similar to: How to solve empty cells in the contingency table?

Hi all, I am trying to analyze the following data. The first three columns are categorical variables (colors of three traits for a peripatus species) and the last column the count of individuals in each three-way classification. I wish to test if the three traits vary independently or if they are correlated across individuals, i.e. - this is a basic three way contingency table question.

Dear Listers: I have a categorical feature selection problem for random forest. Suppose I have a multiple-leveled category variable A, which has m=3 levels: red, green, and blue and the final target is binary classification. I want to evaluate its power in discrimination between 2 classes. We know rf splits multiple-leveled category variable by considering all combinations of its levels. So

I have a contingency table with a 0/1 variable (fwr) that flags a particular condition, a weight variable (cnt) and other variables (e.g. zz) characterizing that observation of the table. I am trying to use GLM. By converting the variables to factors and running the following regression: fwr1<-factor(fwr) zz1<-factor(zz1) res1<-glm(fwr1 ~ zz1,weights=cnt) Is that appropriate? Paul M.

Hello All, Is there any R function out there to turn a multi-way contingency table back to a flat file table of individual rows and attribute columns.? Thanks! marco --------------------------------- [[alternative HTML version deleted]]

Hi all, I have a 3x4 contingency table with row totals all being 100. I want to generate 3 x 4 tables from the null distribution. Which R function can do this? -- Thanks, Jim. [[alternative HTML version deleted]]

Dear R-philes: I am having an issue with exporting contingency tables with xtable(). I set up a contingency and convert it to a matrix for passing to xtable() as shown below. v.cont.table <- table(v_lda$class, grps, dnn=c("predicted", "observed")) v.cont.mat <- as.matrix(v.cont.table) Both produce output as follows: observed predicted uh uh~ uh 201

Hello, Is there anywhere I can find some Bayesian Code for 2 by 2 tables or even the non-central hypergeometric distribution in R? Packages would be helpful but the actual coding in R is much better. Thanks, Thanks Jim [[alternative HTML version deleted]]

Dear R-list, Hi everyone, Im trying to make an analysis of multidimensional contingency tables using R. I' working with the Agresti example where you have the data from 3 categories. The thing is how can I do the analisys using the G2 statistics. Somebody can send me an Idea? I attach the program where you can find the data. Best Regards, > prob1<-

Hi all, is it possible to perform a linear-by-linear association test (lbl_test, package coin) for a SINGLY ordered table ? And, if yes, how to manage the "scores" ? Thanks in advance Gilles - France [[alternative HTML version deleted]]

Hi, I would like to do a multiple correspondence analysis with the mca function in the MASS library on data that I have as a contingency table (which I've tried converting to a data frame). For example, ========= > data(HairEyeColor) > hair.df <- as.data.frame(HairEyeColor) > hair.df Hair Eye Sex Freq 1 Black Brown Male 32 2 Brown Brown Male 38 3 Red

Hi, all. I got a lot of help from the this mailing list, it save me a lot of time. Here is my question. x<-rep(c(2,3,4),20) y<-sample(rep(c(0,1),30)) tr<-rep(c(1:5),6) data1<-data.frame(x,y,tr) data1<-data1[order(data1$tr),] > data1 x y tr1 2 0 16 4 1 111 3 0 116 2 0 121 4 0 126 3 1 131 2 0 136 4 1 141 3 0 146 2 0 151 4 0 156 3 0 12 3 0 27 2 0

How can I draw a 3D perspective of a contingency table, that is showing 3D bars whose height is in proportion of the table cells? --------------- Charles RAUX, Laboratoire d'Economie des Transports CNRS-Universit? Lumi?re Lyon 2-ENTPE email : charles.raux at let.ish-lyon.cnrs.fr http://www.ish-lyon.cnrs.fr/let -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Full_Name: Tao Shi Version: 1.7.0 OS: Windows XP Professional Submission from: (NULL) (149.142.163.65) > x [,1] [,2] [1,] 149 151 [2,] 1 8 > c2x<-chisq.test(x, simulate.p.value=T, B=100000)$p.value > for(i in (1:20)){c2x<-c(c2x,chisq.test(x, simulate.p.value=T,B=100000)$p.value)} > c2tx<-chisq.test(t(x), simulate.p.value=T, B=100000)$p.value > for(i in

>>>>> dmurdoch writes: >> Date: Wed, 16 Jul 2003 01:27:25 +0200 (MET DST) >> From: shitao@ucla.edu >>> x >> [,1] [,2] >> [1,] 149 151 >> [2,] 1 8 >>> c2x<-chisq.test(x, simulate.p.value=T, B=100000)$p.value >>> for(i in (1:20)){c2x<-c(c2x,chisq.test(x, >> simulate.p.value=T,B=100000)$p.value)}

Greetings. I'm having some trouble constructing a contingency table from raw data (actually read in via RPgSQL). Here's the deal: - What I've got: a data frame in the form: Groupcode code1 code2 code3 code4 code5 .. coden where groupcode is one of {P,C,B,S,U,X} and code{1..n} is TRUE or FALSE. code{1..n} are NOT mutually exclusive, so between 0 and n of them can be TRUE. -

Is there any existing function for creating contingency tables that will display counts, row, column, and cell percentages in the same tables....anything similar to crosstabs in S? Marc W. Zodet, MS Health Statistician Center for Cost and Financing Studies Division of Statistical Research and Methods 2101 East Jefferson Street, Suite 500 Rockville, Maryland 20852 Phone: 301-594-7072 Fax:

I'm using R1.7.0 runing with Win XP. Thanks, ...Tao ???????????????????????????????????????????????????????? >x [,1] [,2] [1,] 149 151 [2,] 1 8 >t(x) [,1] [,2] [1,] 149 1 [2,] 151 8 >chisq.test(x, simulate.p.value=T, B=100000) Pearson's Chi-squared test with simulated p-value (based on 1e+05 replicates) data: x X-squared = 5.2001, df =

Hello R-users, I would like to generate two-way contingency tables with zero in one cell. I tried to use the function r2dtable but I could not force one cell to have zero value. Any Idea? Best regards.. Mahmoud [[alternative HTML version deleted]]

Dear Users, How can I obtain a profiles graphic in a CT similar to Excel. Campo Elias PARDO [[alternative HTML version deleted]]

Here is my table > tt A B 1 297 398 2 470 376 3 30 23 4 3 3 5 0 0 b/c two cells are zero, I can't use chisq.test() in R which gives the following output; > chisq.test(tt) Pearson's Chi-squared test data: tt X-squared = NaN, df = 4, p-value = NA Warning message: Chi-squared approximation may be incorrect in: chisq.test(tt) What function should I use