search for: solve

I haven't found a function that directly calculates the matrix inverse, does it exist? Otherwise what would be the fastest way to do it? Patrik Waldmann -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

> Date: Mon, 1 May 2000 16:25:11 +0200 (CEST) > From: gb <gb at stat.umu.se> > > On 1 May 2000, Douglas Bates wrote: > > gb <gb at stat.umu.se> writes: > > > > > How about 'Ainv <- qr.solve(A)'? > > > > > > I happened to read the help page for 'qr.solve' the other day, and there I > > > found that qr.solve(A, b) "is == but much better than solve(A) %*% b". > > > (I guess that 'better than' refers to numerical stability...

Hi Guys, I would like to share my experience when installing the Spatial views packages for R. I could not install 32 packages which are parts of the Spatial views, and I use google-search and search to solve ALL those problems for about 2 days. I hope maybe somebody would benefit from my experience. I admitted that I do not have excellent programming skills at all. So, perhaps some of steps I did are not necessary to solve the problem at hands. But it works and I am happy. I was lucky if the prob...

...s fine), and the second set of code attempts to use mapply but I get a "subscript out of bounds" error. Any guidance would be greatly appreciated. Xj, Yj, and Wj are also lists, and s2, TAU, and GAMMA are scalars.   Thank You.   # THIS WORKS USING THE LOOP for (j in 1:J) { V.tilde.j <- solve((1/s2)*t(Xj[[j]])%*%Xj[[j]] + solve(TAU)) # Not singular case: if(round(det(t(Xj[[j]])%*%Xj[[j]]),8)!=0) {  Beta.hat.j <- solve(t(Xj[[j]])%*%Xj[[j]])%*%t(Xj[[j]])%*%Yj[[j]]  V.j <- s2*solve(t(Xj[[j]])%*%Xj[[j]])  Lambda.j <- solve(solve(V.j) + solve(TAU))%*%solve(V.j)  Beta.tilde.j <- L...

The solve function in r-patched has been changed so that it applies a tolerance when using Lapack routines to calculate the inverse of a matrix or to solve a system of linear equations. A tolerance has always been used with the Linpack routines but not with the Lapack routines in versions 1.7.x and 1.8.0....

Hello! Can anyone explain me what solve() function does: Gaussian elimination or iterative, numeric solve? In addition, I would need both the Gaussian elimination and iterative solution for the course. Are the two built in R? Thanks! PM

I have a situation in R where I would like to find any x (if one exists) that solves the linear system of equations Ax = b, where A is square, sparse, and singular, and b is a vector. Here is some code that mimics my issue with a relatively simple A and b, along with three other methods of solving this system that I found online, two of which give me an error and one of which succ...

Dear all, The background is that I'm trying to fix this bug in the geometry package: https://r-forge.r-project.org/tracker/index.php?func=detail&aid=1993&group_id=1149&atid=4552 Boiled down, the problem is that there exists at least one matrix X for which det(X) != 0 and for which solve(X) fails giving the error "system is computationally singular: reciprocal condition number = ..." (see appended code & attached file). I don't want my function that calls solve(X) to return an error. I can think of two strategies for dealing with this problem: Strategy 1: Some c...

I was playing around with a simple example using solve.qp ( function is in the quadprog package ) and the code is below. ( I'm not even sure there if there is a reasonable solution because I made the problem up ). But, when I try to use solve.QP to solve it, I get the error that D in the quadratic function is not positive definite. This is becau...

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) # They are not equal Here i...

...//dropbox.unl.edu/uploads/20090522/7a0d0313f21fd6a8/Bonacich%20Power.RData > Graph size *Vertices: 20924 Edges: 146938 Directed: FALSE No graph attributes. Vertex attributes: name. No edge attributes. * computer Configuration *WIndows XP service Pack 3 .0 GB RAM* I am using SPARSE matrix to solve the problem This is the code I use to obtain bonpower using Sparse Matrix & alternatively the code is in the following website http://igraph.wikidot.com/r-recipes#toc6 *bonpow.sparse <- function(graph, nodes=V(graph), loops=FALSE, exponent=1, rescale=TRUE, tol=1e-...

To R-helpers, R offers the polyroot function for solving mentioned equations iteratively. However, Dr Math and Mathworld (and other places) show in detail how to solve mentioned equations non-iteratively. Do implementations for R that are non-iterative and that solve mentioned equations exists? Regards, Mads Jeppe

...u really want to proceed. See also ?svd and package irlba. -- Sent from my phone. Please excuse my brevity. On April 20, 2016 4:22:34 AM PDT, A A via R-help <r-help at r-project.org> wrote: > > > >I have a situation in R where I would like to find any x (if one >exists) that solves the linear system of equations Ax = b, where A is >square, sparse, and singular, and b is a vector. Here is some code that >mimics my issue with a relatively simple A and b, along with three >other methods of solving this system that I found online, two of which >give me an error and o...

> On 20 Apr 2016, at 13:22, A A via R-help <r-help at r-project.org> wrote: > > > > > I have a situation in R where I would like to find any x (if one exists) that solves the linear system of equations Ax = b, where A is square, sparse, and singular, and b is a vector. Here is some code that mimics my issue with a relatively simple A and b, along with three other methods of solving this system that I found online, two of which give me an error and one of which succ...

...matrix into a dense one before doing any linear algebra). Bill Dunlap TIBCO Software wdunlap tibco.com On Wed, Apr 20, 2016 at 4:22 AM, A A via R-help <r-help at r-project.org> wrote: > > > > I have a situation in R where I would like to find any x (if one exists) > that solves the linear system of equations Ax = b, where A is square, > sparse, and singular, and b is a vector. Here is some code that mimics my > issue with a relatively simple A and b, along with three other methods of > solving this system that I found online, two of which give me an error and &g...

ABCD are four matrix. A * Inverse((Transpose(A)*Tranpose(B)*B*A+C)) * Transpose(A) * Transpose(B) * D how to write in R in an efficient way? --------------------------------- [[alternative HTML version deleted]]

Why do x<-b%*%ginv(A) and x<-solve(A,b) give different results?. It seems that I am missing some basic feature of matrix indexing. e.g.: A<-matrix(c(0,-4,4,0),nrow=2,ncol=2) b<-c(-16,0) x<-b%*%ginv(A);x x<-solve(A,b);x Thanks in advance, Angel

Full_Name: John Peters Version: 1.5.1 OS: Windows 2000 Submission from: (NULL) (130.155.2.3) solve(a,b) with two arguments gives an error if a is complex and b is a vector: > a_matrix(c(1,2+3i,3,2),ncol=2) > a [,1] [,2] [1,] 1+0i 3+0i [2,] 2+3i 2+0i > b_c(2,2+1i) > solve(a,b) Error in solve.default(a, b) : A must be a complex matrix > is.complex(a) [1] TRUE > is.matri...

...("score.1"),direction='long') long$time<-long$time-1 Now I run this through a gls as follows: fm1 <- gls(score.1~I(time), data=long, correlation=corAR1(form=~1|ID), method='ML') Now, I am trying to replicate the gls using matrix techniques. Here is my code for to solve (X' V^{-1} X) ^{-1} X' V^{-1} y for the point estimates and (X' V^{-1} X) ^{-1} for the standard errors. score<-long$score X.mat<-model.matrix(score~time, long, row.names=F) var.mat<-getVarCov(fm1) I<-diag(sample.size) V <- kronecker(I,var.mat) pe<-solve(crossprod(X.m...

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