similar to: change the class of an object within with()

Dear R-people: François Michonneau's method to obtain the special character plus/minus works on Windows 2000 professional. Many Thanks to François for his work! Phil Smith Centers for Disease Control and Prevention Atlanta, GA _____ From: François MICHONNEAU [mailto:francois.michonneau@gmail.com] Sent: Monday, March 20, 2006 1:58 PM To: r-help@stat.math.ethz.ch; Smith, Phil

Hello all, In one of the packages (phylobase) I'm contributing to, we define a class as follows: setClass("phylo4", representation(edge = "matrix", edge.length = "numeric", label = "character", edge.label = "character", order =

Hello, I would like to know what it means, when the hier.part's output has a data frame of a negative independent contribution for variables (?IJ) and how this can be caused. Thank you very much. Yours sincerely M. Heuner

Dear R-helpers, First I apologize if my question is quite simple I have a large datasets which more 100 variables. For a research I need to change all name of variables with add one or more letters on each variables. For example, > data(Pima.tr) > Pima.tr[1:5,] npreg glu bp skin bmi ped age type 1 5 86 68 28 30.2 0.364 24 No 2 7 195 70 33 25.1 0.163 55 Yes 3 5

Hello wizaRds ! I am looking for a french and recent version of "An introduction to R". Does anybody know where i could find on of these. Thanks. -- Romain Fran??ois 25, avenue Guy Moquet 94 400 Vitry sur seine FRANCE _______________________ _______________________ francoisromain at free.fr 01 46 80 65 60 06 18 39 14 69

Hello. I want to create a list of vectors but each component of the list has a different length. For example: Example=list() Example=list(Example,c(1,2,3)) Example=list(Example,c(11,12,13,14,15)) If I want the first component of the Example list, I have to write: Example[[1]][[2]]. R responses 1 2 3 The second component: Example[[2]]: 11 12 13 14 15 If I iterate the previous example,

Hi ! I have the following: <li id="photo_<%= @photo.id %>" class="photo"> <%= link_to(image_tag(url_for_file_column(:photo, :picture)), {:action => ''photo'', :id => @photo}, {:title => @photo.description}) %> </li> <script type="text/javascript" language="JavaScript"><!--

Hi I was wondering if there is a permutation test available in R for linear models with continuous dependent covariates. I want to do a test like the one shown here. bmi<-rnorm(100,25) x<-c(rep(0,75),rep(1,25)) y<-rnorm(100)+bmi^(1/2)+rnorm(100,2)*x+bmi*x H0<-lm(y~1+x+bmi) H1<-lm(y~1+x+bmi+x*bmi) anova(H0,H1) summary(lm(y~1+x+bmi)) But I want to use permutation testing to

Im doing a project using Rcommander. I have a dataset with 5 variables, BMI being one of them. I would like to find the variance, standard deviation, range, median, etc of BMI. However, everytime I type a command into the script window, (such as var(BMI$BMI,na.rm=TRUE) )an error message pops up saying Object "BMI" not found. I can graph it, work it, do t-test, etc,. It's listed as

As of rev. 61035 in r-devel, setRefClass() now returns a generator function, as setClass() has done since 2.15.0. The convenient style is now: mEdit <- setRefClass("mEdit",......) xx <- mEdit(data = xMat) instead of xx <- mEdit$new(data = xMat) The returned object still has fields and methods accessible as before. See the "Value" and "Reference Class

I am doing a reanalysis of results that have previously been published. My hope was to demonstrate the value of adoption of more modern regression methods in preference to the traditional approach of univariate stratification. I have encountered a puzzle regarding differences between I thought would be two equivalent analyses. Using a single factor, I compare poisson models with and without

Hello, Say I have 2 columns, bmi and gender, the first being all the values and the second being male or female. How would I subset this into males only and females only? I have searched these fora and read endlessly about select[] and split() functions but to no avail. Also the table is not ordered. bmi gender -> bmi gender + bmi gender 1 24.78 male

Hola a todos, -Estoy estudiando el efecto de dos genotipos (~tratamientos) en la aparición de síndrome metabólico (MetS) con datos longitudinales recogidos a tiempo 0,7,10,15,20 y 25 años. -He hecho un dataframe con las siguientes variables MetS: Síndrome Metabólico (Si=1,No=0) bmi: Indice de masa corporal (IMC) cuando se produce la conversión a MetS+ . Para los que permancen MetS-, esta variable

hi everyone, wondering if you could help me with a novice problem. I have a data frame called subjects with a height and weight variable and want to calculate a bmi variable from the two. i have tried: attach(subjects) bmi <- (weight)/((height/100)^2) but it comes up with the error: Warning messages: 1: In Ops.factor(height, 100) : / not meaningful for factors 2: In Ops.factor((weight),

Hello, I'm not sure if I'm in the right place with my question... I'm running R on Windows and wrote a function and saved it as .R file. It looks like this: bmi <- function(weight, height) { bmi <- weight / height^2 bmi } If I want to use this function, I have to mark everything and then press Ctrl-R. But then everything single line is executed on the command line,

Hi All, I have a small dataset named das (43 cases) in which I am trying to create a binary outcome (1/0) based on the following code: if (das$age>65 && das$bmi>30) {das$danger<-1} else das$danger<-0 I am setting a flag called 'danger' to 1 of the subject is over 65 and has a BMI > 30. I find that my statement evaluates the first record in the data.frame and

Hello, I'm looking for a way to get the "name" of an object when it is used within an "sapply". More precisely, with a simple example : I have a named list of objects : myList = list(a=rnorm(10),b=rnorm(10),c=rnorm(10)) I would like to create a new object from each of the components of myList using the "sapply" function, for example to get the mean of all

Hey guys, Does anyone have any wisdom to share on the subject of speccing Rails layouts? Most of it''s plain old view specs stuff, but are there sensible ways to verify things like the yield call? (Mocking doesn''t catch that) Thanks, Matt -- Matt Patterson | Design & Code <matt at reprocessed org> | http://www.reprocessed.org/

I am new to R, and I am trying to cut a continuous variable BMI into different categories and can't figure out how to use it. I would like to cut it into four groups: <20, 20-25, 25-30 and >= 30. I am having difficulty figuring the code for <20 and >=30? Please help. Thank you. -- View this message in context:

Dear R-experts, I am trying to bootstrap (and average) the median squared error evaluation metric for a robust regression. I can't get it. What is going wrong ? Here is the reproducible example. ############################# install.packages( "quantreg" ) library(quantreg) crp <-c(12,14,13,24,25,34,45,56,25,34,47,44,35,24,53,44,55,46,36,67) bmi