similar to: problem in fisher.test

I posted this last night but I think I figured out the problem. I checked on the underlying equation for the Fisher Exact Test for tables greater than 2X2. It looks like I have an insane amount of factorials to be multiplied. Is that problem? Is it an overflow issue? Jeff Last night’s post: I just a ran a fisher.test on a 9x5 table and received the following message

This is a bad bug as reported by Robin Hankin, it is still in "R-patched" ... ##- From: Robin Hankin <r.hankin@auckland.ac.nz> ##- To: r-help@stat.math.ethz.ch ##- Subject: [R] possum sleeping: thanks and fisher.test() FEXACT error ##- Date: Thu, 13 Jun 2002 16:46:26 +1200 ## ..... ## Example slighlty modified (MM) d4 <- matrix(c(0, 0, 0, 0, 0, 0, 3, 0, 1, 0, 0, 0, 0, 0,

Dear helplist Many many thanks to everyone who helped me. The trick was to use tabulate() or, better, tab <- rep(0,50) names(tab) <- 1:50 tab[names(table(sleeps))] <- table(c) My original dataset was a list of 50 trees and a length 12 vector recording which tree a certain possum slept in on 12 nights. As Professor Ripley points out, a Monte-Carlo simulation is easy to set up, and it

(CC'ed to R-bugs ``for the record'') >>>>> "BDR" == Prof Brian D Ripley <ripley@stats.ox.ac.uk> writes: BDR> On Thu, 13 Jun 2002, Martin Maechler wrote: >> >>>>> "MM" == Martin Maechler >> <maechler@stat.math.ethz.ch> writes: >> >> >>>>> "BDR" ==

C?dric Finet wrote: > > I thank you for your answer but I do not understand yet why the Fisher?s exact > test does not work. And why is a "negative key". > > C?dric Finet > Running the original TOMS643 fortran code (R uses an f2c translation of this) says: FEXACT ERROR: 30 Stack length exceeded in f3xact. This problem should not occur. The integer hash key is

Dear all, I’m having an awkward problem in R. When I write the command Fisher.test(school.data,workspace=2e+07), where school.data is the matrix corresponding to the data set, I get the error message: FEXACT error 7. LDSTP is too small for this problem. Try increasing the size of the workspace. Increasing the workspace: Fisher.test(school.data,workspace=1e+10), I

Hi! I am Marta Colombo, student in Mathematical Engineering at "Politecnico di Milano". For my master degree thesis I have to analyze some categorical data. My dataset is composed by 327 individuals and 16 variables. I am using Fisher exact test to test independence on IxJ contingency tables, but I have a problem with one variable. R gives me this error message: FEXACT error 7.

Hello, I have an error message that doesn't seem to make sense in that having read the R documentation I was under the impression that R was able to grab as much memory as it needed, and has been able to do so for some time, so the advice given below about increasing the size of the workspace is redundant. If I'm right, does anyone have a solution to the problem of the size of

Hi, I want to analyse a contigency table (3 x 12) with a fisher.test beacause there are cells that are less than 5. ?mmen Anken Baf Belchen H?chi Hof Porti R?m Schmutz Sch?n Sissa Tann class14 7 26 150 2 46 68 126 66 3 31 7 61 class24 7 6 55 5 49 71 93 90 1 18 16 79 class34 1 1 4 3 19 8 29 61

The hybrid feature in fisher.test looks to me like an excellent way to analyze my two-way tables. The only problem is that it does not seem to be implemented. Am I right about this? An example is pasted below. I note that I get the warning message only when I shouldn't: for a 2x2 table hybrid seems to be ignored without warning. In no case does fisher.test seem to be checking Cochran

Prof Brian D Ripley wrote: > > On Thu, 24 Aug 2000, Emmanuel Paradis wrote: > > > Dear All, > > > > We encountered an unexpected crash with R under Windows when running > > fisher.test(). We had this problem only under this OS and only with the > > following data: > > > > > T > > [,1] [,2] > > [1,] 2 1 > > [2,]

I want to ascertain the basis of the table ranking, i.e. the meaning of "extreme", in Fisher's Exact Test as implemented in 'fisher.test', when applied to RxC tables which are larger than 2x2. One can summarise a strategy for the test as 1) For each table compatible with the margins of the observed table, compute the probability of this table conditional on the

Hi folks, I'm trying to call an R function (fisher.test) in my program for like a billion times! Though my program is in Python and I feel that using rpy2 to interface R to python doesn't give me satisfactory performance. So I looked into R code and found out that fisher.test is actually a wrapper around another function called fexact which is implemented in C. Using Cython I managed to

Hello, I'm using R to apply Fishers exact test to a whole pile of contingency tables, and I've run into the bug shown below. regards, Francis -- > dat1 = matrix(c(0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,0,0, 1,1,0,0,1,0,0,1,0,0,1,0,0,3,0,0,2,0,0,1,0,0,0,1,0,0,2,0,0, 2,0,0,1,0,1,0,0,3,0,0,2,0,0,0,1,0,5,0,0,0,1,0,1,0,0), nrow=3) >

Full_Name: Anja von Heydebreck Version: 1.9.1 OS: Linux Submission from: (NULL) (155.250.128.25) I obtained a negative p-value from the fisher.test() function: > fisher.test(matrix(c(14576,3023,89,68),2), alternative="g")$p.value [1] -8.426593e-13 With R 1.8.1, I got a p-value of -6.239231e-12 for this example. Anja von Heydebreck

Hi folks, Forgive me if this question is a trivial issue. I was doing a series of Fishers' exact test using the fisher.test function in stats package. Since the counts I have were quite large (c(64, 3070, 2868, 4961135)), R suggested me to use *other algorithms* for the test which can be specified through the 'control' argument of the fisher.test function as I understood. But where

Dear all, I have a dataset with qualitative variables (factors) and I want to test the null hypothesis of independance between two variables for each pair by using appropriate tests on contingency tables. I first applied chisq.test and obtained dependance in almost all cases with extremely small p-values and warning messages. > chisq.test(table(data$ins.f, data$ins.st))$p.val [1]

Hi all, I'm sure this is simple but I don't get it. I have a table mytable<-c(rep("Disagree",37),rep("Agree",64)) table(mytable) this gives me Agree Disagree 64 37 but I didn't ask for it to be in alphabetic order. How can I get it in original order? Disagree Agree 37 64 Thanks, Jeff Jeffrey. M. Miller, PhD

I have two vectors (a and b) with counts of animals and wanted to calculate fisher's alpha: library(vegan) a <- c(2043, 1258, 52, 1867, 107, 1624, 2, 157, 210, 402, 5, 107, 267, 2, 13683) b <- c(2043, 1258, 52, 1867, 107, 1624, 2, 157, 210, 402, 5, 107, 267, 2, 3000) fisher.alpha(a) fisher.alpha(b) fisher.alpha(a) gave the following warnings: > fisher.alpha(a) [1] 1.572964

Hi, I received a bug report for my exact2x2 package that seems to apply to fisher.test in the stats package also. Here is some code which creates the error: >x<-factor(c(0,1,1),levels=c(0,1)) > y<-factor(c(1,1,1),levels=c(0,1)) > fisher.test(x,y) Error in fisher.test(x, y) : 'x' and 'y' must have at least 2 levels > The help says that x and y could be