data. For example:
2011 Aug 19
1
Hmisc::rcorr on a 'data.frame'?
Dear all ?Hmisc::rcorr states that it takes as main argument "a numeric matrix". But is it normal that it fails in such an ugly way on a data frame? (See below.) If the function didn't attempt any conversion to a matrix, I would have expected it to state that in the error message that it didn't accept 'data.frame' objects in its input. Also, I vaguely remember having used
2006 Sep 03
2
lm, weights and ...
> lm2 <- function(...) lm(...) > lm2(mpg ~ wt, data=mtcars) Call: lm(formula = ..1, data = ..2) Coefficients: (Intercept) wt 37.285 -5.344 > lm2(mpg ~ wt, weights=cyl, data=mtcars) Error in eval(expr, envir, enclos) : ..2 used in an incorrect context, no ... to look in Can anyone explain why this is happening? (Obviously this is a manufactured example, but it
2020 Apr 20
1
suggestion: "." in [lsv]apply()
Le 19/04/2020 ? 20:46, Gabor Grothendieck a ?crit?: > You can get pretty close to that already using fn$ in the gsubfn package: >> library(gsubfn) fn$sapply(split(mtcars, mtcars$cyl), x ~ >> summary(lm(mpg ~ wt, x))$r.squared) > 4 6 8 0.5086326 0.4645102 0.4229655 Right, I thought about similar syntax but this implementation has similar flaws pointed by Simon, i.e. it reduces
2016 Apr 14
0
Bug in by() function which works for some FUN argument and does not work for others
I think you are not using the best function for what your intentions are. Try: > by(data=mtcars, INDICES=list(as.factor(mtcars$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
2024 Sep 22
2
store list objects in data.table
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
2013 Apr 12
3
Why copying columns of a data.frame becomes numeric?
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
2016 Apr 14
4
Bug in by() function which works for some FUN argument and does not work for others
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: >