similar to: deficient rank question

Hi, I have encountered this error when attempting a One-way Repeated-measure ANOVA with my data. I have read the "Anova in car: SSPE apparently deficient rank" thread by I'm not sure the within-subject interaction has more degrees of freedom than subjects in my case. I have prepared the following testing script: rm(list = ls())

I have design with two repeated-measures factor, and no grouping factor. I can analyze the dataset successfully in other software, including my legacy DOS version BMDP, and R's 'aov' function. I would like to use 'Anova' in 'car' in order to obtain the sphericity tests and the H-F corrected p-values. I do not believe the data are truly deficient in rank. I

Dear: I want to construct a gee model in R. When I ran the program, there was a warning in the output. The warning is "Beginning Cgee S-function, @(#) geeformula.q 4.13 98/01/27 gee(cbind(hyper, nohyper) ~ I(Ethnic) + I(Gender) + I(drink) + : rank-deficient model matrix" What is the rank-deficient model matrix? What should I do to

Take the following example: a <- rnorm(100) b <- trunc(3*runif(100)) g <- factor(trunc(4*runif(100)),labels=c('A','B','C','D')) y <- rnorm(100) + a + (b+1) * (unclass(g)+2) m <- lm(y~a+b*g) summary(m) Here b is discrete but not treated as a factor. I am interested in computing the effect of b within groups defined by the

Dear R-devel, It seems that lsfit incorrectly reports coefficients when the input matrix 'x' is rank-deficient, see the example below: ## here values of 'b' and 'c' are incorrectly swapped > x <- cbind(a=rnorm(100), b=0, c=rnorm(100)); y <- rnorm(100); lsfit(x, y)$coef Intercept a b c -0.0227787 0.1042860 -0.1729261 0.0000000 Warning

I am trying to use lm() for resression followed by stepAIC function. Now when i try to use to predict for some input, predict() gives a warning : prediction from a Rank deficient matrix may be misleading. As I am new to R (or to statistics) How alarming this warning may be? Regards, ____________________________________________________________________________________ The fish are biting.

Hello all, I am getting the following error: Error in linear.hypothesis.mlm(mod, hyp.matrix.1, SSPE = SSPE, V = V, : The error SSP matrix is apparently of deficient rank = 7 < 11 After running: mod.ok <- lm(as.matrix(dat[,-1]) ~ DC, data=dat) (av.ok <- Anova(mod.ok, idata=idata, idesign=~week)) Although if I jitter the data in "dat", the function seems to work. What

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 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

Dear all, I would like to fit an lm in which a subset of the explanatory variables are linearly dependent. Thus I would like to include the restriction that all betas of these variables sum to 1. Is there a way to this in R? Or is this what happens automatically if I set singular.ok= T (which is the default, I believe)? Thanks for your help. Lorenz --

Can anyone help explain to me why the two codes below have different result? I thought I can use log(time)~. to replace log(time)~dist+climb+timef.I am using CVlm from DAAG package. I think nihills is preloaded with the package. Thanks in advance. > CVlm(df=nihills, form.lm=formula(log(time)~.),plotit="Observed",m=2)Analysis of Variance Table Response: log(time) Df Sum Sq

Hi, I am running a simulation and have to perform ANOVA to determine the rank of factors. Used the aov() function and it works great for full factorial design. 1. For a massive set of data, I tried using biglm, while it can create the linear model, all the residuals (for assumption validation) are not recorded and the sum of squares are not there, just the estimated regression coefficient, 95%

Dear Terry, It's not surprising that different modeling functions behave differently in this respect because there's no articulated standard. Please see my response to Martin for my take on the singular.ok argument. For a highly sophisticated user like you, singular.ok=TRUE isn't problematic -- you're not going to fail to notice an NA in the coefficient vector -- but I've

can anybody help with the following error message: Error in 1:object$rank : NA/NaN argument I get it with comparisons of single means in an ANOVA. Example data below. Thanks, Steffen seven subjects participated in each of 6 conditions (intervals). > subject = factor(rep(c(1:7), each = 6)) > interval = factor(rep(c(1:6), 7)) and here is the dependent variable: > dv = c(3.3868,

Dear All, I am trying to perform MANOVA. I have table with 504 columns(species) and 36 rows) with two grouping (season and location) Zx <- Z[c(4:504)] Zxm <- as.matrix(Z) m<- manova(Zxm~Season*location, data=Z) when I do summary.aov, I get respond for each species but summary.manova summary.manova(m) :" residuals have rank" 24<501. What can it be the reason for this error

Dear all; working with a 'fat' data set (700 variables / 50 samples) and trying to run a manova test on it (I'm aware that it's not the best option for this kind of data set) I got the error in the summary.manova function about the rank of the residuals (rank < # variables). Ok. The thing that I don't understand is why I don't get the same type of error in SAS. There

G'day all, I had a look at the GLM code of R (1.4.1) and I believe that there are problems with the function "glm.fit" that may bite in rare circumstances. Note, I have no data set with which I ran into trouble. This report is solely based on having a look at the code. Below I append a listing of the glm.fit function as produced by my system. I have added line numbers so that I

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 <-

Author: cth Repository: /hg/zfs-crypto/gate Revision: 8c3216f8ae4cd2c699db490d2a79406f3c420d36 Log message: 6308380 e_ddi_majorinstance_to_path implementation deficient for /devices/pseudo paths Files: update: usr/src/uts/common/os/sunddi.c

Hi, I understand that the glm.fit calls LINPACK fortran routines instead of LAPACK because it can handle the 'rank deficiency problem'. If my data matrix is not rank deficient, would a glm.fit function which runs on LAPACK be faster? Would this be worthwhile to convert glm.fit to use LAPACK? Has anyone done this already?? What is the best way to do this? I'm looking at very