similar to: Error in structural equation model - "The model has negative degrees of freedom"

Hi John Thanks a lot for the reply. Could you suggest how I can correct this problem? I tried using a correlation matrix instead of raw moments, but still got the same error. I also fixed parameters v1,v2,v3,a1 at 1; then it gave me the error that the system is exatly singular. To answer the points that you raised: 1. x1-x6 are not causes; they are just indicatiors. Does that change my

Dear R-help list members, I forgot to copy my reply to the r-help list. Here's most of it. John -------------------------------- John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox -------------------------------- -----Original Message----- From: John Fox [mailto:jfox at mcmaster.ca] Sent: Sunday,

Hi all, I am attempting to learn my way through the sem package by constructing a simple structural model for some of my data on bird diversity, abundance, and primary productivity. I have constructed a covariance matrix between these variables as per the following: >S_matrix = matrix(c( >+ 0.003083259, 0, 0, >+ 0.143870284, 89.7648490, 0, >+ 0.276950919,

I'm pasting below a working R file featuring a function I'd like to polish up. I'm teaching regression this semester and every time I come to something that is very difficult to explain in class, I try to simplify it by writing an R function (eventually into my package "rockchalk"). Students have a difficult time with predict and newdata objects, so right now I'm

#### I checked through every 3 factor * 3 loading case. #### While, 4 factor * 3 loading failed. #### the data is 6 factor * 3 loading require(sem); cor18<-read.moments(); 1 .68 1 .60 .58 1 .01 .10 .07 1 .12 .04 .06 .29 1 .06 .06 .01 .35 .24 1 .09 .13 .10 .05 .03 .07 1 .04 .08 .16 .10 .12 .06 .25 1 .06 .09 .02 .02 .09 .16 .29 .36 1 .23 .26 .19 .05 .04 .04 .08 .09 .09 1 .11 .13 .12 .03 .05 .03

I'm using specify.model for the sem package. I can't figure out how to represent the residual errors for the observed variables for a CFA model. (Once I get this working I need to add some further constraints.) Here is what I've tried: model.sa <- specify.model() F1 -> X1,l11, NA F1 -> X2,l21, NA F1 -> X3,l31, NA F1 -> X4,l41, NA F1 -> X5, NA, 0.20

I've just found the lavaan package, and I really appreciate it, as it seems to succeed with models that were failing in sem::sem. I need some clarification, however, in the output, and I was hoping the list could help me. I'll go with the standard example from the help documentation, as my problem is much larger but no more complicated than that. My question is, why is there one latent

Dette er en melding med flere deler i MIME-format. --=_alternative 004613C000257091_= Content-Type: text/plain; charset="US-ASCII" And some more informastion I forgot. R does not crash if I write out the formula: set.seed(123) x1 <- runif(1000) x2 <- runif(1000) x3 <- runif(1000) x4 <- runif(1000) x5 <- runif(1000) x6 <- runif(1000) x7 <- runif(1000) x8 <-

Hi All, I'd like to build a package for the community that replicates the output produced by SAS "proc varclus". According to the SAS documentation, the first few steps are: 1. Find the first two principal components. 2. Perform an orthoblique rotation (quartimax rotation) on eigenvectors. 3. Assign each variable to the rotated component with which it has the higher squared

Hi, Using S=1000 and simdata <- replicate(S, generate(3000)) #If you want both "m1" and "m0" #here the missing values are 0 res1<-sapply(seq_len(ncol(simdata.psm1)),function(i) {x1<-merge(simdata.psm0[,i],simdata.psm1[,i],all=TRUE); x1[is.na(x1)]<-0; x1}) res1[,997:1000] #????? [,1]???????? [,2]???????? [,3]???????? [,4]??????? #x1??? Numeric,3000 Numeric,3000

I am trying to calculate the regression for the follwing input data stored in 'data.txt' file.I am reading this and storing it in the variable i .then i am trying to get the predicted value using f1 as dependent and others f2....f10 as independent variables.It is giving the following error. Also i want that i shoul get one predicted value for each row(y). What should i do. Please help me

Just out of curiosity, why might this be occuring: > class(x6) [1] "mcmc" > crosscorr.plot(x6) NULL # Replicable code example(lmer) x6 <- mcmcsamp(fm1, n=1000) crosscorr.plot(x6)

SImplify your call to lm using the "." argument instead of manipulating formulas. > strt <- lm(y1 ~ ., data = dat) and you do not need to explicitly specify the "1+" on the rhs for lm, so > frm2<-as.formula(paste(trg," ~ ", paste(xvars,collapse = "+"))) works fine, too. Anyway, doing this gives (but see end of output)" bst <-

Does anyone know how to pass a list of parameters into a function? for example: somefun=function(x1,x2,x3,x4,x5,x6,x7,x8,x9){ ans=x1+x2+x3+x4+x5+x6+x7+x8+x9 return(ans) } somefun(1,2,3,4,5,6,7,8,9) # I would like this to work: temp=c(x3=3,x4=4,x5=5,x6=6,x7=7,x8=8,x9=9) somefun(x1=1,x2=2,temp) # OR I would like this to work: temp=list(x3=3,x4=4,x5=5,x6=6,x7=7,x8=8,x9=9)

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

Suppose I have the following: x1<-as.vector(rnorm(10)) x2<-as.vector(rnorm(10)) x3<-as.vector(rnorm(10)) x4<-as.vector(rnorm(10)) x5<-as.vector(rnorm(10)) x6<-as.vector(rnorm(10)) x7<-as.vector(rnorm(10)) x8<-as.vector(rnorm(10)) x9<-as.vector(rnorm(10)) x10<-as.vector(rnorm(10)) I would like the mean of x1 to x6 for each vector position. I would do something else

Dear R-help community, If I have a data.frame df as follows: > df x1 x2 x3 x4 x5 x6 1 5 5 1 1 2 1 2 5 5 5 5 1 5 3 1 5 5 5 5 5 4 5 5 1 4 5 5 5 5 1 5 2 4 1 6 5 1 5 4 5 1 7 5 1 5 4 4 5 8 5 1 1 1 1 5 9 1 5 1 1 2 5 10 5 1 5 4 5 5 11 1 5 5 2 1 1 12 5 5 5 4 4 1 13 1 5 1 4 4 1 14 1 1 5 4 5 5 15 1 5 5 4

Yo copio y pego este código y me sale correctamente. Se me ocurre que pueda deberse a la versión de R ¿cuál usas? El 10/09/2020 a las 17:51, Samura . escribió: > Gracias por las respuestas. > > Probé lo de hacer la función y no me salía. Pensaba que hacía algo mal. > Ahora con el código de Marcelino tampoco me sale. > > col1 <- c('x1', 'x2', 'x11',

Hola, me gustar?a hacer algo como en el siguiente ejemplo A un df a?adirle una columna que es la transformaci?n de otra, en plan a todo lo que sea x1, x2, x3 lo llamo prueba 1 todo lo que sea x4,x5,x6 lo llamo prueba 2 el resto de x las dejo como est?n. Ser?a algo as? col1 <- c('x1', 'x2', 'x11', 'x1','x33', 'x1','x4', 'x5',

Hi, You might need to check library(boot).? I have never used that before.? So, I can't comment much.? It is better to post on R-help list.? I had seen your postings on Nabble in the past.? Unfortunately those postings were not accepted in R-help.? You have to directly post at ? r-help at r-project.org after registering at: https://stat.ethz.ch/mailman/listinfo/r-help ?