similar to: lm diagnostics and qr (fwd)

The following snippet suggests that there is either a bug in qr(,LAPACK=T), or some bug in my understanding. Note that the detected rank is correct (= 2) using the default LINPACK qr, but incorrect (=3) using LAPACK. This is running on Linux Redhat 9.0, using the lapack library that comes with the Redhat distribution. I'm running R 1.7.1 compiled from the source. If the bug is in my

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

> 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

Hi, I think there is a problem with the LAPACK version of qr() in version 1.7.0. (version below). 1. The documentation states that LAPACK=TRUE is the default, but the code has LAPACK=FALSE. 2. With LAPACK=TRUE qr() is never pivoting, even in cases where it very clearly should be. e.g. set.seed(0) X<-matrix(rnorm(40),10,4);X[,1]<-X[,2] qrx<-qr(X,LAPACK=TRUE) qrx$pivot # note, no

Hi, I am out of town and will get back to you on the 13th of July. Leo >>> "r-help at stat.math.ethz.ch" 06/27/03 00:32 >>> Send R-help mailing list submissions to r-help at stat.math.ethz.ch To subscribe or unsubscribe via the World Wide Web, visit https://www.stat.math.ethz.ch/mailman/listinfo/r-help or, via email, send a message with subject or body

Hello List, I would like to orthonormalize vectors contained in a matrix X taking into account row weights (matrix diagonal D). ie, I want to obtain Z=XA with t(Z)%*%D%*%Z=diag(1) I can do the Gram-Schmidt orthogonalization with subsequent weighted regressions. I know that in the case of uniform weights, qr can do the trick. I wonder if there is a way to do it in the case of non uniform

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 need to perform a QR-decomposition of a sparse matrix, so I've been trying to use the Matrix package. Unfortunately I don't seem to be getting exactly the same results as if I had used the qr() command from the base package. Here is an example of what I'm doing. > spdata <-rpois(50,1) > y <- rnorm(10,0,1) > S <-

Dear R people, I do not have much knowledge about linear algebra but currently I need to understand what the function qr.qty is actually doing. The documentation states that it calculates t(Q) %*% y via a previously performed QR matrix decomposition. In order to do that, I tried following basic example: m<-matrix(c(1,0,0,0,1,0,0,0,1,0,0,1),ncol=3) # 4x3 matrix

With this example set.seed(123) A <- matrix(runif(40), nrow = 8) y <- 1:nrow(A) A.laqr <- qr(A, LAPACK=TRUE) both qr.qy(A.laqr,y) and qr.qty(A.laqr,y) give the respective error messages Error in qr.qy(A.laqr, y) : 'b' must be a numeric matrix Error in qr.qty(A.laqr, y) : 'b' must be a numeric matrix However when Lapack is not used as in A.liqr <- qr(A,

Dear maintainers, I'm reporting a bug in qr.coef that mishandles the colnames of matrix. A minimal reproducible example is as follows: x <- cbind(rep(0, 10), rep(0, 10)) y <- rep(0, 10) q <- qr.default(x) qr.coef(q, y) [1] NA NA If x has colnames, then qr.coef will end up with an error: x <- cbind(x1 = rep(0, 10), x2 = rep(0, 10)) y <- rep(0, 10) q <- qr.default(x)

Hi, I need help regarding estimating a linear model where restrictions are imposed on the coefficients. An example is as follows: Y_{t+2}=a1Y_{t+1} + a2 Y_t + b x_t + e_t restriction a1+ a2 =1 Is there a function or a package that can estimate the coefficient of a model like this? I want to estimate the coefficients rather than test them. Thank you for your help Ahmad Abu Hammour --------------

Hi, Why the qr() produces a negative Q compared with Gram-Schmidt? (note example below, except Q[2,3]) Here is an example, I calculate the Q by Gram-Schmidt process and compare the output with qr.Q() a <- c(1,0,1) b <- c(1,0,0) c <- c(2,1,0) x <- matrix(c(a,b,c),3,3) ########################## # Gram-Schmidt ########################## A <- matrix(a,3,1) q1 <-

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

The bug I submitted yesterday (It's not entered in the bug data base, so I have no ID for it) included a suggested fix that is not correct. It worked for the examples I gave because there was no pivoting in fact, or only pivot permutations that were idempotent. A correction that works in general on the examples I gave makes these two changes in qr.coef(): ## coef[qr$pivot, ]

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, I have noticed different rank values calculated by qr() depending on LAPACK parameter. When it is FALSE (default) a true rank is estimated and returned. Unfortunately, when LAPACK is set to TRUE, the min(nrow(A), ncol(A)) is returned which is only occasionally a true rank. Would not it be more consistent to replace the rank in the latter case by something based on the following pseudo code ?

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 ```

Hi mailinglist members, I’m actually working on a time series prediction and my current approach is to decompose the series first into a trend, a seasonal component and a remainder. Therefore I’m using the stl() function. But I’m wondering how to get the single components in order to predict the particular fitted series’. This code snippet illustrates my problem: series <-

Hello, I don't understand what went wrong or how to fix this. How do I set qr=TRUE for gam? When I produce a fit using gam like this: fit = gam(y~s(x),data=as.data.frame(l_yx),family=family,control = list(keepData=T)) ...then try to use predict: (see #1 below in the traceback() ) > traceback() 6: stop("lm object does not have a proper 'qr' component.\n Rank zero or should