similar to: proportional matrix rows

Hi I have a square matrix Ainv of size N-by-N where N ~ 1000 I have a rectangular matrix H of size N by n where n ~ 4. I have a vector d of length N. I need X = solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*% d and H %*% X. It is possible to rewrite X in the recommended crossprod way: X <- solve(quad.form(Ainv, H), crossprod(crossprod(Ainv, H), d)) where quad.form() is a little

Hi the manpage says that crossprod(x,y) is formally equivalent to, but faster than, the call 't(x) %*% y'. I have a vector 'a' and a matrix 'A', and need to evaluate 't(a) %*% A %*% a' many many times, and performance is becoming crucial. With f1 <- function(a,X){ ignore <- t(a) %*% X %*% a } f2 <- function(a,X){ ignore <-

Hi [this after some considerable thought as to R-help vs R-devel] I want to write a (S3) method for as.function(); toy example follows. Given a matrix "a", I need to evaluate trace(ax) as a function of (matrix) "x". Here's a trace function: tr <- function (a) { i <- seq_len(nrow(a)) return(sum(a[cbind(i, i)])) } How do I accomplish the following:

I found out the other day that crossprod() will take a single matrix argument; crossprod(x) notionally returns crossprod(x,x). The two forms do not return identical matrices: x <- matrix(rnorm(3000000),ncol=3) M1 <- crossprod(x) M2 <- crossprod(x,x) R> max(abs(M1-M2)) [1] 1.932494e-08 But what really surprised me is that crossprod(x) is slower than crossprod(x,x): R>

On page 22 of the R-introduction guide it's written: the quadratic form x^{'} A^{-1} x which is used in multivariate computations, should be computed by something like x%*%solve(A,x), rather than computing the inverse of A. Why isn't it good to compute t(x) %*% solve(A) %*% x? Thanks a lot for help!

Hi. macosx 10.6.8 With R-2.13.1 and also revision 56948 I get the following repeatable segfault: wt118:~% R --vanilla --quiet > R.Version() $platform [1] "x86_64-apple-darwin9.8.0" $arch [1] "x86_64" $os [1] "darwin9.8.0" $system [1] "x86_64, darwin9.8.0" $status [1] "" $major [1] "2" $minor [1] "13.1" $year [1]

R-3.0.1 rev 62743, binary downloaded from CRAN just now; macosx 10.8.3 Hello, eigen(symmetric=TRUE) behaves strangely when given complex matrices. The following two lines define 'A', a 100x100 (real) symmetric matrix which theoretical considerations [Bochner's theorem] show to be positive definite: jj <- matrix(0,100,100) A <- exp(-0.1*(row(jj)-col(jj))^2) A's being

Hi how do I extract those components of a list that satisfy a certain requirement? If jj <- list(list(a=1,b=4:7),list(a=5,b=3:6),list(a=10,b=4:5)) I want just the components of jj that have b[1] ==4 which in this case would be the first and third of jj, viz list (jj[[1]],jj[[3]]). How to do this efficiently? My only idea was to loop through jj, and set unwanted components to NULL,

Dear All, Is there any function in R for computing "multivariate chi-square distribution"? How about "multivariate gamma distribution"? I appreciate any comment on this subject. Thank you, Amin Zollanvari PhD student Department of Electrical and Computer Engineering, Texas A&M University, College Station, TX

I got the following error: a = read.csv("mat.csv") b = as.matrix(a) tb = t(b) bb = tb %*% b dim(bb) ibb = solve(bb) bb %*% ibb > ibb = solve(bb) Error in solve.default(bb) : system is computationally singular: reciprocal condition number = 1.77573e-19 > Are there any ways to find more information about why it is singular? Thanks.

Hi again everyone I have an arbitrarily dimensional array "a" and a list "jj" of length length(dim(a)). The elements of jj are vectors of indexes. How do I use do.call() to extract a[ jj[[1]], jj[[2]], jj[[3]], ...] ? Toy example follows: a <- matrix(1:30,5,6) jj <- list(5:1,6:1) I want the following a[ jj[[1]],jj[[2]] ] How do I do this? OBAttempts:

Hello everybody When one is working with complex matrices, "transpose" very nearly always means *Hermitian* transpose, that is, A[i,j] <- Conj(A[j,i]). One often writes A^* for the Hermitian transpose. I have only once seen a "real-life" case where transposition does not occur simultaneously with complex conjugation. And I'm not 100% sure that that wasn't a

Hi, I am sure this has come up before, but my searches of the archive didn't give any results (maybe I didn't use the right keywords, but if I use too many, the search times out). I have a vector of dimensions n, length is not fixed, eg n <- c(4,5,7) or n <- c(19,4,5,7) and a function f that takes a vector of indices, same length of n, and gives a scalar. I would like to fill

how can I import/export data from ms excel? also is there a way to import/export data of the hdf format? is there a way to import/export hdf format for ms excel as well? thank you very much in advance. /js ###################################################################### This e-mail message has been scanned and cleared by MailMarshal http://www.marshalsoftware.com

Hi How do I test for an object having a particular virtual class? In the following, "onion" is a virtual class, and "octonion" is a non-virtual class contained in onion. The last call to inherits() gives FALSE [R-2.5.0], when inherits.Rd led me to expect TRUE. setClass("onion", representation = "VIRTUAL" )

Hello everyone I am trying to get one of my packages through R's QC. The package is clean for me under R-2.4.1, R-2.5.0, and R-devel, but Kurt gets > > * checking R code for possible problems ... WARNING > hypercube: no visible binding for global variable ?f? Function hypercube() [cut-&-pasted below] is intended to return an adjacency matrix for an n-dimensional

Hi I'm having difficulty with a multiple plot. What I want is 12 plots, all of which are the same size and shape, differing only in colour. Each one is a square, so there is an asp=1 in the plot call. I'm working in an Sweave environment so I am free to choose the height and width of the plot. I want the columns to be labelled at the head (here t=1,2,3) and the rows to be labelled at

Hi has anyone coded up domain colouring for visualizing complex analytic functions (such as elliptic functions)? [ the idea is to depict a complex function f(z) using a filled.contour() variant in which the hue is given by Arg(f(z)), and the saturation by Mod(f(z)). ] -- Robin Hankin Uncertainty Analyst Southampton Oceanography Centre European Way, Southampton SO14 3ZH, UK tel

Hello everybody I'm using contour() to draw streamlines of potential flow, eg jj <- seq(from= -4, to=4,len=20) jj <- outer(jj,jj,function(x,y){x})+1i*outer(jj,jj,function(x,y){y}) f <- function(x){x^2} contour(Im(f(jj)), nlevels=44 , labels="") How best to put arrows on the contour lines to show the direction of flow? (ie I want contour lines looking like

Hi I am having difficulty with apply(). I want apply() to return a matrix, but sometimes a vector is returned. Toy example follows. Function jj() takes a couple of matrices m1 and m2 as arguments and returns a matrix with r rows and c columns where r=nrow(m2) and c=nrow(m1). jj <- function(m1,m2,f,...){ apply(m1, 1, function(y) { apply(m2, 1, function(x) { f(x, y, ...)