similar to: Cannot Load A Package

Dear John, Dear Joseph, Thank you for your quick answers and the pointer to semnet. I try to clarify on my assumptions: - yes, I am willing to assume multivariate normality - no, I don't want to assume a single factor model - I assume there is an unknown number of factors, and I do not know which items belong to which factors but I still want to estimate single item reliabilities Is this

Hi Team, I using the syntax as: data.df<- data.frame( city= c(rep(c("Delhi", "Bangalore","Chandigarh"),each=5)), population= c(4000:6000,3500:4300,3000:3200) ) But i am getting the error as arguments imply differing number of rows: 15, 3003. Tried searching google but could not understand & find the solution. Thanks, Shivi [[alternative HTML version

Hi Sarah, Thank you for your help. I tried using CR1<-as.matrix(CR1) but gives error Error in corrplot(CR1, method = "circle") : The matrix is not in [-1, 1]!. I am using a corrplot library. Please find the reproducible example: dput(head(CR1,10)) structure(c(26L, 46L, 39L, 38L, 47L, 59L, 56L, 61L, 43L, 60L, 78L, 63L, 2L, 58L, 8L, 1L, 1L, 9L, 11L, 2L, 1037500L, 46747L, 346300L,

That does clarify for me that you're missing a step: I didn't clearly follow your description at first. corrplot expects a correlation matrix, not your original data. You need to use cor() first. That's pretty clear in the documentation. See for instance the examples: data(mtcars) M <- cor(mtcars) corrplot(M) Sarah On Sat, Mar 17, 2018 at 12:00 PM Shivi Bhatia <shivipmp82 at

Created a new data set with 3 numeric variable to find correlation CR1<- mar%>% as_data_frame%>% select(AGE, OLD_CAR_PURCHASE_YRS, Total.Spend.With.AA) had to convert it to a data frame, code: as.matrix(as.data.frame(CR1)) Now i need to run a correlation plot for these 3 variables: corrplot(CR1, method = "circle") But i am getting this error: Error in

4000:6000 gives you 4000, 4001, ..., 6000. I suspect you want population= c(seq(4000, 6000, length=5), seq(3500, 4300, length=5), seq(3000, 3200, length=5)) Bob On 20 September 2017 at 17:07, Shivi Bhatia <shivipmp82 at gmail.com> wrote: > Hi Team, > > I using the syntax as: > > data.df<- data.frame( > city= c(rep(c("Delhi",

I'm assuming you are using the corrplot package. If so, your data object does need to be a matrix, not a data frame. Since it's already a data frame, your line of code: as.matrix(as.data.frame(CR1)) doesn't need the as.data.frame function, but more importantly, you didn't assign the result to anything: as.matrix() does not work in place. CR1 <- as.matrix(CR1) Now try. If

Jens I'm not sure what you intend by "predefined assumptions". 1. If you merely want to conduct an exploratory rather than confirmatory analysis for the relevant paths, there are ways within SEM to do this. (In this case you could use John Fox's SEM package). 2. If you do not wish to assume multivariate normality, then you may use a variety of alternative (to maximum likelihood)

Dear R-List, (I am not sure whether this list is the right place for my question...) I have a dataframe df.cfa

Sorry if this question has been asked previously, I searched but found little. There also doesn't seem to be a dedicated SEM list-serv so hopefully this will find its way to the appropriate audience. In discussing SEM with a colleague I mentioned that a model they were fitting in AMOS was equivalent to a linear regression and that the coefficients would be the same. This of course was the

Someone can help me? I tried several things and always don't converge # Model library(sem) dados40.cov <- cov(dados40,method="spearman") model.dados40 <- specify.model() F1 -> Item11, lam11, NA F1 -> Item31, lam31, NA F1 -> Item36, lam36, NA F1 -> Item54, lam54, NA F1 -> Item63, lam63, NA F1 -> Item65, lam55, NA F1 -> Item67, lam67, NA F1 ->