search for: emmean

...hilesh Singh wrote: > I am a Professor of Statistics at Indira Gandhi Krishi Vishwavidyalaya, > Raipur, India. While teaching in class about analysis of variance using R, > I was doing a one-way analysis for the two data-sets given below in the > R-class. I got a typical error in "emmeans" package, please help: > > Data-set-1: > -------------- > Medley and Clements (1998) investigated the impact of zinc contamination > (and other heavy metals) on the diversity of diatom species in the USA > Rocky Mountains. The diversity of diatoms (number of species) and de...

...-Jan-2018 8:18 PM, "Sal Mangiafico" <salvatore.s.mangiafico at gmail.com> wrote: One way to avoid this error is to create the aov without using the with function, but instead use the data= option in the aov function. That is, medley2 = aov(diversity ~ zinc, data=medley.clementis) emmeans::emmeans(medley2, "zinc") You can see the difference in the calls: medley2$call medley.clementis.aov$call This works for the other data set as well, e.g. keough2 = aov(serpulid.ln ~ biofilm, data=keough.raimondi.ln) ~ Sal Mangiafico On 1/8/2018 4:44 PM, Rolf Turner wrote: On 07/0...

I am a Professor of Statistics at Indira Gandhi Krishi Vishwavidyalaya, Raipur, India. While teaching in class about analysis of variance using R, I was doing a one-way analysis for the two data-sets given below in the R-class. I got a typical error in "emmeans" package, please help: Data-set-1: -------------- Medley and Clements (1998) investigated the impact of zinc contamination (and other heavy metals) on the diversity of diatom species in the USA Rocky Mountains. The diversity of diatoms (number of species) and degree of zinc contamination (ca...

I am estimating fairly simple zero-inflated negative binomial models in glmmTMB. The models have just two factors and their interaction, a total of 4 levels. I was trying to use emmeans() to obtain estimates and 95% CI for these four levels. However, when I use emmeans() with the arguments component="response", type ="response", I do not get estimates that look like the negative binomial (conditional) part of the model multiplied by 1 - zero-inflation probabi...

...ad their own dataset, select columns they want an ANOVA analysis run on, and generate graphs that will allow users to view their results. However, I am getting the following error: *"Argument is of length zero."* Being new to Shiny, I am having trouble passing the user input column to an emmeans argument in order to do a post hoc analysis, and using that information to produce a graph. Can somebody help me with this? *The code for my dataset and application are provided below; copying and pasting directly into R should generate the reproducible example.* In my application, the following...

...also tried: (by (GROUP) cor.test (WK1FREQ,WK2FREQ)) (2) I also wanted to run the following analyses, written in SPSS syntax. I have searched the R archives and gather I may need to use lm instead of glm, but got lost. UNIANOVA avfreq BY sex WITH blik /METHOD = SSTYPE(3) /INTERCEPT = INCLUDE /EMMEANS = TABLES(OVERALL) /EMMEANS = TABLES(sex) /PRINT = DESCRIPTIVE ETASQ OPOWER HOMOGENEITY /CRITERIA = ALPHA(.05) /DESIGN = blik sex blik*sex . GLM avmth3fq avfreq BY SEX /WSFACTOR = timex 2 Polynomial /METHOD = SSTYPE (3) /PRINT = DESCRIPTIVE ETASQ TEST(MMATRIX) HOMOGENEITY /CRITER...

...something like this; # Function to check and install missing packages install_if_missing <- function(pkg) { if (!require(pkg, character.only = TRUE)) { install.packages(pkg, dependencies = TRUE) } library(pkg, character.only = TRUE) } # List of required packages. packages <- c("emmeans", "multcomp", "sandwich", "multcompView", "ggplot2", "stringr", "dplyr", "openxlsx", "tidyr") # Check and install each package lapply(packages, install_if_missing) Regards, Tim -----Original Message----- From: R...

Dear R-Devel, I seem to no longer be able to access the bug-reporting system, so am doing this by e-mail. My report concerns models where variables are explicitly referenced (or is it "dereferenced"?), such as: cars.lm <- lm(mtcars[[1]] ~ factor(mtcars$cyl) + mtcars[["disp"]]) I have found that it is not possible to predict such models with new data. For example:

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA512 Hi folks: Curious question. I've added Rcmdr to my setup (R 4.4.1). I would like to add all of the plugins. Is there a way to get install.packages() to gather up everything starting 'rcmdrplugin', or do I have to list each package individually between the brackets?? ?? Regards... - -- Brian Lunergan Russell, Ontario Canada -----BEGIN

...oom" "caper" "car" "CDM" [9] "CLME" "coin" "crosstable" "dclone" [13] "doBy" "drc" "Ecfun" "emmeans" [17] "epigrowthfit" "evd" "Exact" "fitode" [21] "fixest" "ggfortify" "ggplot2" "GLMMadaptive" [25] "glmmTMB" "gratia" "hdm...

I read through the editorial. This is the one of the most mega-ultra-super-biased articles I've ever read. e.g. The authors encourage Baysian methods, and literally encourage subjective approaches. However, there's only one reference to robust methods and one reference to nonparametric methods, both of which are labelled as purely exploratory methods, which I regard as extremely

? Tue, 21 May 2024 08:00:11 +0000 Dario Strbenac via R-devel <r-devel at r-project.org> ?????: > Would a less resource-intensive value, such as 1, be a safer default > CPU value for confint? Which confint() method do you have in mind? There is at least four of them by default in R, and many additional classes could make use of stats:::confint.default by implementing vcov(). >