similar to: How to get the pseudo left inverse of a singular square m atrix?

>I'm rusty, but not *that* rusty here, I hope. > >If W (=Z*Z' in your case) is singular, it can not have >inverse, which by >definition also mean that nothing multiply by it will >produce the identity >matrix (for otherwise it would have an inverse and >thus nonsingular). > >The definition of a generalized inverse is something >like: If A is a >non-null

Dear R-listers, I have a dxr matrix Z, where d > r. And the product Z*Z' is a singular square matrix. The problem is how to get the left inverse U of this singular matrix Z*Z', such that U*(Z*Z') = I? Is there any to figure it out using matrix decomposition method? Thanks a lot for your help. Fred

R version: 1.7.1 OS: Red Hat Linux 7.2 In this example, I would expect an error for the overly long variable name. This is always reproducable for me. > formula(paste("y~",paste(rep("x",50000),collapse=""))) Segmentation fault Sincerely, Jerome Asselin -- Jerome Asselin (Jérôme), Statistical Analyst British Columbia Centre for Excellence in HIV/AIDS St.

I am trying to fit a lognormal distribution on interval-censored data. Some of my intervals have a lower bound of zero. Unfortunately, it seems like survreg() cannot deal with lower bounds of zero, despite the fact that plnorm(0)==0 and pnorm(-Inf)==0 are well defined. Below is a short example to reproduce the problem. Does anyone know why survreg() must behave that way? Is there an alternate

Hi, I would like to recommend a minor modification in axis() which I believe can simplify the making of plots for publications. I am trying to define default values for par() in order to make labels bigger and lines thicker, so that the resulting plots look good when resized for publication purposes. I ran into the following problem... axis() does not use par() values as default for

R version: 1.7.1 OS: Red Hat Linux 7.2 When "covmat" is supplied in princomp(), the output value "center" is all NA's, even though the input matrix was indeed centered. I haven't read anything about this in the help file for princomp(). See code below for an example: pc2$center is all NA's. Jerome Asselin x <- rnorm(6) y <- rnorm(6) X <- cbind(x,y)

I am confused on how the log-likelihood is calculated in a parametric survival problem with frailty. I see a contradiction in the frailty() help file vs. the source code of frailty.gamma(), frailty.gaussian() and frailty.t(). The function frailty.gaussian() appears to calculate the penalty as the negative log-density of independent Gaussian variables, as one would expect: >

I would like to document my findings (with a potential FIX) regarding the issue of calculation of the degrees of freedom with nlme(). The program given at the bottom of this email generates and fit 20 data sets with a mixed-effects LINEAR model, but using the function nlme() instead of lme(). In each case, the correct number of degrees of freedom for the intercept parameter is 12. However, in

Is there a simpler way then the solution to the one that was posted here? I'm not very proficient with legend, and I don't understand this solution. All I have is two or more lines on one plot that I want to put a legend on and I can't figure out how to do it from the examples. Can you give a very simple example? It does not have to be fancy!! I have never worked with a

Hi there, I want to justify to right the text of my legend. Consider this short reproducable example. x <- 1:5 y1 <- 1/x y2 <- 2/x plot(rep(x,2),c(y1,y2),type="n",xlab="x",ylab="y") lines(x,y1) lines(x,y2,lty=2) legend(5,2,c("1,000","1,000,000"),lty=1:2,xjust=1,yjust=1)

I am using the ginv function from MASS and have run across this problem that I do not understand. If I define the matrix A as below, its g-inverse does not satisfy the Moore-Penrose condition A %*% ginv(A) %*% A = A. The matrix A is X'WX in a quadratic regression using some very large dollar values. The much simpler matrix B does satisfy the MP condition. Am I doing something wrong? Is

The problem occurs when the sample odds ratio is Inf, such as in the following example. Given the fact that both upper bounds of the two 95% confidence intervals are Inf, I would have expected that the two lower bounds be equal, but they aren't. x <- matrix(c(9,4,0,2),2,2) x # [,1] [,2] #[1,] 9 0 #[2,] 4 2 rbind("two.sided.95CI"=fisher.test(x)$conf.int,

R version: 1.7.1 OS: Red Hat Linux 7.2 Hi all, The formula object in model.frame() is not retrieved properly when model.frame() is called from within a function and the "subset" argument is supplied. foo <- function(formula,data,subset=NULL) { cat("\n*****Does formula[-3] == ~y ?**** TRUE *****\n") print(formula[-3] == ~y) cat("\n*****Result of model.frame()

Dear Fred, If x1 and x2 are *not* normally distributed, you can use independent component analysis (ICA) which is based on the idea that x will be "more normal" than either x1 and x2 following the central limit theorem. See package(fastICA) by JL Marchini, C Heaton, and BD Ripley for details. HTH Thomas > -----Original Message----- > From: Feng Zhang [mailto:f0z6305 at

Generalised Inverse: The Moore-Penrose Generalisied Inverse is probably better defined as a pseudo-Inverse that arises in solving least squares problems. Another well known pseudo-Inverse is the so-called Drazin pseudo-Inverse. If memory serves (and it's been 10-12 years!) it can be obtained via a diagonalisation. Anyway, I dare say Prof. Ripley (among others) probably has "all the

Thorsten is right. There is a direct formula for computing the Moore-Penrose inverse using the singular value composition of a matrix. This is incorporated in the following: mpinv <- function(A, eps = 1e-13) { s <- svd(A) e <- s$d e[e > eps] <- 1/e[e > eps] return(s$v %*% diag(e) %*% t(s$u)) } Hope it helps. Dietrich

Tracking down this bug was joint work with Jermoe Asselin (jerome at hivnet.ubc.ca) and Patrick Connolly (p.connolly at hortresearch.co.nz). We collectively were able to determine that this is a problem in Windows 2000 but not in Linux. Timezones of the form GMT-5, GMT+3, etc. do not work properly in Windows 2000 for nearby dates in daylight savings time although they do work for nearby dates

Full_Name: Jerome Asselin Version: 1.7.1 OS: Red Hat Linux 7.2 Submission from: (NULL) (24.77.125.119) Setting the parameter na.action=na.omit should remove incomplete records in princomp. However this does not seem to work as expected. See example below. Sincerely, Jerome Asselin data(USArrests) princomp(USArrests, cor = TRUE) #THIS WORKS USArrests[1,3] <- NA princomp(USArrests, cor =

Full_Name: Jerome Asselin Version: 1.6.1 OS: RedHat Linux 7.2 Submission from: (NULL) (142.103.173.179) Hello, I get a segmentation fault when I run the following code. I wouldn't expect meaningful results because my response variable contains only missing values. However, I would expect something like a regular error (not a segmentation fault). library(survival) data <-

Full_Name: Jerome Asselin Version: 1.6.2 OS: RedHat Linux 7.2 Submission from: (NULL) (142.103.173.179) In the following example, the line type lty=2 does not show properly across the entire line. x <- c(seq(0,.5,.001),seq(.6,1,.1)) y <- rep(1,length(x)) plot(x,y,type="s",lty=2) Sincerely, Jerome Asselin