search for: solved

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

...r 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 problem I have was already solved by someone else, or if the problem is just a matter of installing another R packages that depends on other packages. Sometimes it has something to do not with R package but with a package that has to be installed in my operating system (in my case LinuxMint 8, based on ubuntu karmic). so I just...

Hello, I would like to use mapply to avoid using a loop but for some reason, I can't seem to get it to work. I've included copies of my code below. The first set of code uses a loop (and it works 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,

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. (You can use the optional argument tol = 0 to override this check for computational

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 succeeds on the

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

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 because Dmat is zero because I don't have a

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 All, > Please help me with this problem.I have been having this problem for over a month now and I could not find any information.I later realised that error is with MATRIX package. I am working on graph object using IGRAPH version 0.5.2-2 package & wanted to do Bonacich Power. What I have is undirected GRAPH. This is my graph object. The file 'Bonacich

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

This is kind of like asking for a solution to x+1=x+1. Go back to linear algebra and look up Singular Value Decomposition, and decide if you 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

> 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

This is not a solution but your lsfit attempt #Error in lsfit(A, b) : only 3 cases, but 4 variables lsfit(A,b) gave that error because lsfit adds a column of 1 to its first argument unless you use intercept=FALSE. Then it will give you an answer (but I think it converts your sparse matrix into a dense one before doing any linear algebra). Bill Dunlap TIBCO Software wdunlap tibco.com On

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 >

Dear List: I am having some trouble reproducing some GLS estimates using matrix operations that I am not having with other R procedures. Here are some sample data to see what I am doing along with all code: mu<-c(100,150,200,250) Sigma<-matrix(c(400,80,16,3.2,80,400,80,16,16,80,400,80,3.2,16,80,400),n c=4) sample.size<-100 temp <-

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