similar to: how to (combine / recode / merge) two factor.

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 ->

Dear R-list, I would like to use the lattice library to show several groups on the same graph. Here's my example : ## the data f1 <- factor(c("mod1","mod2","mod3"),levels=c("mod1","mod2","mod3")) f1 <- rep(f1,3) f2 <-

#I have a dataset with two factor. I want to combine those factors into a single factor and count the number of data values for each new factor. The following gives a comparable dataframe: a <- rep(c("a", "b"), c(6,6)) b <- rep(c("c", "d"), c(6,6)) df <- data.frame(f1=a, f2=b, d=rnorm(12)) df # I use the 'interaction' function to combine

An indicator matrix is a binary matrix with orthogonal columns whose rows sum to 1. A row of this matrix could be [0 1 0 0]. My problem is to group the similar rows (profiles) so that to create a compact form of the matrix. Is there an R function that deals with this problem or do I have to write it from scratch? Thanks, Angelos Markos Dr. Applied Informatics, University of Macedonia, Greece

Hi, I've written a script that checks all bivariate correlations for variables in a matrix. I'm now trying to run a logistic regression on each pair (x,y) where y is a factor with 2 levels. I don't know how (or whether I want) to try to fathom what's up with glm. What I wrote is attached. Here's what I get. *****************************************************

If I have a column with 2 levels, but one level has no remaining observations. Can I remove the level? Had intended to do it as listed below, but soon realized that even though there are no observations, the level is still there. For instance summary(dbs3.train.sans.influential.obs$HAC) yields 0 ,1 4685,0 nlevels(dbs3.train.sans.influential.obs$HAC) yields [1] 2 drop.list <- NULL

Hi, I have a lattice lot conditioned on two variables. Example code is: library(lattice) x <- data.frame(d=runif(100), f1=sample(c('yes', 'no'),100,replace=TRUE), f2=c(rep('Run1',30),rep('Run2',30),rep('Run3',40))) histogram(~d | f1 + f2, x) In the plot, for a given value of f2, there are two panels, one for 'n'

Dear R helpers: Sorry to bother for a basic question about model.matrix. Basically, I want to apply the dummy coding rule in a dataframe with complete factor levels to another dataframe with incomplete factor levels. I used model.matrix, but could not get what I want. The following is an example. #Suppose I have two dataframe A and B

I have a list of data frames: > str(data) List of 4 $ :'data.frame': 700773 obs. of 3 variables: ..$ V1: chr [1:700773] "200130446465779" "200070050127778" "200030633708779" "200010587002779" ... ..$ V2: int [1:700773] 0 0 0 0 0 0 0 0 0 0 ... ..$ V3: num [1:700773] 1 1 1 1 1 ... $ :'data.frame': 700773 obs. of 3 variables: ..$

Hi I know this is a basic question and I know I have done it before but I can't find the answer any more. I have a data set, say: F1, F2, Value F1 and F2 are Factors. I would like to plot plot(TheCombinationOf(F1, F2), Value) I remember there was a function for TheCombinationOf() but I don't remember the name. I could create a new factor based on F1 and F2, but I would prefer the

In the almost current development version (2009-05-22 r48594) and also in https://svn.r-project.org/R/trunk/src/library/base/man/factor.Rd ?factor contains (compare the formulations marked by ^^^^^^) \section{Warning}{ The interpretation of a factor depends on both the codes and the \code{"levels"} attribute. Be careful only to compare factors with the same set of levels (in

Is this a bug? > f1 <- factor(c("a", NA), levels = c("a", "NA") ) > f2 <- f1[, drop = TRUE] > f2 [1] a <NA> Levels: a <NA> I would have expected f2 to have only one level, "a". It seems to me that the code in [.factor does not follow the advice in help("factor") on how to set factor codes to be

I have two nested data frames: a<-rnorm(6) b<-rnorm(9) f1<-c("x1","x2","x3")) f2<-c("y1","y2") id<-c(1:6) a_df<-data.frame(cbind(id,f1,"y1",a)) id<-c(1:9) b_df<-data.frame(cbind(id,f1,"y2",b)) I want to preserve id and f1, but want to collapse f2 and take the corresponding mean values of a and b.

Dear list, after fitting an lrm with the Design package (stored as "mymodel") I try running a summary, but I get the following error: dim(mydata) [1] 235 9 names(mydata) [1] "id" "VAR1" "VAR2" "VAR3" "VAR4" "VAR5" "VAR6" "VAR7" "VAR8" summary(mymodel) Error in `contrasts<-`(`*tmp*`,

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, I have a problem with the "help.start()" and the search engine on a AMD64 : The search engine use the java plugin but the java plugin is not available for the jre.1.5 ... I have a linux box : fedora 4 on 64bit and R2.1.1 When I start the search engine, I had the message that the java plugin was missing. So I try to install manual the java run time jre-1.5 for the AMD64, but the

Ahh yes, sorry about that. Here's the corrected snippet: # Create an Example Data Frame Containing Car x Color data carnames <- c("bmw","renault","mercedes","seat") carcolors <- c("red","white","silver","green") datavals <- round(rnorm(16, mean=10, sd=4),1) data <- data.frame(Car=rep(carnames,4),

This is just my curiousity working. Suppose I write: f1 <- function(x) x + 1 f2 <- function(x) 2 * f1(x) f2(10) # 22 f1 <- function(x) x - 1 f2(10) # 18 This is quite obvious. But is there any way to define f2 in such a way that we "freeze" the definition of f1? f1 <- function(x) x + 1 f2 <- function(x) # put something here 2 * f1(x) # probably put something else here

While I don't know anything about Box's M test, I googled around and found a Matlab M-file that computes it. Below is my straight-forward translation of the code, without knowing Matlab or the formula (and done in a few minutes). I hope this demonstrates one of Prof. Ripley's point: If you really want to shoot yourself in the foot, you can probably program R to do that for you. [BTW:

Dear R-helpers, Happy New Year to all the helpful members of the list. Here is the behavior I'm looking for: > v1 <- c("a","b","c") > n1 <- c(0, 1, 2) > v2 <- c("c", "a", "b") > n2 <- c(0, 1 , 2) > (f1 <- data.frame(v1, n1)) v1 n1 1 a 0 2 b 1 3 c 2 > (f2 <- data.frame(v2, n2))