similar to: Question about computing offsets automatically

Hi, I'm having a problem using the 'SHLIB' command in R. I keep getting syntax errors. The help gives the following usage for it: Rcmd SHLIB [-o dllname] files 'files' is supposed to be a list of files, but whether I set up a list or a character vector, I still get a syntax error. Does anyone know if they're are supposed to be quotes somewhere in there? If someone has an

I'm using version 2.0.0 of R on a Windows 2000 OS. Packages that I installed in 1.9.1 are not installing properly in 2.0.0. Specifically, I can install packages from local zip files with the following code (which I execute through the GUI): > install.packages(choose.files('',filters=Filters[c('zip','All'),]), .libPaths()[1], CRAN = NULL) updating HTML package

Hi, I'm working with a programmer on translating code I wrote for a simulation in S into FORTRAN. Since the S source code is not available, I decided to use the random number generator (RNG) in R. The problem is that the R code is complex enough that we can't isolate the proper source files to include. Although we can link to the C code and call the RNG, we haven't been able to find

Dear List, I would appreciate help on the following matter: I am aware that higher dimensional contingency tables can be analysed using either log-linear models or as a poisson regression using a generalized linear model: log-linear: loglm(~Age+Site, data=xtabs(~Age+Site, data=SSites.Rev, drop.unused.levels=T)) GLM: glm.table <- as.data.frame(xtabs(~Age+Site, data=SSites.Rev,

Hello everyone, I'm using R 1.5.1 on Windows. I chose the option 'Install Source Files' when installing R. The code below shows my use of the command SHLIB and the error I get. C:\R\rw1051\bin>Rcmd SHLIB C:\Stage 9 (Phase 2)\cir3int.f 'perl' is not recognized as an internal or external command, operable program or batch file. I'm assuming I'm missing some perl

I have a sparse contingency table (most cells are 0): > xtabs(~.,data[,idx:(idx+4)]) , , x3 = 1, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 31 2 0 0 112 3 0 0 94 , , x3 = 2, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 3, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 1, x4

Full_Name: Paul Louisell Version: 1.6.2 OS: Windows NT Submission from: (NULL) (192.249.47.9) I've found a problem with either the 'apply' or 'is.factor' functions, and it seems like it might be cause for concern. Here are the commands taken directly from the R session: > is.factor(test.frame[, 1]) [1] TRUE > is.factor(test.frame[, 2]) [1] TRUE >

Dear list, I am trying to make a contingency table with xtabs but I am getting a 0 where I expect a 'NA'. Here is a simple example: options(stringsAsFactors = FALSE) rn <- LETTERS[1:4] df1 <- data.frame(r07 = rep(rn, each=4), r08 = rep(rn, 4), value = 1:16) xtabs(value ~ r07 + r08, df1) # Delete the combination [A, C] df1 <- df1[-3,] # Set 'value'

Hi, I've been trying to sort the values of the factors of contingency tables generated with xtabs. For example, I have a factor called "artic" with three possible values that I would like to order in a specific way -- but I'm not sure how to go about this. > test <- read.table("test.tab",header=TRUE,sep="\t") > artic =

Hi R users I have serveral binary variables (e.g., X1, X2, X3, X4, X5, X,6, and X7) and one continuous variable (e.g., Y1). I combined these variables using data.frame() mydata <- data.frame(X1,X2,X3,X4,X5,X6,X7,Y1) after that, I sorted this data.frame rank.by.Y1<-order(mydata[,8]) sorted.mydata<-mydata[rank.by.Y1,] after that, I replaced Y1's values with values ranging from 1

Full_Name: Paul Louisell Version: 1.6.2 OS: Windows NT Submission from: (NULL) (192.249.47.9) There is a slight bug in the function 'any'. Given a logical vector (possibly including values of 'NA'), the default action of any should be as follows: (i) at least one value = T should return T (ii) NO values = T (a) at least one value = 'NA' should return 'NA'

Hi, I am constructing a contingency table using xtabs. The function works great: mo yr Sep Oct Nov Dec 1950 -7.164486e-02 3.152674e-02 -1.283389e-02 1.570382e-01 1951 3.054293e-02 4.665234e-02 -2.445499e-04 8.720204e-02 1952 3.937034e-02 -4.790636e-02 5.022616e-02 1.180279e-01 but I wonder if there is an argument I can pass to xtabs

Does anyone know of another version of the Mcnemar test that provides: 1. Odds Ratios 2. 95% Confidence intervals of the Odds Ratios 3. Sample probability 4. 95% Confidence intervals of the sample probability Obviously the Odds Ratios and Sample probabilities are easy to calculate from the contingency table, but I would appreciate any help on how to calculate the confidence

How does one ideally handle and display multidimenstional contingency tables in R v. 2.6.2? E.g.: > prob1<- data.frame(victim=c(rep('white',4),rep('black',4)), + perp=c(rep('white',2),rep('black',2),rep('white',2),rep('black',2)), + death=rep(c('yes','no'),4), count=c(19,132,11,52,0,9,6,97)) > prob1 victim perp

Dear Friends. I am just starting to use R. And in this occasion I want to construct a high-dimensional contingency table, because I want to crate a mosaic plot with the vcd package. My table is in this format: año ac.rep cat.gru conteos 1 2005 R parejas 253 2 2005 N parejas 23 3 2006 R parejas 347 4 2006 N parejas 39 5 2007 R

Full_Name: Daniel Frey Version: 1.7.1 OS: Windows 2000 Submission from: (NULL) (80.254.164.242) Generating a data frame out of a xtabs result acts unusual. Take the following sample to reproduce it: > a.a <- c("a","a","a","b","b") > a.b <- c("c","c","d","e","f") >

Before I reinvent the wheel, I have need for a relatively straightforward crosstabulation (2 x n) function. I know that R has table(), ftable(), xtabs(), and summary(xtabs()), but none of these produce a fully "tricked" out cross-tabulation with marginal totals, expected cell frequencies, and an array of statistics about the contingency table. Is there a more complete (something

Hello, Does anyone know of an R version of loess that allows more than 4 predictors and/or allows the specification of offsets? For that matter, does anyone know of _any_ version of loess that does either of the things I mention? Thanks, Paul Louisell 650-833-6254 ploua@allstate.com Research Associate (Statistician) Modeling & Data Analytics ARPC [[alternative HTML version deleted]]

Hello there! I am still struggling with a binomial response over all categorical variables (some of them with 3 levels, most with 2 levels). After initial struggles with glm's (struggle coming from the data, not the actual analysis) I have decided to prefer contingency tables. I have my data such as: response:

Hello All R Users, Function loglm() in library MASS can be cajoled to accomodate structural zeros in a cross-classification table. An example from Fienberg demonstrates how this can be done. My question is: Can the function glm() perform the same task? Can glm() estimate a log-linear model with fixed zeros like loglm()? Thanks for your help, Andrew ## Fienberg, The Analysis of Cross-Classified