similar to: probable bugs in stats::loglin calculation of pearson chisq

The null loglinear model is an intercept-only model for log frequency, log(f) = \mu For a one-way table the test of the null model is the same as the chisq.test. This can be fit using loglin(), but I don't think there is any way to specify this using MASS::loglm > t1<- margin.table(Titanic,1) > t1 Class 1st 2nd 3rd Crew 325 285 706 885 > loglin(t1, NULL) 0

Kai, Apologies for the double message, it didn't go to the list last time. On Thu, Oct 22, 2015 at 9:59 AM, Kai Nitschke < kai.nitschke at uniklinik-freiburg.de> wrote: > > ?.Call? calls the C/C++ function ?C_LogLin?. But when I am running it line > by line I get > the following error on line 23/24: > Error: object 'C_LogLin' not found > > Hence, my

Summary: loglin will fail if there is a global variable called c. I have fixed this for 0.64.2 and 0.65. On Sat, 22 May 1999, Cor en Aylin wrote: > Dear Brian, > > Using the mass library for R (the most recent port dated 16 may 1999) I Could you give version numbers in such reports please: at the moment there is an official version (5.3pl037) of MASS for R, plus a pre-release of MASS

> tab <- array( sample(3^5), rep(3,5) ) > loglin( tab, list(1,2,3,4,5) )[[1]] # AOK 2 iterations: deviation 5.456968e-12 [1] 10909.89 > loglin( tab, list(1,2,3,4,5), c(1,2,3) )[[1]] # OUCH! Process R bus error at Wed Jul 21 17:03:55 2004 this is inconsistent - sometimes issuing this msg several times before barfing: Error in switch(z$ifault, stop("This should not

Dear R gurus, I am looking for a way to fit a predictive model for a contingency table which has counts. I found that glm( family=poisson) is very good for figuring out which of several alternative models I should select. But once I select a model it is hard to present and interpret it, especially when it has interactions, because everything is done "relative to reference cell". This

Hi everyone, I have a question regarding a C function of the "stats" package in R. I tried to understand the ?loglin? basic function of the ?stats? package implemented in R. The implemented function itself runs without any problem (perhaps see sample). When I ran it line by line it stopped at the lines 23-24 of the loglin-function; (the following line): z <- .Call(C_LogLin,

Hello, I'm currently trying to model the movement of a slope (v.obs) with a regression model. The data can be found following the given links: either http://www.sendspace.com/file/dnugwc or http://rapidshare.com/files/420569660/sel.day.txt I want to use the Box-Cox transformation to normalize the response as well as the predictor variables. The scatterplot looks like this: library(zoo)

Dear List, recently tried to reproduce the results of some custom model selection function after updating R, which unfortunately failed. However, I ultimately found the issue to be that testing with pchisq() in drop1() seems to have changed. In the below example, earlier versions (e.g. R 2.4.1) produce a missing P-value for the variable x, while newer versions (e.g. R 2.7.1) produce 0 (2.2e-16).

Good day all, My statisticians want an R procedure that will produce grouped stacked barplots. Barplot will stack or group, but not both. The ftable function can produce a table of the exact form they want, but the barplot doesn't show all the divisions we want. For an example, here's the sample from the help file for "ftable:" data(Titanic) ftable(Titanic, row.vars = 1:3)

I sent this in with an old version, but it's in latest version as well. The fix is simple. In the summary.table function, the parameter is calculated incorrectly for a test of independence among all cells when the table is more than 2-way table. Example: Consider X: > X a b c 1 A1 B2 C1 2 A3 BA3 C2 3 A2 B1 C4 4 A1 B2 C3 5 A3 BA3 C2 6 A1 BA3 C1 7 A2 BA3 C2 8 A1

Hello all, I would like to know what the difference is between chisq.test and fisher.test when using the Monte Carlo method with simulate.p.value=TRUE? Thank you -- View this message in context: http://r.789695.n4.nabble.com/Difference-in-Monte-Carlo-calculation-between-chisq-test-and-fisher-test-tp2322494p2322494.html Sent from the R help mailing list archive at Nabble.com.

Hello, I am trying to fit gamma, negative exponential and inverse power functions to a dataset, and then test whether the fit of each curve is good. To do this I have been advised to calculate predicted values for bins of data (I have grouped a continuous range of distances into 1km bins), and then apply a chi-squared test. Example: > data <- data.frame(distance=c(1,2,3,4,5,6,7),

Hi list, i've got a question about the chisq.test function. in the use of the "given probabilities" method (p= ...), normally there should be typed in probabilities in the range of 0 to 1 with the absolute sum of 1.0 (r-help) But it is possible to use probabilities > than 1. or the sum <1.! without any warning message Ok, now the question, what does r calcutate in these

Hi Tao: The P-values for 2x2 table are generated based on a random (discrete uniform distribution) sampling of all possible 2x2 tables, conditioning on the observed margin totals. If one of the cells is extremely small, as in your case, you get a big difference in P-values. Suppose, you changed the cell with value 1 to, say, 5 or 6, then the two P-values are nearly the same. However, I

R has amazing capabilities, but percentage tables are a weak spot IMHO. There's prop.table but that's rather unwieldly, especially for multiway tables. CrossTable by Marc Schwartz in the gregmisc library makes percentage tables a breeze but is limited to two-way tables. So I decided to try my own hand at writing an R-function that would make it easy to produce nicely formatted percentage

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

Hello everybody, I'm seeing some strange behavior on R 1.8.1 on Intel/Linux compiled with gcc 3.2.2. The p-value calculated from the chisq.test function is incorrect for some input values: > chisq.test(matrix(c(0, 1, 1, 12555), 2, 2), simulate.p.value=TRUE) Pearson's Chi-squared test with simulated p-value (based on 2000 replicates) data: matrix(c(0, 1, 1,

>>>>> 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)}

Hi all, I'm trying to do an example in Deepayan Sarkar's Lattice book. It involves making a barchart based on the Titanic dataset. I can get the barchart to plot fine; however, when I try to edit panel.grid, I get an error: > titan<-barchart(Class ~ Freq | Age + Sex, data = as.data.frame(Titanic), groups=Survived, stack=TRUE, layout=c(4,1),

Hi, I have a text mining project and currently I am working on feature generation/selection part. My plan is selecting a set of words or word combinations which have better discriminant capability than other words in telling the group id's (2 classes in this case) for a dataset which has 2,000,000 documents. One approach is using "contrast-set association rule mining" while the