similar to: r: eviews and r // eigen analysis

Hi R-Wizards: I've looking through the R docs and have yet to find what I'm looking for and have tried a few intermediate steps to now avail yet and rather than spend another few hours looking for the solution, I figured I would post a message. I have a matrix (actually a set of them) that I want to pull all the names of the non-zero columns into a vector/list for further processing:

As usual, the R newsgroup set me straight (thanks to Douglas Bates, Robert Balshaw and Albyn Jones). There is really no difference between using orthogonal polynomials of the form: Linear -3 -1 1 3 Quadratic 1 -1 -1 1 Cubic -1 3 -3 1 Versus > poly(c(1:4),3) 1 2 3 [1,] -0.6708204 0.5 -0.2236068 [2,] -0.2236068 -0.5 0.6708204 [3,] 0.2236068

Looking to the wonderful statistical advice that this group can offer. In behavioral science applications of stats, we are often introduced to coefficients for orthogonal polynomials that are nice integers. For instance, Kirk's experimental design book presents the following coefficients for p=4: Linear -3 -1 1 3 Quadratic 1 -1 -1 1 Cubic -1 3 -3 1 In R orthogonal

Full_Name: David Pleydell Version: 2.2.1 OS: Debian Etch Submission from: (NULL) (193.55.70.206) There appears to be a conflict between the chol functions from the Matrix and the SparseM packages. chol() can only be applied to a matrix of class dspMatrix if SparseM is not in the path. with gratitude David > library(Matrix) > sm <- as(as(Matrix(diag(5) + 1), "dsyMatrix"),

Hi I have the problem with fitting curve to data with lm and glm. When I use polynominal dependiency, fitted values from model are OK, but I cannot recive proper values when I use coefficents to caltulate this. Let me present simple example: I have simple data.frame: (dd) a: 1 2 3 4 5 6 b: 3 5 6 7 9 10 I try to fit it to model: model=glm(b~poly(a,3),data=dd) I have following data

I am trying to use lm for a simple linear fit with weights. The results I get from IDL (which I am more familiar with) seem correct and intuitive, but the "lm" function in R gives outputs that seem strange to me. Unweighted case: > x<-1:4 > y<-(1:4)^2 > summary(lm(y~x)) Call: lm(formula = y ~ x) Residuals: 1 2 3 4 1 -1 -1 1 Coefficients:

Dear All i have a little puzzle about eigenvector in the R. As we know that the eigenvector can be displayed on several form. For example A=matrix(c(1,2,4,3),2,2) if we want to get the eigenvalue and eigenvector, the code followed eigen(A) $values [1] 5 -1 $vectors [,1] [,2] [1,] -0.7071068 -0.8944272 [2,] -0.7071068 0.4472136 however, we also can calculate the vector matrix

Ladies and Gentlemen, using > A <- structure(c(1, 0, 0, 3, 2, 1, 4, 5, -3, -2, 1, 0), .Dim = as.integer(c(3,4))) I get > dim(A) [1] 3 4 > qr.R(qr(A),complete=TRUE) [,1] [,2] [,3] [,4] [1,] -1 -3.000000 -4.000000 2.0000000 [2,] 0 -2.236068 -3.130495 -0.8944272 [3,] 0 0.000000 -4.919350 -0.4472136 > qr.R(qr(A),complete=FALSE) [,1]

Hi, All: What tools exist for diagnosing singular gradient problems with 'nls'? Consider the following toy example: DF1 <- data.frame(y=1:9, one=rep(1,9)) nlsToyProblem <- nls(y~(a+2*b)*one, DF1, start=list(a=1, b=1), control=nls.control(warnOnly=TRUE)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial

Hi, with small matrices eigen works as expected: > eigen(cbind(c(1,4),c(4,7)), only.values = TRUE) $values [1] 9 -1 $vectors NULL > eigen(cbind(c(1,4),c(4,7))) $values [1] 9 -1 $vectors [,1] [,2] [1,] 0.4472136 -0.8944272 [2,] 0.8944272 0.4472136 > eigen(cbind(c(1,-1),c(1,-1))) $values [1] -3.25177e-17+1.570092e-16i -3.25177e-17-1.570092e-16i $vectors

I am trying to find the distance between a vector and each row of a dataframe. I am using the function "distancevector" in the package "hopach" as follows: mydata<-as.data.frame(matrix(c(1,1,1,1,0,1,1,1,1,0),nrow=2)) V1 V2 V3 V4 V5 1 1 1 0 1 1 2 1 1 1 1 0 vec <- c(1,1,1,1,1) d2<-distancevector(mydata,vec,d="euclid") The Euclidean distance

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

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

Hello, I'm using VAR models in R in order to obtain impulse responses of stock market shock on US economy. I have series of quarterly changes in real gdp, S&P 500 and quarterly level of unemployment for 1985 - 2012 period. My series are stationary. So I did all the steps below. However I don't understand what do irf function results mean. These are the cumulative orthogonal responses

Dear HelpeRs, I wonder if anyone knows of ways to read EViews file types. I did not find a function in the package 'foreign' and a search query submitted to http://search.r-project.org was not successful. Any hint is very much welcome. Dietrich Trenkler -- Dietrich Trenkler c/o Universitaet Osnabrueck Rolandstr. 8; D-49069 Osnabrueck, Germany email: Dietrich.Trenkler at

Dear all, I met with some problems when dealing with a time series with serial correlation. FIRST, I generate a series with correlated errors set.seed(1) x=1:50 y=x+arima.sim(n = 50, list(ar = c(0.47))) SECOND, I estimate three constants (a, b and rho) in the model Y=a+b*X+u, where u=rho*u(-1)+eps library(nlme) gls(y~x,correlation = corAR1(0.5)) # Is it the right procedure?

Dear Sir, First of all Happy Holidays!,... I am writing to you because I am a bit confused about ARCH estimation. Is there a way to find what garch() exactly does, without the need of reading the source code (because I cannot understand it)? In Eviews (the results at the end) I am getting different results than in R (for those that have the program I do: Quick -> Estimage Equation ->

Hi, It is my understanding that the eigenvectors of a circulant matrix are given as follows: 1,omega,omega^2,....,omega^{p-1} where the matrix has dimension given by p x p and omega is one of p complex roots of unity. (See Bellman for an excellent discussion on this). The matrix created by the attached row and obtained using the following commands indicates no imaginary parts for the

The following patch adds a trivial family method for linear models. -k --- src/library/base/funs/lm.orig Tue May 27 02:53:58 1997 +++ src/library/base/funs/lm Sun Aug 10 22:00:46 1997 @@ -412,3 +412,5 @@ } formula.lm<-function(x)formula(x$terms) + +family.lm <- function(x) { gaussian() } =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- r-devel mailing list

R-1.7.0 built on NetBSD 1.6, but the validation test suite failed: Machinetype: Intel Pentium III (600 MHz); NetBSD 1.6 (GENERIC) Remote gcc version: gcc (GCC) 3.2.2 Remote g++ version: g++ (GCC) 3.2.2 Configure environment: CC=gcc CXX=g++ LDFLAGS=-Wl,-rpath,/usr/local/lib make[5]: Entering directory `/local/build/R-1.7.0/src/library' >>> Building/Updating