similar to: Storing data from a test as a vector or matrix

It looks from str(SA) that Response IPS1 is a data.frame of class "anova", which probably cannot be coerced to vector. Maybe you can use unlist() instead of as.vector() Or something like SA[["Response IPS1"]]["as.factor(WSD)",] ## to select the first row only, even maybe with unlist() Without a better REPRODUCIBLE example, I cannot tell more (maybe some others

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

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 list member, My question is related to input file format to an Anova from car package. Here is an example of what I did: My file format is like this (and I dislike the idea that I will need to recode it): Hormone day Block Treatment Plant Diameter High N.Leaves SH 23 1 1 1 3.19 25.3 2 SH 23 1 1 2 3.42 5.5 1 SH 23 1 2 1 2.19 5.2 2 SH 23 1 2 2 2.17 7.6 2 CH 23 1 1 1 3.64 6.5 2 CH 23 1 1 2

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, Suppose I have: > summary(manova(plank.man)) Df Pillai approx F num Df den Df Pr(>F) plankton.new[, 1] 1 0.5267 9.8316 6 53 2.849e-07 *** Residuals 58 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 My understanding is the MANOVA summary returns a list.

Dear list, I use Sweave almost exclusively for writing papers, and I have become quite spoiled by the excellent xtable export facilities. Has anybody written an xtable method for the Anova function in CAR, or has anybody used a different set of functions to import Anova results into a table in an Sweave document? If not, any handy hints on how to write a good homebrew based on the output of Anova

Hi, Does anyone know what the error message mean? > Boot2.Pillai <- function(x, ind) { + x <- as.matrix(x[,2:ncol(x)]) + boot.x <- as.factor(x[ind, 1]) + boot.man <- manova(x ~ boot.x) + summary(manova(boot.man))[[4]][[3]] + } > > man.res <- manova(as.matrix(pl.nosite) ~ + as.factor(plankton.new[,1]))$residuals > boot2.plank <-

Hi all, I have received the following error from summary.manova: Error in summary.manova(manova.test, test = "Pillai") : residuals have rank 36 < 64 The data is simulated data for 64 variables. The design is a 2*2 factorial with 10 replicates per treatment. Looking at the code for summary.manova, the error involves a problem with qr(). Does anyone have a suggestion as to how to

Hello all, I have data: Genes time rep vart dye y trt 130911 sa1-d07 030min 1 col g 9.636244 o 145771 sa1-d07 030min 1 col r 8.107577 c 93335 sa1-d07 030min 1 ler g 7.409566 o 94821 sa1-d07 030min 1 ler r 5.107160 c 10119101 sa1-d07 030min 2 col g 8.336862 o 11605101 sa1-d07 030min 2 col r 7.824530 c 725313 sa1-d07

I don't think it will be easy to identify the problem remotely. You can try logging in with password (if not disabled), or sshing with some other key, or logging into some other user. If you are able to get access to the machine, post the server's error log here. On Mon, Oct 24, 2016 at 9:55 PM, Jin Li <lijin.abc at gmail.com> wrote: > Hi Tanmay, > > The server is not

Hi All, I use Ubuntu server 10.04 LTS as virtualization platform. Actually running kernel 2.6.32-31-server #61-Ubuntu S root at jupiter:~# uname -a Linux jupiter 2.6.32-31-server #61-Ubuntu SMP Fri Apr 8 19:44:42 UTC 2011 x86_64 GNU/Linux We have on running virtual root at jupiter:~# virsh list --all Id Name State ---------------------------------- 1 kvmtik.4safety.cz

Can R do a repeated measures MANOVA and tell what dimensionality the statistical variance occupies? I have been using MATLAB and SPSS to do my statistics. MATLAB can do ANOVAs and MANOVAs. When it performs a MANOVA, it returns a parameter d that estimates the dimensionality in which the means lie. It also returns a vector of p-values, where each p_n tests the null hypothesis that the mean

Tom, Is not this query worth answering? -Siva -----Original Message----- From: Sivamurugu K. Pillai Sent: Friday, April 08, 2005 3:14 PM To: ''Mailing List for Shorewall Users'' Subject: ProxyARP in a Routed environment Hi, In a routed network setup , is it possible to use ProxyARP given the condition that the shorewall external interface and the DMZ interface are in a

Hello everybody ! I try to obtain in R eigenvectors and canonical analysis on MANOVA results, but I don't find how to process? In particular, I would be interesting to obtain "standardized canonical coefficients" of the canonical variates. There analysis give some information on the correlation between response variates. My data are organised in 2 terms (one is continu, one is a

Hi all, I'm trying to run a repeated measures ANOVA on some univariate ecological data that was collected over two growing seasons. I ran the test using the methodology found on this website: http://rtutorialseries.blogspot.com/2011/02/r-tutorial-series-one-way-repeated.html Upon running the actual ANOVA I got this error message: "> rmanova=anova(yearmodel, idata=yearframe,

Thanks for both the mapply and array approaches! However, although intended to generate the same result, they don't: # mapply approach n = 3 sa <- rnorm(n,0.8,0.1) so <- rnorm(n,0.5,0.1) m <- rnorm(n,1.2,0.1) mats = mapply(function(sa1, so1, m1) matrix(c(0,sa1*m1,so1,sa1),2,2,byrow=T), sa, so, m, SIMPLIFY = FALSE) print(mats) [[1]] ????????? [,1]????? [,2] [1,] 0.0000000

Dear all mailing listers, Does Anyone have the same problem as mine when using the fast99 (extended-FAST method) to perform SA of model with norm distribution inputs? See the simple example given following. Any suggestion will be greatly appreciated. Thank you! Marino # Simple example # 1. uniform version (It works well) library(sensitivity) Myfun<-function(x){return(rowSums(x))} SA1

Suppose I have interest in a matrix with the following symbolic structure (specified by 3 parameters: sa, so, m): matrix(c(0,sa*m,so,sa),2,2,byrow=T) What I can't figure out is how to construct a series of matrices, where the elements/parameters are rnorm values. I'd like to construct separate matrices, with each matrix in the series using the 'next random parameter value'.

Dear all, is there anyone who can help me extracting p-values from an Anova object from the car library? I can't seem to locate the p-values using str(result) or str(summary(result)) in the example below > A <- factor( rep(1:2,each=3) ) > B <- factor( rep(1:3,times=2) ) > idata <- data.frame(A,B) > fit <- lm( cbind(a1_b1,a1_b2,a1_b3,a2_b1,a2_b2,a2_b3) ? sex,