similar to: easy way to fit saturated model in sem package?

Is there a way to use full information maximum likelihood (FIML) to estimate missing data in the sem package? For example, suppose I have a dataset with complete information on X1-X3, but missing data (MAR) on X4. Is there a way to use FIML in this case? I know lavaan and openmx allow you to do it, but I couldn't find anything in the documentation for the sem package. Thanks! -- Dustin Fife

Does anyone know if someone is developing full-information maximum likelihood (FIML) estimation algorithms for basic regression functions, like glm()? I think that having FIML options for workhorse functions that typically use ML would give R an edge over other statistical software, given how well FIML performs in missing data situations compared to ML. While my current level of programming

Dear all, I am trying to estimate simultaneous equation model concerning growth in russian regions. I run the analysis by means of FIML in R sem package. I am not familiar with SEM yet, but I've just got several suitable estimated specifications. Nevertheless, sometimes R gives the following warning message: Warning message: Negative parameter variances. Model is probably underidentified.

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 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 R users/developers I requested help to solve the problem of formulating Multivariate Sample selection model by using Full Information Maximum Likelihood (FIML)estimation method. I could not get any response. I formulated the following code of FIML to analyse univariate sample selection problem. Would you please advise me where is my problem library (sem) library(nrmlepln) Selection

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, 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

> 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

Is the following a bug in your opinion? I think so. This works as expected: ``` with(mtcars, plot(wt, mpg, plot.first = { plot.window(range(wt), range(mpg)) arrows(3, 15, 4, 30) })) ``` This does not. ``` plot(mpg ~ wt, data = mtcars, plot.first = { plot.window(range(wt), range(mpg)) arrows(3, 15, 4, 30) }) ``` With error: ``` Error in arrows(3, 15, 4, 30) : plot.new has not

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

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

That definitely looks like a bug, but not one that anyone will be eager to fix. It's very old code that tried to be clever, and that's the hardest kind of code to fix. Remember Kernighan's Law: "Everyone knows that debugging is twice as hard as writing a program in the first place. So if you?re as clever as you can be when you write it, how will you ever debug it?"

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

Dear all, I would like to know if it is possible to estimate multi-group SEM by using R... Thank you _________________________________________________________________ ¿Cuánto espacio necesitas para guardar tus emails? Con Hotmail tienes 5GB y puede ampliarse a más. [[alternative HTML version deleted]]

Dear R developers, Currently many (all?) test functions in R describe the alternative hypothesis, but not the the null hypothesis being tested. For example, cor.test: > require(boot) > data(mtcars) > with(mtcars, cor.test(mpg, wt, met="kendall")) Kendall's rank correlation tau data: mpg and wt z = -5.7981, p-value = 0.000000006706 alternative hypothesis: true tau is not

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(),

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

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