search for: factorize

I am getting the following build break on changeset 13662. I am compiling on x86_64 SLES10 with gcc 4.1.0. Is there a fix for this? Thanks, Aravindh Puthiyaparambil Xen Development Team Unisys, Tredyffrin PA make[1]: Entering directory `/root/xen/xen-unstable.hg/tools/firmware'' make[2]: Entering directory `/root/xen/xen-unstable.hg/tools/firmware/rombios'' gcc -o biossums

Hi, I am trying to run ANOVA with an interaction term on 2 factors (treat has 7 levels, group has 2 levels). I found the coefficient for the last interaction term is always 0, see attached dataset and the code below: > test<-read.table("test.txt",sep='\t',header=T,row.names=NULL) > lm(y~factor(treat)*factor(group),test) Call: lm(formula = y ~ factor(treat) *

Dear R users, I have posted this question several days ago and received not a single suggestion. I believe I have provided sufficient information for at least some help. Here I repost the question with several modifications. I want to run nested fixed-factor Anova in R on different experiments. I have 48 levels of the main factor x1 and 242 levels of the nested factor z1, and continuous response

Hi all, Is there are reason that there is no c.factor method? Analogous to c.Date, I'd expect something like the following to be useful: c.factor <- function(...) { factors <- list(...) levels <- unique(unlist(lapply(factors, levels))) char <- unlist(lapply(factors, as.character)) factor(char, levels = levels) } c(factor("a"), factor("b"),

The expressions > x<-factor(x) and > class(x)<-"factor" behave differently when x is already an ordered factor. It may not be a bug but it caught me out when I was trying to remove the "orderedness" from a factor variable. The following R code illustrates the difference. Is this difference between the 2 commands desirable? > x<-1:3 > class(x) NULL

I have a matrix which I wanted to convert to a data frame. As I could not succeed and resorted to export to csv and reimport it again. Why did I fail in the attempt and how can I achieve what I wanted without this roundabouts? The original matrix: > str(comb_model0) num [1:90, 1:4] 3.5938 0.0274 0.0342 0.0135 0.0207 ... - attr(*, "dimnames")=List of 2 ..$ : chr [1:90]

hello I need a clarification. in logistic regression, saturated model having all combinations and interactions of variables should be constructed in this way ? : > rsat2=glm(cbind(landp,landa) ~ as.factor(rlito)*as.factor(rslp)*as.factor(rasp)*as.factor(rplc)*as.factor(rwi),family= binomial(link=logit),data=rdf46) then using stepAIC to eliminate some models having high AIC values. >

Hi I am new to R so I apologize if this is trivial. I am trying to predict the resistance or susceptibility of my sequences to a certain drug with a randomForest function from a file with amino acids on each of the positions in the protein. I ran the following: > library(randomForest) > > path <- "C:\\..." > path2 <- "..." > name <-

as.factor / as.ordered is not written as a generic. This differs from as.numeric, as.matrix, and other as.*. The following seems to address this and does not break make check-all. FWIW, the patch is against r55563, because with r55564 I see /home/mtmorgan/src/R-devel/src/main/dounzip.c:75:15: error: storage size of ?dt? isn?t known /home/mtmorgan/src/R-devel/src/main/dounzip.c:88:5: warning:

I got two questions on factors in regression: Q1. In a table, there a few categorical/factor variables, a few numerical variables and the response variable is numeric. Some factors are important but others not. How to determine which categorical variables are significant to the response variable? Q2. As we knew, lm can deal with categorical variables. I thought, when there is a categorical

Dear R users, I want to run nested fixed-factor Anova in R on different experiments. In this toy example I have 3 levels of the main factor x1 and 7 levels of the nested factor z1 x1 and continuous response variable y1. x1 [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 [38] 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 [75] 3 3 3

I need to look through a dataset with two factor variables, and depending on certain criteria, create a new variable containing the data from one of those other variables. The problem is, R keeps making my new variable an integer and saving the data as a 1 or 2 (I believe the levels of the factor). I've tried using as.factor in the IF output statement, but that doesn't seem to work. Any

Hi, I'm a PhD student and I want to use the function 'daisy' from the package 'cluster' to compute dissimilarities. My variables are of mixed types so I use the argument 'stand' in daisy to define the type of my variables. I have the following error message : Warning message: binary variable(s) 13, 16, 17, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,

Hi Ansely, As Jim points out we really need some sample data to go with the code. Have a look at ?dput which is the best way to supply sample data here or have a look at http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example and/or http://adv-r.had.co.nz/Reproducibility.html for some general suggestions on asking questions here---including discussions of using

HI I use glm in R to do logistic regression. and treat both response and predictor as factor In my first try: ******************************************************************************* Call: glm(formula = as.factor(diagnostic) ~ as.factor(7161521) + as.factor(2281517), family = binomial()) Deviance Residuals: Min 1Q Median 3Q Max -1.5370 -1.0431 -0.9416 1.3065 1.4331 Coefficients:

Hi, Given > sec98 <- factor(rep(1:2,3), labels=c("A", "B")) > sec99 <- factor(rep(2:1,3), labels=c("A", "B")) > sec99[c(2,5)] <- NA > sec00 <- factor( c( rep(1,3), rep(2,3) ), labels=c("A", "B")) > sec00[c(2,4)] <- NA > sec1 <- ifelse(!is.na(sec99), sec99, ifelse(!is.na(sec00), sec00, NA

Hello everyone, I wanted to ask if there is an R-package to fit the following Poisson regression model log(\lambda_{ijk}) = \phi_{i} + \alpha_{j} + \beta_{k} i=1,\cdots,N (subjects) j=0,1 (two levels) k=0,1 (two levels) treating the \phi_{i} as nuinsance parameters. Thank you very much -- -Tony [[alternative HTML version deleted]]

In R 3.3.0 (also in R 2.7.2), the documentation on 'factor', in "Details" section, has this statement. 'factor(x, exclude = NULL)' applied to a factor is a no-operation unless there are unused levels: in that case, a factor with the reduced level set is returned. It is not true for a factor 'x' that has NA. In that case, if levels of 'x' doesn't

Hello! I have a data frame with a factor and a numeric variable: x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,2,10,20,100,200)) For each level of "factor" - I would like to divide each value of "values" by the mean of "values" that corresponds to the level of "factor" In other

Full_Name: Franz Quehenberger Version: 2.10.1 OS: Windows XP Submission from: (NULL) (145.244.10.3) aggregate is supposed to produce a data.frame that contains a line for each combination of levels of the variables in the by list. The first columns of the result contain these combinations of levels. With two by variables the second by-variable takes always only one value. However, it works fine