similar to: How can we predict the value of dependent variable using independent variable

Dear all, I have a data frame in which the columns need to be ordered. The first column X is at the right position, but the remaining columns X1-Xn should be ordered like this: X1, X2, X3 etc instead of like below. > colnames(pos1) [1] "X" "X1" "X10" "X11" "X12" "X13" "X14" "X15" "X16"

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

Hi All, I'm using the subset function to select a list of variables, some of which are contiguous in the data frame, and others of which are not. It works fine when I use the form: subset(mydata,select=c(x1,x3:x5,x7) ) In reality, my list is far more complex. So I would like to store it in a variable to substitute in for c(x1,x3:x5,x7) but cannot get it to work. That use of the c function

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, 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'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

Hi, Please could someone explain how this element of predict.lm works? From the help file ` newdata An optional data frame in which to look for variables with which to predict. If omitted, the fitted values are used. ' Does this dataframe (newdata) need to have the same variable names as was used in the original data frame used to fit the model? Or will R just look across consecutive

#### 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

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)

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 ?

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

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

Hi, Suppose I have n variables, X1, X2, ..., Xn and want to visualize the pairwise correlations on a 2-dimmensional plot. In particular, I plan to mark down n dots on a piece of paper and dot i represents Xi, in a way such that the highly correlated variables are placed closer; that is, shorter the distance between two dots the higher the correlation are between two corresponding variables. Do

Dear All, I am new to R, I have one question which might be easy. I have a large data with more than 250 variable, i am reducing number of variables by redun function as in the example below, n <- 100 x1 <- runif(n) x2 <- runif(n) x3 <- x1 + x2 + runif(n)/10 x4 <- x1 + x2 + x3 + runif(n)/10 x5 <- factor(sample(c('a','b','c'),n,replace=TRUE)) x6 <-

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

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',

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',

Dear friends, I'm doing a simulation on logistic regression model, but the programs can't work well,please help me to correct it and give some suggestions. My programs: data<-matrix(rnorm(400),ncol=8) #sample size is 50 data<-data.frame(data) names(data)<-c(paste("x",1:8,sep="")) #8 independent variables,x1-x8; #logistic regression model is

Hi, Suppose I have a vector in real number (x1, x2, x3, x4, x5, x6) My question is how I can get x5*x3*x1 + x6*x4*x2 ? Thanks a lot. Dot. -- View this message in context: http://www.nabble.com/how-can-I-do-this-sum--tp18931693p18931693.html Sent from the R help mailing list archive at Nabble.com.