similar to: aggregate and ordered factors, feature?

For ordered factor the natural contrast coding would be to parametrize by the succsessive differences between levels, which does not assume equal spacing of factor levels as does the polynomial contrasts (implicitly at least). This requires the contr.cum, which could be: contr.cum <- function (n, contrasts = TRUE) { if (is.numeric(n) && length(n) == 1) levs <- 1:n

Hi Dimitris, thank you for your reply, > does the following work in your data: > > levs <- unique.default(Names) > factor(Names, levs[order(unique.default(Weights))]) your solution is really shorter, but has two issues to be meant here: 1. "unique.default" is applied twice, what might be a bit expensive for strings. 2. your solution brings an implicit

When attempting to merge 3 data frames, one of which has fewer columns than the others, rbind.data.frame correctly refuses to perform the bind. However, the error message given is a bit obscure due to a logical bug in the match.names() internal function to rbind.data.frame. Illustration: ## Three data frames with same column variable names: > foo <- data.frame(v1 = c('a',

Hi, what would be a good way of enhancing the ancova function from the HH package when using a 2 factor ANCOVA? The current behaviour for the "ancova" function from package HH is: ---------------------------------------------- | P1 || P1 || P3 || P4 | | PS | <- the lattice strip ---------------------------------------------- | x|| x ||x ||x | |

Pursuant to a prior "on topic" thread (http://tolstoy.newcastle.edu.au/R/e4/help/08/07/17192.html ) where I found I could not use AOV to perform an anova on my large data set, I'm now trying to code the analysis "by hand" so to speak. However, as demonstrated below, when comparing my attempt to aov() using a smaller data set, I seem to betray some sort of

[I originally posted this on the R-help mailing list, and it was suggested that R-devel would be a better place to dicuss it.] Running ?table? on a factor with levels containing non-ASCII characters seems to result in extremely bad performance on Windows. Here?s a simple example with benchmark results (I?ve reduced the number of replications to make the function finish within reasonable time):

Hi, the following R lines work fine in R 2.4.0, but not in R 2.4.1 or any devel versions of R 2.5.0 (see below for details). library(drc) # to load the dataset 'PestSci' library(nlme) ## Setting starting values sv <- c(0.43355869, 2.49963220, 0.05861799, 1.73290589, 0.38153146, 0.24316978) ## No error m1 <- nlme(SLOPE ~ c + (d-c)/(1+exp(b*(log(DOSE)-log(e)))), fixed =

Dear R-help list, I am trying to construct a lattice histogram using 3 factors. My dataframe looks like this: (simulating a waterbalance over groundwater with different salinities) s days model EC EC_max 0.4 1 "A" 10 9 0.42 2 "A" 10 9 0.44 3 "A" 10 9 : : : :

Dear List, creating factors in a given non-default orders is notoriously difficult to explain in a course. Students love the ifelse construct given below most, but I remember some comment from Martin M?chler (?) that ifelse should be banned from courses. Any better idea? Not necessarily short, easy to remember is important. Dieter data = c(1,7,10,50,70) levs =

Full_Name: Lutz Prechelt Version: 2.4.1 OS: Windows XP Submission from: (NULL) (160.45.111.67) aggregate (from package stats) should preserve the ordering of levels of factors it works on and also their 'ordered' attribute if present. But it does not. Here is an example: ff =

>From commits to R devel, I saw attempts to speed up subsetting and 'match', and to cache results of conversion of small nonnegative integers to character string. That's good. I am sorry for pushing, still. Is the partial new behavior of function 'factor' with respect to NA really worthy? match(xlevs, nlevs)[f] looks nice, too. - Using f <- match(xlevs, nlevs)[f]

>>>>> Suharto Anggono Suharto Anggono via R-devel <r-devel at r-project.org> >>>>> on Sun, 15 Oct 2017 16:03:48 +0000 writes: > In R devel, function 'factor' has been changed, allowing and merging duplicated 'labels'. Indeed. That had been asked for and discussed a bit on this list from June 14 to June 23, starting at

My idea (like in https://bugs.r-project.org/bugzilla/attachment.cgi?id=1540 ): - For remapping, use f <- match(xlevs, nlevs)[f] instead of f <- match(xlevs[f], nlevs) (I have mentioned it). - Remap only if length(nlevs) differs from length(xlevs) . On use of 'order' in function 'factor' in R devel, factor.Rd still says 'sort.list' in "Details" section. My

I am trying once again. By just changing f <- match(xlevs[f], nlevs) to f <- match(xlevs, nlevs)[f] , function 'factor' in R devel could be made more consistent and back-compatible. Why not picking it? -------------------------------------------- On Sat, 25/11/17, Suharto Anggono Suharto Anggono <suharto_anggono at yahoo.com> wrote: Subject: Re: [Rd] Function

In R devel, function 'factor' has been changed, allowing and merging duplicated 'labels'. Issue 1: Handling of specified 'labels' without duplicates is slower than before. Example: x <- rep(1:26, 40000) system.time(factor(x, levels=1:26, labels=letters)) Function 'factor' is already rather slow because of conversion to character. Please don't add slowdown.

Hallo all Please help me. I am lost and do not know what is the problem. I have a factor called kvartaly. > attributes(kvartaly) $levels [1] "1Q.04" "2Q.04" "3Q.04" "4Q.04" "1Q.05" "2Q.05" "3Q.05" "4Q.05" $class [1] "factor" > mode(kvartaly) [1] "numeric" > str(kvartaly) Factor w/ 8

Suppose I have a dataframe 'd' defined as L3 <- LETTERS[1:3] d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace = TRUE)) (d <- d0[d0$fac %in% c('A', 'B'),]) x y fac 2 1 2 B 3 1 3 A 4 1 4 A 5 1 5 A 6 1 6 B 8 1 8 A Even though factor 'fac' in 'd' only has 2 levels, but it seems to bear the birthmark

Hi all, I had some trouble in?regrouping factor levels for a variable. After some experiments, I have figured out how I can recode to modify the factor levels. I would now like some help to understand why some methods work and others don't. Here's my code : rm(list=ls()) ###some trials in recoding factor levels char<-letters[1:10] fac<-factor(char) levels(fac) print(fac) ##first

Hello (and Happy New Year), When I create a factor with labels in the order I want, write the data as a text file, and then retrieve them, the factor levels are no longer in the proper order. Here is what I do (I tried many variations): # educ is a numeric vector with 1,001 observations. # There is one NA # Use educ to create a factor feducord <- factor(educ, labels = c('Elem',

With 2.7.0 patched (not tested with 2.0.0), I get an error message in a program that ran correctly in R 2.6.2 when the grouping factor of a stripplot contains an Umlaut: I am aware that there are a few locale-changes in R 2.7.0, but I could not easily locate who's at fault Dieter library(lattice) dt = data.frame(x=rnorm(100),y=1:100,levs= as.factor(c("Gru","Gr?")))