similar to: Matrix Problem

I have a matrix 2900X2900 and I have to solve it. But R says that it can't allocate 67899kb. How can I do? Appreciate your kindly help! Salvati Nicola Dipartimento di Statistica "G.Parenti" salvati@ds.unifi.it tel. 055.4237.224 fax 055.4223560 [[alternative HTML version deleted]]

Hi.. I have to invert a 15000 x 15000 matrix (generalized inverse). I do run the process on a fairly powerful computer. but still complains indufficient memory. Is there a way one can invert a large matrix in some other efficient manner. Thanks Harsh --------------------------------- [[alternative HTML version deleted]]

We have a problem inverting a matrix which has the following eigenvalues: > eigen(tcross, only.values=TRUE) $values [1] 7.917775e+20 2.130980e+16 7.961620e+13 8.241041e+12 2.258325e+12 [6] 3.869428e+11 6.791041e+10 2.485352e+09 9.863098e+08 9.819373e+05 [11] 3.263408e+05 2.929853e+05 2.920419e+05 2.714355e+05 8.733435e+04 [16] 8.127136e+04 6.543883e+04 5.335074e+04

Can R invert a 160000x160000 matrix with all positive numbers? Thanks a lot!

Hello, I am trying to generate random survival times by inverting the function,  S(t)= exp(b*F(t)), where b is constant and F(t) is some cumulative distribution function, let say that F(t) is cdf of normal distribution or any others distributions.   as we know that S(t) has uniform distribution on  (0,1) so we can write that U= exp(b*F(t)), where U is uniform (0,1). Now to generat the time t, we

Hi; I am having problems inverting matrices using the function solve() For example R can not invert the following matrix [,1] [,2] [,3] [,4] [,5] [1,] 25 500 11250 275000 7.106250e+06 [2,] 500 11250 275000 7106250 1.906250e+08

Hi, I wonder if anyone has written any code to implement the suggestions of Smart et al (2004) in the Bulletin of the Ecological Society of America for a new way of graphically presenting the results of logistic regression (see www.esapubs.org/bulletin/backissues/085-3/bulletinjuly2004_2column.htm#t ools1 for the full text)? I couldn't find anything relating to this sort of graphical

Full_Name: Ravi Varadhan Version: 2.8.1 OS: Windows Submission from: (NULL) (162.129.251.19) Inverting a matrix with solve(), but using LAPACK=TRUE, gives erroneous results: Here is an example: hilbert <- function(n) { i <- 1:n; 1 / outer(i - 1, i, "+") } h5 <- hilbert(5) hinv1 <- solve(qr(h5)) hinv2 <- solve(qr(h5, LAPACK=TRUE)) all.equal(hinv1, hinv2) #

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

Refiling this. The actual fix was slightly more complicated. Will soon be committed to R-Patched (aka 2.9.1 beta). -p rvaradhan at jhmi.edu wrote: > Full_Name: Ravi Varadhan > Version: 2.8.1 > OS: Windows > Submission from: (NULL) (162.129.251.19) >=20 >=20 > Inverting a matrix with solve(), but using LAPACK=3DTRUE, gives erroneo= us > results: Thanks, but there seems

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

Dear All, My question is simple but I need someone to help me out. Suppose I have a positive definite matrix A. The funtion chol() gives matrix L, such that A = L'L. The inverse of A, say A.inv, is also positive definite and can be factorized as A.inv = M'M. Then A = inverse of (A.inv) = inverse of (M'M) = (inverse of M) %*% (inverse of M)' = ((inverse of

I'm having the most curious error while using the sem package. For the model I'm working with, I keep getting the following error: Error in J[cbind(1:n, observed)] <- 1 : subscript out of bounds I''ve used debug=TRUE with sem, and there don't appear to be any problems with model - there are no latent variables in this model. The variables in the covariance matrix

Hello, We have booked a couple of slots during EuroLLVM this year that we would like to dedicate to real hacking!!! Therefore, we would like to offer to the attendees this year an opportunity to escape from the presentation sessions and dive into fun coding to learn something new or to solve some interesting problems. The current proposal is to have 2 x 45 mins on Monday afternoon and 2 x 45

Howdy, I am trying to invert a matrix for the purposes of least squares. I have tried a number of things, and the variety of results has me confused. 1. When I try solve() I get the following: >Error in solve.default(t(X) %*% X) : system is computationally singular: reciprocal condition number = 3.76391e-20 2. When I try qr.solve(), I get: >Error in qr.solve(t(X) %*% X) : singular matrix

Had a try. The reason why a simple delay is not that good is mainly due to the initialization of the filter parameter that still takes a few seconds (if they are perfectly in sync, you sort of get lucky). Otherwise, you real recording seems to have something odd in it. Are you sampling from a different card then the one that's playing the sound? or maybe the mic (or something else) in the room

How to invert match parameter? Like this: tc filter add dev eth0 protocol ip parent 1: prio 1 u32 \ match ip src 10.20.30.40 \ match ip dst !10.30.0/24 \ match ip dst !10.40.0/24 \ flowid 1:20 _______________________________________________ LARTC mailing list / LARTC@mailman.ds9a.nl http://mailman.ds9a.nl/mailman/listinfo/lartc HOWTO: http://lartc.org/

Dear all, I encounter some covariance matrix with quite small eigenvalues (around 1e-18), which are smaller than the machine precision. The dimension of my matrix is 17. Here I just fake some small matrix for illustration. a<-diag(c(rep(3,4),1e-18)) # a matrix with small eigenvalues b<-matrix(1:25,ncol=5) # define b to get an orthogonal matrix b<-b+t(b) bb<-eigen(b,symmetric=T)

Hi, I am writing some basic smoothers in R for cleaning some spectral data. I wanted to see if I could get close to matlab for speed, so I was trying to compare SparseM with Matrix to see which could do the choleski decomposition the fastest. Here is the function using SparseM difsm <- function(y, lambda, d){ # Smoothing with a finite difference penalty # y: signal to be smoothed #

Harmen, your suggestion of inverting the mapping almost worked (and Andrew was correct that the function I need is the same as the one in which poolinit appears). Unfortunately, it appears that this mapping only considers the original function and not any of its clones. Since the pool descriptor in question may very well only exist in a clone, I can't use this. Is there another way?