similar to: bootstrap

heya, i am fitting linear mixed effect model to a response Y. Y shows an s-shaped distribution when using QQ-plots (some zero values and some very high values). hence, which transformation should i apply that Y follows a normal distribution? any r-function/package available to do this? thanks for any hint, regards, lukas ??? Lukas Indermaur, PhD student eawag / Swiss Federal Institute of

Hi, Fitting all possible models (GLM) with 10 predictors will result in loads of (2^10 - 1) models. I want to do that in order to get the importance of variables (having an unbalanced variable design) by summing the up the AIC-weights of models including the same variable, for every variable separately. It's time consuming and annoying to define all possible models by hand. Is there a

Hello Instead of testing against 0 i would like to test regression slopes against -1. Any idea if there's an R script (package?) available. Thanks for any hint. Cheers Lukas ??? Lukas Indermaur, PhD student eawag / Swiss Federal Institute of Aquatic Science and Technology ECO - Department of Aquatic Ecology ?berlandstrasse 133 CH-8600 D?bendorf Switzerland Phone: +41 (0) 71 220

Hello I want to batch job the calculation of many GLM-models, extract some values and store them in a file. Almost everything in the script below works (read file, extract values and write them to file) except I fail in indexing the GLM with the modelstructure it should run. Running GLM's conventionally is no problem. Conventionally a GLM is calculated as:

Heya I am fitting linear mixed effects model in R and want to assess the model fit (with Animal number as random factor; repeated measures for Animals): ts.model <- lme(LOG_FOC_MW ~ R_DN_SUM + ANIMAL + SEX+ YY, data = t.data, random = ~ 1 | ANIMAL, correlation=corCAR1(0.2, form = ~1 | ANIMAL ), method='ML', na.action=na.omit)). Is there a possability to easly compute an

Estimate region of highest probabilty density Dear R-community I have data consisting of x and y. To each pair (x,y) a z value (weight) is assigned. With kde2d I can estimate the densities on a regular grid and based on this make a contour plot (not considering the z-values). According to an earlier post in the list I adjusted the kde2d to kde2d.weighted (see code below) to estimate the

Dear R-helpers I am running R2.0.0 under Windows 2000. I am compiling a number of file paths into a simple text file that will be read by some other software we use. Unfortunately, it can only handle file paths with back slashes (MS Windows convention), and from R, I get file paths with forward slashes. The following didn't work. > gsub('/', '\\',

Hi, I am facing a hierarchically structured dataset, and I am not sure of the right way to analyses it with cforest, if their is one. - - BACKGROUND & PROBLEM We are analyzing the behavior of some social birds facing different temperature conditions. The behaviors of the birds were recorder during many sessions of 2 hours. Conditional RF (cforest) are quite useful for this analysis

Dear All Where can I lookup good methods to compute p from q=bin(m,n)p^n*(1-p)^(m-n) such that q<=alfa, alfa small. Are there such libs, code and source in R? Best Regards -- Jan Burse SIAM, EAWAG Scheuchzerstr. 67 ?berlandstr. 133 8006 Z?rich 8600 D?bendorf tel: +41-1-364 17 66 tel: +41-1-823 55 34

Hi, I originally posted this on the Rcpp github tracker, but it was suggested I post it here. I tried to compile the package https://github.com/khabbazian/sparseAHC/ under Windows. The package requires C++11 so I had to install the R devel build with gcc 4.9.3, and the latest Rtools. I got compilation and installation to work using Rcpp (0.12.3, from CRAN source). Package loads fine. However,

Hi, I am trying to make a package B that extends another package A. Package A uses Rcpp, and I want to extend a class X used there. So package A has src/X.h and inst/include/X.h class X { ... } src/X.cpp X::X( arguments ) { ... } My package B wants to do this: DESCRIPTION [...] LinkingTo: Rcpp, RcppArmadillo, A Imports: Rcpp, RcppArmadillo, A src/Y.h class Y: public X { ... } src/Y.cpp

Hi all, I want to modify a large number of text files (ca 4000) by replacing a value found on a particular line in them with a value from an R object. For a single file I would normally use: con<-file ("foo.txt", open="r+") content<-readLines(con) content[n]<-"test" writeLines(content,con) close(con) For repeating this for several files I can

A dataframe holds 3 vars, each checked true or false (1, 0). Another var holds the grouping, r and s: ### start:example set.seed(20) d <- data.frame(sample(c(0, 1), 20, replace=T), sample(c(0, 1), 20, replace=T), sample(c(0, 1), 20, replace=T)) names(d) <- c("A", "B", "C") e <- rep(c("r", "s"), 10) ### end:example How do I get the

A Likert scale may have produced counts of answers per category. According to theory I may expect equality over the categories. A statistical test shall reveal the actual equality in my sample. When applying a chi square test with increasing number of repetitions (simulate.p.value) over a fixed sample, the p-value decreases dramatically (looks as if converge to zero). (1) Why? (2) (If

I have created a function to do something: i <- factor(sample(c("A", "B", "C", NA), 793, rep=T, prob=c(8, 7, 5, 1))) k <- factor(sample(c("X", "Y", "Z", NA), 793, rep=T, prob=c(12, 7, 9, 1))) mytable <- function(x){ xtb <- x btx <- x # do more with x, not relevant here cat("The table has been created,

How to I "recode" a factor into a binary data frame according to the factor levels: ### example:start set.seed(20) l <- sample(rep.int(c("locA", "locB", "locC", "locD"), 100), 10, replace=T) # [1] "locD" "locD" "locD" "locD" "locB" "locA" "locA" "locA"

A list contains several matrices. Over all matrices (list elements) I'd like to access one matrix cell: m <- matrix(1:9, nrow=3, dimnames=list(LETTERS[1:3], letters[1:3])) l <- list(m1=m, m2=m*2, m3=m*3) l[[3]] # works l[[3]][1:2, ] # works l[[1:3]][1, 1] # does not work How can I slice all C-c combinations in the list? S?ren -- S?ren Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag,

Hello a <- c("Mama", "Papa", "Papa; Mama", "", "Sammy; Mama; Papa") a <- strsplit(a, "; ") mama <- rep(F, length(a)) mama[sapply(a, function(x) { sum(x=="Mama") }, simplify=T) > 0] <- T papa <- rep(F, length(a)) papa[sapply(a, function(x) { sum(x=="Papa") }, simplify=T) > 0] <- T # ... more

In https://stat.ethz.ch/pipermail/r-help/2008-March/156868.html I found what linear separability means. But what can I do if I find such a situation in my data? Field (2005) suggest to reduce the number of predictors or increase the number of cases. But I am not sure whether I can, as an alternative, take the findings from my analysis and report them. And if so, how can I find the linear

Dear R-ers, I conducted a regression analysis, and then intended to add the regression function (y=4.33+1.07x) in a graph. But the following code can only give me a text like y=a+bx. Who can help me out? Thank you very much in advance. CODE # Read data x <- c(1, 2, 3, 4, 5, 6, 7, 8, 9) y <- c(6, 5, 8, 9, 11, 10, 11, 12, 15) data01 <- data.frame(x, y) #