similar to: QR decomposition question

In summary.manova the qr decomposition of a NxN matrix is calculated and for some cases is giving me a rank < N. However, following suggestions of professor Ripley to calculate the rank of a Matrix On 7 Jun 2002, Brian Ripley wrote: > For a more reliable answer, look at the SVD > (function svd) and look at the > singular values. For example (from lda.default) X.s <-

Dear All, I luckily found the following feature (or problem) when tried to apply ifelse-function to an ordered data. > test <- c(TRUE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE) > ifelse(test, 0, 1:4) [1] 0 0 0 4 1 2 3 > It roots into the ifelse-syntax: ans[!test & !nas] <- rep(no, length.out = length(ans))[!test & !nas] Would it be possible to disable this feature in the

Hello dear r-gurus! I have a question about the logit-model. I think I have misunderstood something and I'm trying to find a bug from my code or even better from my head. Any help is appreciated. The question is shortly: why I'm not having same coefficients from the logit-regression when using a link-function and an explicite transformation of the dependent. Below some details. I'm

Hi, I would like to generate a correlation matrix with a particular structure. For example, a 3n x 3n matrix : A_(nxn) aI_(nxn) bI_(nxn) aI_(nxn) A_(nxn) cI_(nxn) aI_(nxn) cI_(nxn) A_(nxn) where - A_(nxn) is a *specified* symmetric, positive definite nxn matrix. - I_(nxn) is an identity matrix of order n - a, b, c are (any) real numbers Many attempts have been unsuccessful because a

> Thomas Lumley wrote: > > On Tue, 31 May 2005, Duncan Murdoch wrote: > > > > > >>M??kinen Jussi wrote: > >> > >>>Dear All, > >>> > >>>I luckily found the following feature (or problem) when tried to > >>>apply > >>>ifelse-function to an ordered data. > >>> > >>> >

Hi mighty R-gurus and other enthusiastics, I just encountered this: library(its) x <- its(sort(rnorm(10)), as.POSIXct(Sys.time() + 1:10)) plot(x, type = "p", pch = c(rep("A", 5), rep("B", 5))) Am I missing something if I expect that all the points labeled as 'A' should be below all those labeled as 'B'? Thanks, Jussi M?kinen platform

Hello, I had a couple questions about manova modeling in R. I have calculated a manova model, and generated a summary.manova output using both the Wilks test and Pillai test. The output is essentially the same, except that the Wilks lambda = 1 - Pillai. Is this normal? (The output from both is appended below.) My other question is about the use of MANOVA. If I have one variable which has a

Dear R-users; Previously I posted a question about the problem of rank deficiency in summary.manova. As somebody suggested, I'm attaching a small part of the data set. #*************************************************** "test" <- structure(.Data = list(structure(.Data = c(rep(1,3),rep(2,18),rep(3,10)), levels = c("1", "2", "3"), class =

Hi.? There appears to be a bug in R function manova.? My friend and I both ran it the same way as shown below (his run) with the shown data set. His results are shown below. we both got the same results.? I was running with R 2.3.1. I'm not sure what version he used. Thanks very much, David Booth Kent State University -----Original Message----- From: dvdbooth at cs.com To: kberk at

Hi all, I am experimenting the function "manova" in R. I tried it on a few data sets, but I did not understand the result: I used "summary(manova_result)" and "summary(manova_result, test='Wilks')" and they gave a bunch of numbers... But I need the Sum-of-Squares of BETWEEN and WITHIN matrices... How do I read off from the R's manova results? Any

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

Dear all, I need to do a Manova but I have an unbalanced design. I have morphological measurements similar to the iris dataset, but I don't have the same number of measurements for all species. Does anyone know a procedure to do Manova with this kind of input in R? Thank you very much, Naiara. -------------------------------------------- Naiara S. Pinto Ecology, Evolution and Behavior 1

Dear All, I would be very appreciative of your help with the following 1). I am running multivariate multiple regression through the manova() function (kindly suggested by Professor Venables) and getting two different answers for test=c("Wilks","Roy","Pillai") and tests=c("Wilks","Roy",'"Pillai") as shown below. In the

Dear friends, Sorry for this somewhat generically titled posting but I had a question with using contrasts in a manova context. So here is my question: Suppose I am interested in doing inference on \beta in the case of the model given by: Y = X %*% \beta + e where Y is a n x p matrix of observations, X is a n x m design matrix, \beta is m x p matrix of parameters, and e is a

Hi R-users I'm trying to do multivariate analysis of variance of a experiment with 3 treatments, 2 variables and 5 replicates. The procedure adopted in SAS is as follow, but I'm having difficulty in to implement the contrasts for comparison of all treatments in R. I have already read manuals and other materials about manova in R, but nothing about specific contrasts were found in them,

Dear helpeRs, the following data set comes from Johnson/Wichern: Applied Multivariate Statistical Analysis, 6th ed, pp. 304-306. /X <- structure(c(9, 6, 9, 3, 2, 7), .Dim = as.integer(c(3, 2))) Y <- structure(c(0, 2, 4, 0), .Dim = as.integer(c(2, 2))) Z <- structure(c(3, 1, 2, 8, 9, 7), .Dim = as.integer(c(3, 2)))/ I would like to compute Wilk's Lambda in R, which I know is 0.0385.

I would like to test the hypothesis that the difference between pairs, for several variables, is zero. This is easily done separately for each variable with: lm(Y ~ rep(0, nrow(Y))) where Y is a matrix whose columns are the differences for each variable between pair members. However, I would like to get an overall probability across all variables from a Wilks or Pillai-Bartlett statistic as in

Hi Sergio Doing those tests 30 times is going to give you a huge Type I error rate, even if there was a function that did that. There is a reason why TukeyHSD doesn't make it easy. In general, if there are useful comparisons among the species, you are better off setting up and testing contrasts than doing all-pairwise Tukey tests. Also, the PCA scores are ordered in terms of variance

Hi, I would like to conduct a MANOVA. I know that there 's the manova() funciton and the summary.manova() function to get the appropriate summary of test statistics. I just don't manage to specify my model in the manova() call. How to specify a model with multiple responses and one explanatory factor? If I type:

I wonder whether any of you know of an efficient way to calculate the approximate degrees of freedom of a gamm() fit. Calculating the smoother/projection matrix S: y -> \hat y and then its trace by sum(eigen(S))$values is what I've been doing so far- but I was hoping there might be a more efficient way than doing the spectral decomposition of an NxN-matrix. The degrees of freedom