similar to: Names of list members in a plot using sapply

Hi R users: I will like to know if somebody works on insurance statistics (actuarial problems) and had use TRICAST, and can tell me if with all the R tools it can be build a solution like TRICAST or similar. In a word: Do you think that R has all the statistical tools (I mean modeling tools) to make a job similar to TRICAST? Does TRICAST has modeling tools that are not implemented on R yet?

Hello All: I am trying to replicate the results of an example found in Alan Agresti's "Categorical Data Analysis" on pages 267-269. The example is one of a 2 x 2 cross-classification table of ordinal counts: job satisfaction and income. I am able to get Agresti's results for the independence model (G^2 = 12.03 with df = 9) assuming as he does that the data is nominal, but

Is names(model1$coef) what you're looking for? -----Original Message----- From: Kenneth Cabrera [mailto:krcabrer at epm.net.co] Sent: 29 November 2002 10:36 Cc: R-help at stat.math.ethz.ch Subject: [R] Obtaining the variable names of a glm object Hi, R users! Suppose I make a model like this:

Hi R users: I will like to have a legend with math symbols and also with the value of a variable. But I cannot obtain both at the same time (symbol + value of a variable): Here is a reproducible example: m1<-5 m2<-12 plot(1:5,1:5,type="n") legend("topleft",legend=c(paste(expression(mu),"=",m1),expression(paste(mu,"=",m2))),lty=1:2) Thank you for

Hi R users: I will like to have a legend with math symbols and also with the value of a variable. But I cannot obtain both at the same time (symbol + value of a variable): Here is a reproducible example: m1<-5 m2<-12 plot(1:5,1:5,type="n") legend("topleft",legend=c(paste(expression(mu),"=",m1),expression(paste(mu,"=",m2))),lty=1:2) Thank you for

Hi R users: I got an error that "l1" is not found. This is a sample code: f1<-function(i,l1){ print(l1[[i]]) l1[[i+1]]<<-i return(i) } z1<-list(x=100,NULL,a=c(1,1)) lapply(1:3,f1,l1=z1) My final goal is to update the value of each list component to use it in the following step. I see that print function founds l1[[1]], but the assign instruction not. Thank you

Hallo all Suppose I have a summary amount of various colours used in different areas colour plastic paint black 15.5 173.8 brown 6.0 523.2 green 2.5 6.4 red 77.1 237.4 yellow 144.6 77.3 It seems that there are some preferable colours in these two areas (yellow in plastic, brown an red in paint). The problem seems to me similar to chi square test but it is intended for testing counts or

I have a probably basic problem concerning the R language. I will write some small R programs but don't really know how to start. Do I have to create my files in some editor like emacs or can I write my code in the R prompt? How to compile the code? I haven't found any good manual parts about this subject. regards A.P

Hi, I'm using: platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 1 minor 5.1 year 2002 month 06 day 17 language R and I would like to apply: > update.packages() trying URL

Hi R users: I want to know if any one of you had the problem with Tinn-R, when you call the library svIDE on the new R 2.5.0, (because in the old R 2.4.1 works with out any problem). I got this message: > library(svIDE) Loading required package: tcltk Loading Tcl/Tk interface ... done Warning messages: 1: '\A' is an unrecognized escape in a character string 2: no se reconoce el valor

Hi R users: I'm trying to make a function where two of the parameters are functions, but I don't know how to put each set of parameters for each function. What am I missing? I try this code: f2<-function(n=2,nsim=100,fun1=rnorm,par1=list(),fun2=rnorm,par2=list()){ force(fun1) force(fun2) force(n) p1<-unlist(par1) p2<-unlist(par2) force(p1) force(p2)

Hi R users: I have de following data frame (called "Sx") Descripcion Nitratos Cont85g 72.40 Cont85g 100.50 Cont85g 138.30 Cont80g 178.33 Cont80g 79.01 Cont80g 74.16 Cont75g 23.70 Cont75g 15.80 Cont75g 16.20 Patron80g

Hello gurus, I have some variables, and i am creating combinations for analysis in the end i need these variables to be displayed like "LEPTIN+SAA+PTH". currently i am using loop to perform this. I would appreciate any pointers to do it without the loop. > mols=c("LEPTIN","SAA","PTH","sEGFR") > samples=mols[1:3] > samples [1]

Hi R users. I am experimenting with ubuntu 7.04 Feisty. I install the ggobi package with apt-get. I got almost all the packages, but when I try to obtain rggobi, I got this message: ------------------------------------------------------------------------------------------------- install.packages("rggobi") Aviso en install.packages("rggobi") : argument 'lib' is

Hi R users: Thank you very much for you help, you are very kind. I am again asking for your help to make a function that returns the numbers of elements of each category given in another vector. I mean, I have a vector "x" with several elements (most of them more that once) and other vector "y" with the elements that I want to count in the first vector (is not a exhaustive

Hi, let's see, if someone can help my with this one: I have the string as follows: > str<-("one","two","three") Now I want to concatenate the items to one string, seperateted by space or something else, >str >"one, two, three" If possible without a loop. My actual goal ist to create string like >str.names >"female = names1, male

Hi, May be this helps: m1<- as.matrix(read.table(text=" y1 g24 y1 0 1 g24 1 0 ",sep="",header=TRUE)) m2<-as.matrix(read.table(text="y1 c1 c2 l17 ?y1 0 1 1 1 ?c1 1 0 1 1 ?c2 1 1 0 1 ?l17 1 1 1 0",sep="",header=TRUE)) m3<- as.matrix(read.table(text="y1 h4??? s2???? s30 ?y1 0 1 1 1 ?h4 1 0 1 1 ?s2 1 1 0 1 ?s30 1 1 1

Hi R users: How can I obtain with bwplot boxplots with bwplot whose box width will vary acording to other variable. bwplot(categ1~continuos|categ2,box.ratio=continuos2,data=data.base) But it doesn't work as I expected. Thank you for your help. Kenneth

Hi R users: I found that I cannot stack() a data.frame with factors. db1<-data.frame(replicate(6,factor(sample(c("A","B"),6,replace=TRUE)))) str(db1) db2<-stack(db1) db2 "db2" does not have any row. How can I stack them by the variables X1,X2,...,X6? Thank you for your help. Kenneth