similar to: "is.qr" definition (PR#465)

Hi everyone, I do a lot of work with large variance matrices, and I like to use "crossprod" for speed and to keep everything symmetric, i.e. I often compute "crossprod(Q %*% t(A))" for "A %*% Sigma %*% t(A)", where "Sigma" decomposes as "t(Q) %*% Q". I notice in the code that "crossprod", current definition > crossprod function (x,

Hi everyone, Can I suggest an addition to the help page for logical AND and OR, indicating the behaviour of NA (which is currently not mentioned at all). I think something simple like an extra paragraph in Details, eg \code{NA} is a valid logical object. Where a component of \code{x} or \code{y} is \code{NA}, the result will be \code{NA} if the outcome is ambiguous. In other words \code{NA

Hi everyone, Can I suggest the following cosmetic change to cut.default (R-1.3.1), to adjust the labels in the case that include.lowest = TRUE: arrow:dma0jcr% diff -u cut.R my.cut.R --- cut.R Tue Nov 13 12:28:35 2001 +++ my.cut.R Tue Nov 13 12:35:16 2001 @@ -25,6 +25,11 @@ ch.br[-nb], ",", ch.br[-1], if(right)"]"

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

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 ```

I find this a bit puzzling ... > solve(matrix(c(5, 2, 3, 1), 2, 2), c(Inf, 3)) [1] NaN Inf > solve(matrix(c(5, 2, 3, 1), 2, 2)) %*% c(Inf, 3) [,1] [1,] -Inf [2,] Inf I would expect the answer to be c(-Inf, Inf), so why has the -Inf been replaced by NaN in solve? Cheers, Jonathan. --please do not edit the information below-- Version: platform = sparc-sun-solaris2.7 arch = sparc

> 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, 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 <- >

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

Hi everyone, Can I suggest an additional output component in cancor, from package mva? It would be useful to have the number of canonical correlation vectors, equivalently the rank of the covariance between x and y (label "rank"). This would usually be min(dx, dy), where dx and dy have already been computed for the svd function, but there might be situations where it was less than

I was surprised to find that the variance of a scalar, using var(), is NA. Surely this should be zero? Cheers, Jonathan. --please do not edit the information below-- Version: platform = sparc-sun-solaris2.7 arch = sparc os = solaris2.7 system = sparc, solaris2.7 status = Patched major = 0 minor = 99.0 year = 2000 month = February day = 9 language = R Search Path: .GlobalEnv,

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

I would like to suggest a slight modification to diag(). In the case where x is a matrix with both row names and column names the same, it would be reasonable if the resulting vector also had those names. I often use diag() on variance matrices, where this modification is helpful. The modification requires replacing if (is.matrix(x) && nargs() == 1) return(c(x)[1 +

Just a minor point, but the adj argument in ?text is described wrongly (0.64.1). Perhaps the following might be added to the details. Values of 0, 0.5, and 1 specify left/top, middle and right/bottom, respectively. The default is for centred text, ie \code{adj = c(0.5, 0.5)}. Cheers, Jonathan. Jonathan Rougier Science Laboratories Department of

Hi Everyone, To complete the outer() and kronecker() functions in the base, may I suggest the following tensor() function, which allows the multiplication of arrays through sets of conformable dimensions. I am happy to write a help page if required. The code also needs a setdiff() function which prompts me to ask: what about simple set functions? I expect many of us have written our own

I suggest adding a 'pivot' argument to qr.R, to obtain columns in the same order as the original x, so that a <- qr(x) qr.Q(a) %*% qr.R(a, pivot=TRUE) returns x. -------------------------------------------------- # File src/library/base/R/qr.R qr.R <- function(qr, complete = FALSE, pivot = FALSE) { # Args: # qr: a QR decomposition, produced by qr() # complete:

The following commands z <- matrix(seq(0, 1, len=1000)) image(z, 1, z, zlim=c(0, 1), col=gray(1:10/10)) appear to violate the image functionality as described on the help page. As I understand it, by specifying a zlim of c(0, 1) and a colour vector of length 10, the colours should correspond to the intervals (0, 0.1), (0.1, 0.2), ..., (0.9, 1.0), this being 10 equally-spaced intervals on (0,

In legend() is it possibly to mix lines and boxes? I would like to do something like legend(locator(1), legend = c("al", "ben", "chris", "dave"), lty = c(1, 2, 3, NA), fill = c(NA, NA, NA, "gray")) to indicate that "al", "ben" and "chris" have lines, and "dave" has a gray shading. Many thanks,

Hi everyone, As "intensities" is deprecated as a synonym for "density" in the output of hist it would be a good idea to replace its occurence with "density" in plot.histogram, where we currently have y <- if (freq) x$counts else x$intensities Note that, to add to the confusion, x$density is also referred to in plot.histogram. Cheers, Jonathan.

I'm just back from a computer practical where we ran into > "f" <- function(x) 2(x^2) # for 2 * x^2 This definition was taken as legitimate, although using the function generated the error message > f(3) Error in f(3) : attempt to apply non-function Am I right in thinking that in prevous versions of R this definition would have been declared as a syntax error? Surely a