similar to: Is there a bug in qr(..,LAPACK=T)

Hello. I am trying to invert a matrix, and I am finding that I can get different answers depending on whether I set LAPACK true or false using "qr". I had understood that LAPACK is, in general more robust and faster than LINPACK, so I am confused as to why I am getting what seems to be invalid answers. The matrix is ostensibly the Hessian for a function I am optimizing. I want to get

I have been struggling to find some informaation on what lm exactly does. I know it uses the QR decomp. However, I was recently faced with a somewhat badly scaled matrix and summary(lm) said Coefficients: ( 4 not defined because of singularities) does anyone know how lm chooses these 4 coef. is it forward building of the model --> drop last when qr sends a non full rank design matrix? My

Hi, all. I want to write some functions like glm() so i studied it. In glm.fit(), it calls a fortran subroutine named "dqrfit" to compute least squares solutions to the system x * b = y To learn how "dqrfit" works, I just follow how glm() calls "dqrfit" by my own example, my codes are given below: > qr <- >

man for `qr` says that the function uses LINPACK's DQRDC, while it in fact uses DQRDC2. ``` The QR decomposition of the matrix as computed by LINPACK or LAPACK. The components in the returned value correspond directly to the values returned by DQRDC/DGEQP3/ZGEQP3 ```

> Date: Wed, 30 Jun 1999 11:12:24 +0200 (MET DST) > From: Torsten Hothorn <hothorn at amadeus.statistik.uni-dortmund.de> > > yesterday I had a little shock using qr (or lm). having a matrix > > X <- cbind(1,diag(3)) > y <- 1:3 > > the qr.coef returns one NA (because X is singular). So I computed the > Moore-Penrose inverse of X (just from the

Might it be possible to tighten the definition of "is.qr". I noticed that after I mistakenly typed example(lm) # make lm object named lm.D9 qr.Q(lm.D9) which exhausted the heap memory and produced two warning messages. As an object of class "lm" has a "qr" component, "is.qr" failed to detect that "lm.D9" was not a "qr" object. The

Hi all, Regular and avid readers of this column will know that Don Driscoll and I have recently posted two messages requesting assistance concerning an apparent failure of "se.contrast" to produce an se for a contrast. So far, an ominous silence rings in our ears, but read on Gentle Reader, and see if even the machinations of "debug" doesn't stimulate you to respond with a

I am testing 'qr' with an admittedly contrived matrix and I am getting different results than I am from another package. The matrix that I am using is: x <- matrix(seq(.1, by=.1, length.out=12), 4) So the whole test is: x <- matrix(seq(.1, by=.1, length.out=12), 4) qr(x) And the output from 'R' is: $qr [,1] [,2] [,3] [1,] -0.5477226 -1.2780193

Dear All i have a little puzzle about eigenvector in the R. As we know that the eigenvector can be displayed on several form. For example A=matrix(c(1,2,4,3),2,2) if we want to get the eigenvalue and eigenvector, the code followed eigen(A) $values [1] 5 -1 $vectors [,1] [,2] [1,] -0.7071068 -0.8944272 [2,] -0.7071068 0.4472136 however, we also can calculate the vector matrix

hi all i have a question that about the eigen analysis found in R and in eviews. i used the same data set in the two packages and found different answers. which is incorrect? the data is: aa ( a correlation matrix) 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 now > svd(aa) $d [1] 4.9204

Hi, All: What tools exist for diagnosing singular gradient problems with 'nls'? Consider the following toy example: DF1 <- data.frame(y=1:9, one=rep(1,9)) nlsToyProblem <- nls(y~(a+2*b)*one, DF1, start=list(a=1, b=1), control=nls.control(warnOnly=TRUE)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial

R-users E-mail: r-help@r-project.org I have a quenstion on "gam()" in "gam" package. The help of gam() says: 'gam' uses the _backfitting algorithm_ to combine different smoothing or fitting methods. On the other hand, lm.wfit(), which is a routine of gam.fit() contains: z <- .Fortran("dqrls", qr = x * wts, n = n, p = p, y = y *

Hello, Are there any access functions to the various residual variables that should result from a repeated measures ANOVA ? MyAOVObject$residuals does not exist, and simply printing MyAOVObject gives a very long print of all fields in the result list, many of which I can't see what they are exactly : $error.qr$qraux, for instance. What I would like basically is to inspect those residuals

Ladies and Gentlemen, using > A <- structure(c(1, 0, 0, 3, 2, 1, 4, 5, -3, -2, 1, 0), .Dim = as.integer(c(3,4))) I get > dim(A) [1] 3 4 > qr.R(qr(A),complete=TRUE) [,1] [,2] [,3] [,4] [1,] -1 -3.000000 -4.000000 2.0000000 [2,] 0 -2.236068 -3.130495 -0.8944272 [3,] 0 0.000000 -4.919350 -0.4472136 > qr.R(qr(A),complete=FALSE) [,1]

Hi, with small matrices eigen works as expected: > eigen(cbind(c(1,4),c(4,7)), only.values = TRUE) $values [1] 9 -1 $vectors NULL > eigen(cbind(c(1,4),c(4,7))) $values [1] 9 -1 $vectors [,1] [,2] [1,] 0.4472136 -0.8944272 [2,] 0.8944272 0.4472136 > eigen(cbind(c(1,-1),c(1,-1))) $values [1] -3.25177e-17+1.570092e-16i -3.25177e-17-1.570092e-16i $vectors

Dear R Core, working on my dynamic factor modelling package, which requires several subroutines to create and update several system matrices, I come back to the issue of being annoyed by R not supporting multiple assignment out of the box like Matlab, Python and julia. e.g. something like A, C, Q, R = init_matrices(X, Y, Z) would be a great addition to the language. I know there are several

Hi there, Given a positive definite symmetric matrix, I can use chol(x) to obtain U where U is upper triangular and x=U'U. For example, x=matrix(c(5,1,2,1,3,1,2,1,4),3,3) U=chol(x) U # [,1] [,2] [,3] #[1,] 2.236068 0.4472136 0.8944272 #[2,] 0.000000 1.6733201 0.3585686 #[3,] 0.000000 0.0000000 1.7525492 t(U)%*%U # this is exactly x Does anyone know how to obtain L such

Below is a SIMEX object that was generated with the "simex" function from the "simex" package applied to a logistic regression fit. From this mountain of information I would like to extract all of the values summarized in this line: .. ..$ variance.jackknife: num [1:5, 1:4] 1.684 1.144 0.85 0.624 0.519 ... Can someone suggest how to go about doing this? I can extract the

Hi All: I'm new to R and have a few questions about getting R to run efficiently with large datasets. I'm running R on Windows XP with 1Gb ram (so about 600mb-700mb after the usual windows overhead). I have a dataset that has 4 million observations and about 20 variables. I want to run probit regressions on this data, but can't do this with more than about 500,000 observations before

Dear Friends, I had performed anova test on certain data frame (Health Care Management) and got results [summary(aov)]. I am new to R and also some extent to Statistics. Can somebody help me how should I interpret these figures. I feel difficulty in interpreting values and respective rows and columns. The following is the result to which I request interpretation: > anova.stress$effects