similar to: Fast version of Fisher's Exact Test

I have a matrix say, 1 4 23 30 and I want to find the previously attainable fisher's exact test p-value. Is there a way to do this in R? -- Thanks, Jim. [[alternative HTML version deleted]]

Hi, R-help, I have a group of data from RNA-seq want to be analyzed by Fisher's exact test in R. I want to compare the significant difference of about 30,0000 individuals in two different samples, and I have no idea how to use R, so could you please give me some suggestions or the scripts for Fisher's exact test? Thank you very much. Best, Guanfeng Wang [[alternative HTML version

Hi, I'm try to compute the minimum sample size needed to have at least an 80% of power, with alpha=0.05. The problem is that empirical proportions are really small: 0.00154 in one case and 0.00234. These are the estimated failure proportion of two medical treatments. Thomas and Conlon (1992) suggested Fisher's exact test and proposed a computational method, which according to their table

Full_Name: Martha Nason Version: 2.0.1 OS: Windows XP Submission from: (NULL) (137.187.154.154) I am running simulations using fisher's test on 2 x c tables and a very small p.value from fisher's test (<2.2e-16) is returned as a negative number. Code follows. > set.seed(0) > nreps.outer <-7 > pvalue.fisher <- rep(NA,nreps.outer) > > population1 <- c(

Hi, I'm trying to run Fisher's Exact test on the data below, but I'm not sure how interpret the data shown. Can someone tell me what this is saying? Looking at the numbers it should be that there's no significant difference between the HDL and LDL, but a p-value of 1 seems high. Is the low value in the LDL unbound making the test unstable and should I be using an alternative? Best

I have a matrix with 2 columns and I want to do fishers exact test for these with the totals for each row being 100 say. The data has the form: 23 12 32 21 12 2 and these represents the tables: 23 12 77 88 32 21 78 79 12 2 88 98 How do I use apply to speed up aclculation of the fisher.exact test? -- Thanks, Jim. [[alternative HTML version deleted]]

Timothy, I believe you are mistaken. Fisher's exact test give the correct answer even in the face of small expected values for the cell counts. Pearson's Chi-square approximates Fisher's exact test and can give the wrong answer when expected cell counts are low. Chi-square was developed because it is computationally "simple". Fisher's exact test, particularly with tables

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

Appology in advance if this has been asked and answered. I am getting different answers using Fisher's exact test in the package ctest. For example: x_cbind(c(8, 22), c(31, 29)) fisher.test(x) Fisher's Exact Test for Count Data data: x p-value = 0.04024 alternative hypothesis: two.sided However, in SAS or other packages on the net the answer I get is p-value= 0.02664. Is this

Hi, I have a table of observed counts for various genetic markers. Instead of doing Fisher's exact test for each marker one at a time and recording the P value manually, is there a script to go through the whole list and generate the P value column automatically? An example of my data: Counts_CHB and Counts_AA are the observed counts for one allele. 2N_CHB and 2N_AA are the total number

Full_Name: Lixin Han Version: 2.4.0 OS: Windows 2000 Submission from: (NULL) (155.94.110.222) A character vector c('a','b') is supplied to rm(). As a result, 'c' is deleted unintentionally. > a <- 1:5 > b <- 'abc' > c <- letters > ls() [1] "a" "b" "c" > rm(c('a','b')) > ls() character(0)

Dears Sirs During my computational work I encountered unexpected behavior when calling "ar" function, namely # time series x<-ts(c(-0.2052083,-0.3764986,-0.3762448,0.3740089,0.2737568,2.8235722,- 1.7783313,0.2728676,-0.3273164),start=c(1978,3),frequency=4,end=c(1980,3)) # ar function res.ar<-ar(x,aic=TRUE,demean=F) # call "ar" again and ............

Hi all, I have been having problems generating png files on a Mac running OS X 10.6.8. Here's a simple example > png("foo.png", type = "cairo"); plot(1:10); dev.off(); null device 1 libpng warning: Application built with libpng-1.2.26 but running with 1.5.2 The resultant file is of size 0 Kb. Is this the proper place to report this issue? Or should

Hi all, What are current methods people use in R to identify mis-spelled column names when selecting columns from a data frame? Alice Johnson recently tackled this issue (see [BioC] posting below). Due to a mis-spelled column name ("FileName" instead of "Filename") which produced no warning, Alice spent a fair amount of time tracking down this bug. With my fumbling fingers

Hi I was wondering whether anyone can help me with this problem....it's been driving me nuts, I've been trying to figure it out for months and months without success!! Basically I have a group of participants who attended 2 experimental sessions a few months apart. I took measures of the way they approach two tasks at Time 1 and the same two tasks at Time 2. I have categorical data (a

Hi useRs, A recent coding infelicity along these lines yielded a corrupt data frame. foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4] > bar X1 X2 X3 X4 NewCol 1 1 4 7 10 <NA> 2 2 5 8 11

Dear All, I have a problem understanding the difference between the outcome of a fisher exact test and a chi-square test (with simulated p.value). For some sample data (see below), fisher reports p=.02337. The normal chi-square test complains about "approximation may be incorrect", because there is a column with cells with very small values. I therefore tried the chi-square with

Hi all, I believe this is a bug in the model.matrix function. I'd like a second opinion before filing a bug report. If I have a nested covariate B with multiple values for just one level of A, I can not get a non-singular design matrix out of model.matrix > df <- data.frame(A = factor(c("a", "a", "x", "x"), levels = c("x",

In an effort to select the most appropriate number of clusters in a mixture analysis I am comparing the expected and actual membership of individuals in various clusters using the Fisher?s exact test. I aim for the model with the lowest possible p-value, but I frequently get p-values below 2.2e-16 and therefore does not get exact p-values with standard Fisher?s exact tests in R. Does anybody know

Full_Name: Steven McKinney Version: 2.9.0 OS: Mac OS X 10.5.6 Submission from: (NULL) (142.103.207.10) A corrupt data frame can be constructed as follows: foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4]