similar to: Bug in FEXACT: gave negative key (PR#6986)

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

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 All, I am using fexact.c on a C++ program I wrote. To minimize dependencies on other files (e.g., to not need to include R.h and ctest.h ---now I only include the R files Boolean.h, Constants.h, and Memory.h), I have re-written all declarations of Sints as ints and, what is potentially more serious, I have re-written the line (lines 329 and 330, in fexact.c on R-1.6.1) /* IMAX is the

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

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

Dear R experts: Here is my problem: #Data 1 Y <- c(0.5, 0.1, 0.5, 1.3, 1.4, 1.6, 1.65, 2.4, 2.6, 3.4, 3.6, 4.3, 4.42, 4.8, 4.7, 3.4, 3.3, 2.8, 2.8, 1.2, 1.1, 0.5, 0.2, 0.1, -0.2, -1.5, -2.5, -1.3, -0.5, -0.1) X <- seq(1:30) X1 <- c(rep("T1", 24), rep("T2", 6)) dat1 <- data.frame(Y, X, X1) require(lattice) mcol <- c("green", "red")

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 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

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.

I am trying to get R 1.0 running on the Mac. The main target is MacX. Anyone else working on that? For the recent Macintosh system, I am trying to compile R using MachTen. The R core compiles and runs without any changes, using the Unix make files. The libraries give some problems, presumably due to handling of shared libraries. It seems I am missing something to link. Before I spend my time

Sarah's solutions are good, and here's another, even more basic: tmp1 <- unique(dat1$B) tmp2 <- seq_along(tmp1) dat1$C <- tmp2[ match( dat1$B, tmp1) ] > dat1 N B C 1 1 29_log 1 2 2 29_log 1 3 3 29_log 1 4 4 27_cat 2 5 5 27_cat 2 6 6 1_log 3 7 7 1_log 3 8 8 1_log 3 9 9 1_log 3 10 10 1_log 3 11 11 3_cat 4 12 12 3_cat 4 As a single line

I need to split a given matrix in a sequential order. Let my matrix is : > dat <- cbind(sample(c(100,200), 10, T), sample(c(50,100, 150, 180), 10, > T), sample(seq(20, 200, by=20), 10, T)); dat [,1] [,2] [,3] [1,] 200 100 80 [2,] 100 180 80 [3,] 200 150 180 [4,] 200 50 140 [5,] 100 150 60 [6,] 100 50 60 [7,] 100 100 100 [8,] 200 150 100 [9,]

Buenas tardes a todos, tengo un problema con R: ejecuto el mismo script en el ordenador del trabajo y en mi portátil con los mismos datos y obtengo resultados diferentes (siendo los correctos, los obtenidos en el trabajo): rm(list=ls()) directorio<-"C:\\Documents and Settings\\BDs\\" library(RODBC) library(car) library(gdata) ### DATOS A ###

Hi, Here's one way to approach it, using the coercion of factor to numeric. Note that I changed your data.frame() statement to avoid coercing strings to factors, just to make it simpler to set the levels. dat1 <-data.frame(N=seq(1, 12,1), B=c("29_log","29_log", "29_log", "27_cat", "27_cat", "1_log", "1_log",

Hi Sarah, Thank you so much!! I got your good ideas. Ding -----Original Message----- From: Sarah Goslee [mailto:sarah.goslee at gmail.com] Sent: Friday, May 11, 2018 11:40 AM To: Ding, Yuan Chun Cc: r-help mailing list Subject: Re: [R] add one variable to a data frame [Attention: This email came from an external source. Do not open attachments or click on links from unknown senders or

Dear all, let suppose I have following vector:   > dat1 <- c(rep("asd", 5), rep("xyz", 12), rep("erd", 17)) > dat1 <- dat1[sample(1:length(dat1), length(dat1), replace=F)] > dat1  [1] "erd" "xyz" "erd" "asd" "asd" "erd" "xyz" "asd" "erd" "erd"

HI R help, I was trying to get identical data frame from a list using two methods. #Suppose my list is: listdat1<-list(rnorm(10,20),rep(LETTERS[1:2],5),rep(1:5,2)) #Creating dataframe using cbind dat1<-data.frame(do.call("cbind",listdat1)) colnames(dat1)<-c("Var1","Var2","Var3") #Second dataframe conversion

Dear all, I have following two list object, both are basically collection of matrices : dat1 <- matrix(rnorm(25*6), ncol=6) dat1 <- split(dat1, seq(5,25,by=5)) dat1 <- lapply(dat1, matrix, ncol=6) dat2 <- matrix(rnorm(25*4), ncol=4) dat2 <- split(dat2, seq(5,25,by=5)) dat2 <- lapply(dat2, matrix, ncol=4) Now I want to replace last 4 columns of each matrix at "dat1"

Sarah et. al.: As a matter of aesthetics (i.e. my personal ocd-ness) I prefer using the public API of an object, i.e. *not* to makes use of the representation of a factor as essentially an integer vector with labels, but rather to use its documented behavior. (Feel free to ignore this remark!) Anyway, >cumsum(!duplicated(dat1$B)) [1] 1 1 1 2 2 3 3 3 3 3 4 4 will do it. This is very

Dear all, Please forgive me if there is a duplicate post; my previous mail perhaps didnt reach the list....... Let say I have following time series library(zoo) > dat1 <- zooreg(rnorm(10), start=as.Date("2010-01-01"), frequency=1) > dat1[c(3, 7,8)] = NA > dat1 2010-01-01 2010-01-02 2010-01-03 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 2010-01-09