search for: lvls

Hello, I can get the median for each factor, but I'd like another column to go with each factor. The nm column is a long name for the lvls column. So unique work except for the order can get messed up. Example: x = data.frame(val=1:10,lvls=c('cat2',rep("cat1",4),rep("cat2",4),'cat1'),nm=c('longname2',rep("longname1",4),rep("longname2",4),'longname1')) x val lvls...

...n works fine. agg2 <- function(test=FALSE) { # create a factor y dat <- data.frame(y=sample(LETTERS[1:3],100,replace=TRUE)) # create a factor x dat$x <- sample(letters[1:4],100,replace=TRUE) # make a column of 1s and zeros # 1 when that row has that level of y # 0 otherwise lvls <- levels(dat$y) dat$ya <- 1*(dat[,1] == lvls[1]) dat$yb <- 1*(dat[,1] == lvls[2]) dat$yc <- 1*(dat[,1] == lvls[3]) # this works fine if you give the exact function agg1 <- aggregate(cbind(ya,yb,yc)~x,data=dat,sum) # and fine if you accept fo <- as.formula("cbin...

I'm working on an R package in C and can't seem to get the same level information about a factor that the R console displays. If I define a factor as: lvls <- factor(c('red','blue','blue','green','red'), c('blue','green','red'), ordered=TRUE) When I get the "levels" attribute in C, I get back the the first vector, not the second. If I run attr(lvls,"levels") in R,...

...ollowing previous correspondence . Dataset <- read.table("C:/Program Files/R/rw2011/data/miss/model1a.dat", header=TRUE, sep="\t", na.strings="NA", dec=".", strip.white=TRUE) attach(Dataset) require (nlme) with(Dataset, table(runnb, id, grp)) b.lvls <- table(Dataset$runnb) nb <- length(b.lvls) fit <- vector(mode="list", nb) for(i in 1:nb) fit[[i]]<- lme (trans1 ~ Index1 + grp, random = ~ 1 | id / grp , data = Dataset, na.action = "na.exclude") This (abov...

...is automatically if your cross classifying variables are factors with levels all the cities (sorted, if you like): > x <- sample(letters[1:5],8, rep=TRUE) > y <- sample(letters[1:5],8,rep=TRUE) > xtabs(~ x + y) y x c d e a 1 0 0 b 0 0 1 c 1 0 0 d 1 1 1 e 1 1 0 > lvls <- sort(union(x,y)) > x <- factor(x, levels = lvls) > y <- factor(y, levels = lvls) > xtabs( ~ x + y) y x a b c d e a 0 0 1 0 0 b 0 0 0 0 1 c 0 0 1 0 0 d 0 0 1 1 1 e 0 0 1 1 0 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people k...

I have two factors l1, l2, and I'd like to merge them. (Remark: The factors can not be converted to charaters) Function c() does not give me the result I want: > l1 = factor(c('aaaa', 'bbbb')) > l2 = factor(c('ccc', 'dd')) > lMerge = factor(c(l1, l2)) > lMerge [1] 1 2 1 2 Levels: 1 2 > I'd like to merge l1 and l2 and to get lMerge

...t; > Spencer Graves wrote: > > > What about the following: > > > F1 <- factor(c("b", "a")) > > > F2 <- factor(c("c", "b")) > > > k1 <- length(F1) > > > k2 <- length(F2) > > > F12.lvls <- unique(c(levels(F1), levels(F2))) > > > F. <- factor(rep(F12.lvls[1], k1+k1), levels=F12.lvls) > > > F.[1:k1] <- F1 > > > F.[-(1:k1)] <- F2 > > > F. > > [1] b a c b > > Levels: a b c > > > > This saves converting t...

...th levels all the cities (sorted, if you like): > > > x <- sample(letters[1:5],8, rep=TRUE) > > y <- sample(letters[1:5],8,rep=TRUE) > > > xtabs(~ x + y) > y > x c d e > a 1 0 0 > b 0 0 1 > c 1 0 0 > d 1 1 1 > e 1 1 0 > > > lvls <- sort(union(x,y)) > > x <- factor(x, levels = lvls) > > y <- factor(y, levels = lvls) > > > xtabs( ~ x + y) > y > x a b c d e > a 0 0 1 0 0 > b 0 0 0 0 1 > c 0 0 1 0 0 > d 0 0 1 1 1 > e 0 0 1 1 0 > > Cheers, > Bert > &g...

Hello, I am trying to map a factor variable within a data frame to a new variable whose entries are derived from the content of the original variable and there are fewer factors in the new variable. That is, I'm trying to set up a surjection. After first thinking that this would be a common operation and would have a quite simple interface, I can not seem to find one, nor any similar posts

Hi, I have a character vector as below: a<-c('10','3R','4','4R','5','5R','6','6R','7','8','9','7R','1','10R','11'

Hi! This is my first time posting. I've read the general rules and guidelines, but please bear with me if I make some fatal error in posting. Anyway, I have a continuous response and 29 predictors made up of continuous variables and nominal and ordinal categorical variables. I'd like to do lasso on these, but I get an error. The way I am using "lars" doesn't allow for the

I know that I can get a count of histogram bins in base R with plot=FALSE. However, I'd like to do the same thing with lattice. The problem is that I've set up shingles, and I'd like to get the count within each bin within each shingle. plot=FALSE doesn't seem to do it. [[alternative HTML version deleted]]

Dear Bert and Huzefa, Apologies for the late reply, my account got hacked and I have just managed to recover it. Thank you very much for your replies and the solutions. Both work well. I was wondering if there was any way to ensure (force) that all possible combinations show up in the output. The full dataset has 25 cities but of course people have not moved from Boston to all the other 24

Hello all, I have problem with setting up random effects. I have a model: y=x1+x2+x1*x2+z1+z1*x2 where x1, x2, x1*x2 are fixed effects and z1, z1*x2 are random effects (crossed effects) I use library(nlme) 'lme' function. My question is: how I should set up random effects? I did lme(y~x1+x2+x1:x2, data=DATA, random=~z1+z1:x2, na.action='na.omit') but it did not work.

or in base R : ?xtabs ?? as in: xtabs(~previous_location + current_location,data=x) (You can convert the 0s to NA's if you like) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, May 8, 2018 at 9:21 AM, Huzefa