similar to: help on permutation/randomization test

Dear listers, I was used to package pgirmess under Windows with everything OK, but, for the first time, I had a trial this afternoon on Ubuntu 7.10 gutsy (I have a double boot computer and work more and more under unix) and R 2.7.0. Everything went OK except this: sudo R CMD check pgirmess ..... * checking examples ... ERROR Running examples in 'pgirmess-Ex.R' failed. The error most

Hi, I've got a dataset with 7 variables for 8 different species. I'd like to test the null hypothesis of no difference among species for these variables. MANOVA seems like the appropriate test, but since I'm unsure of how well the data fit the assumptions of equal variance/covariance and multivariate normality, I want to use a permutation test. I've been through CRAN looking at

Hi all, I'm not familiar with R programming and I'm trying to reproduce a result from a paper. Basically, I have a dataset which I would like to model in terms of successive increments, i.e. (y denote empirical values of y) y_1 = y1, y_2 = y1 + delta1, y_3 = y1 + delta1 + delta2. ... y_m = y1 + sum_2^m delta j where delta_j donote successive increments in the y-values, i.e. delta

Hi - I am fitting a probit model using glm(), and the deviance and residual degrees of freedom are different depending on whether I use a binary response vector of length 80 or a two-column matrix response (10 rows) with the number of success and failures in each column. I would think that these would be just two different ways of specifying the same model, but this does not appear to be the case.

Hello, I have to compute a multiple summation (not an integration because the independent variables a are discrete) for all the values of a function of several variables f (x_1,...,x_n), that is sum ... sum f(x_1,...,x_n) x_1 x_n have you some suggestion? Is it possible? I know that for multiple integration there is the function adapt, but it has at most n=20. In my case n depends on the

>>>>> jing hua zhao >>>>> on Mon, 24 Jun 2019 08:51:43 +0000 writes: > Hi All, > Thanks for all your comments which allows me to appreciate more of these in Python and R. > I just came across the matrixStats package, > ## EXAMPLE #1 > lx <- c(1000.01, 1000.02) > y0 <- log(sum(exp(lx))) > print(y0) ## Inf

Does this give correct results for special floats (0, infs)? We tried to improve (for single floats) x86 rcp in llvmpipe with newton-raphson, but unfortunately not being able to give correct results for these two cases (without even more additional code) meant it got all disabled in the end (you can still see that code in the driver) since the problems are at least as bad as those due to bad

Hi, I want to ue a randomisation test, to compare two sample means, for a small set of data, I've looked at towt.permutation, perm.test and permtest, as well as permTS(), which is the best one to use for such a simple purpose? I know perm.test package isn't used anymore and coins recommended, but reading the manual I couldnt find the function most like it, ore more likely I

Signed-off-by: Ilia Mirkin <imirkin at alum.mit.edu> --- Untested beyond compiling a few shaders to see if they look like they might work. nvdisasm agrees with envydis's decoding of these things. Will definitely get ahold of a G200 to run tests on before pushing this. .../drivers/nouveau/codegen/nv50_ir_emit_nv50.cpp | 94 ++++++++++++++++++---

Hi Hadley, Thanks for the counterpoint. Response below. On Thu, May 16, 2019 at 1:59 PM Hadley Wickham <h.wickham at gmail.com> wrote: > The existing behaviour seems inutitive to me. I would consider these > invariants for n vector x_i's each with size m: > > * nrow(rbind(x_1, x_2, ..., x_n)) equals n > Personally, no I wouldn't. I would consider m==0 a degenerate

Hi, guys I am Wenjin from Graduate School of Chinese Academy of Science, pursing a master degree and my current research interests including using Data mining and Information retrieve technology to analysis software engineering (SE) data and support SE. I have great interested in "Weight Schemes" project. and in the last few days I have learnt some detail about DFR model family by

If I use the Ineq library and the Gini function in this way: >Gini(c(100,0,0,0)) I obtain the result 0.75 instead of 1 (that is the perfect inequality). I think Gini's formula in Ineq is based on a formula as reported here: http://mathworld.wolfram.com/GiniCoefficient.html but in the case of perfect inequality: x_1=.......=x_n-1 =0 x_n>0 these formula are equal to 1 - 1/n, not to

Hi, This is most likely only a minor technicality, but I saw the following: On page 6 of the 'parallel' vignette (http://stat.ethz.ch/R-manual/R-devel/library/parallel/doc/parallel.pdf), the random-number generator "L'Ecuyer-CMRG" is said to have seed "(x_n, x_{n-1}, x_{n-2}, y_n, y_{n-1}, y_{n-2})". However, in L'Ecuyer et al. (2002), the seed is given with

I am now writing an R code to do the systematic permutation test and the random permutation test. And I encountered a couple of problems which I cannot figure them out. At first, whenever I type "permtest" in my R2.6.1, it always shows:"there is no such a function", I tried again and again and eventually I found I have to use "library (BHH2)" before I use

Hi guRus, My discipline (experimental psychology) is gradually moving away from Null Hypothesis Testing and towards measures of evidence. One measure of evidence that has been popular of late is the likelihood ratio. Glover & Dixon (2005) demonstrate the calculation of the likelihood ratio from ANOVA tables, but I'm also interested in non-parametric statistics and wonder if anyone has any

Dear All, Suppose I have a vector X = (x_1, x_2, ...., x_n), X_sort = sort(X) = (x_(1), x_(2), ... , x(n) ), and I would like to know the original position of these ordered x_(i) in X, how can I do it? case 1: all values are unique x <- c( 3, 5, 4, 6) x.sort <- sort(x) # # I would like to obtain a vector (1, 3, 2, 4) which indicates that 3 in x is still the 1st element in x.sort, 5 is at

Hello, Say I want to make a multiple regression model with the following expression: lm(y~x1 + x2 + x3 + ... + x_n,data=mydata) It gets boring to type in the whole independent variables, in this case x_i. Is there any simple way to do the metaprogramming for this? (There are different cases where the names of the independent variables might sometimes have apparent patterns or not)

Hi, I have to draw samples from an asymmetric-Laplace-Normal distribution: f(u|y, x, beta, phi, sigma, tau) \propto exp( - sum( ( abs(lo) + (2*tau-1)*lo )/(2*sigma) ) - 0.5/phi*u^2), where lo = (y - x*beta) and y=(y_1, ..., y_n), x=(x_1, ..., x_n) -- sorry for this huge formula -- A WinBUGS Gibbs sampler and the HI package arms sampler were used with the same initial data for all parameters. I

Hello, I have what I suspect might be an easy problem but I am new to R and stumped. I have a data set that looks something like this b<-c(2,3,4,5,6,7,8,9) x<-c(2,3,4,5,6,7,8,9) y<-c(9,8,7,6,5,4,3,2) z<-c(9,8,7,6,1,2,3,4) data<-cbind(x,y,z) row.names(data)<-c('a','b','c','d','e','f','g','h') which gives: x y z

Hi, I am currently attempting to write a small program for a randomization test (based on rank/combination) for matched pairs. If you will please allow me to introduce you to some background information regarding the test prior to my question at hand, or you may skip down to the bold portion for my issue. There are two sample sizes; the data, as I am sure you guessed, is matched into pairs and