search for: inverting

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

In our architecture (TCE) we have predicate registers, not flags. But our predicates work in a way that in the predicated instruction do not contain only the predicate register, but a predicate register and boolean value pair, the boolean value stating if the instruction is to be executed only if the register is true(normal behavior of predicate regs) or only if the predicate reg is

Dear List, I am using the wireframe function in the lattice package, and I am wondering if it is possible to invert the default axes orientation for x and y axes... what parameter should I look for? Best regards, Simone Gabbriellini

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

Hi uRsers, when inverting a 2 by 2 matrix using solve, I encountered a error message: solve.default(sigma, tol = 1e-07) : system is computationally singular: reciprocal condition number = 1.7671e-017 and then I test the determinant of this matrix: 6.341393e-06. In my program, I have a condition block that whether...

Hello, just a simple question How can I do to invert the sign of a number? (for example: -4 to 4, 2 to -2 and so on..) I was looking for a specific function in R but I didn't found it... thank you Francesco -- View this message in context: http://www.nabble.com/Invert-the-sign-of-a-number-tp26110267p26110267.html Sent from the R help mailing list archive at Nabble.com.

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...

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)

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,...

Hello R-Users, I have to invert a matrix 3000X3000 and the solve method doesn't work or it is too slow. Are there any methods to invert a big matrix? Regards Stefan --------------------------------- [[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]]

Is there an easy way to invert a table? (not to solve for the inverted matrix, just swap rows for columns & vice versa). I've gone through my data manipulation bible (Phil Spector's book), but to no avail. [[alternative HTML version deleted]]

Hi all. As the subject has come up. Powerware ups may contain several cpu's and different firmware versions. Five of them is always ordered in a specific way. 1 control panel 2 inverter 3 rectifier 4 net interface card 5 display control So if a ups have a inverter cpu but not a control panel it sets the firmware revision to 00.00 to the control panel cpu, and get the version for the

Hello everyone I was wondering how would one simulate from a Normal Wishart Distribution in R. A normal inverted Wishart distribution is denoted by NIW (M,C,d,S), where X/(Sigma) ~ N( M,C,(Sigma) ) -> a matrix normal distribution, (Sigma) -> column dispersion matrix (Sigma) ~ IW (d,S) -> inverted Wishart distribution Thanks a lot ! Best Shantanu [[alternative HTML version

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 a...

I've looked around but I can't find the method in R for testing whether the resulting estimated coefficients of an AR model imply that the model is invertible. To quote from eric zivot's blue book : " the AR(p) is invertible provided the rots of the characteristic equation Phi(z) = 1 - phi_1*z - phi_2*z^2 = phi_3*z^3 - ..... Phi_p*z^p = 0 lie outside the complex circle".

I'm having a problem with both the 5.8 and 5.9 kernels using the nouveau DRM driver. I have a laptop with a VGA card (specs below) connected to a 5120x1440 screen. At boot time, the card correctly detects the screen, tries to allocate fbdev fb0, then the video hangs completely for 15-30 seconds until it goes blank. This used to work in Linux 5.7 and earlier, although it allocated a 3840x1080

When I execute xrandr I receive, omitting the details of resolution options, Screen 0: minimum 320 x 200, current 1920 x 1080, maximum 8192 x 8192 LVDS-1 unknown connection (normal left inverted right x axis y axis) DP-1 disconnected (normal left inverted right x axis y axis) HDMI-1 disconnected (normal left inverted right x axis y axis) HDMI-2 connected 1920x1080+0+0 (normal left inverted right

Dear Rers, I have threes series, x, y, z and I want to fit a model z ~ x + y. First of all, I fit a lm. I found the residuals are correlated, by looking at the acf() and pacf(). Then I tried to fit a gls model allowing residuals to be correlated (correlation = corARMA(p=5, q=1)): y.na <- as.data.frame(y[complete.cases(y),]) y.gls <- gls(z ~ x + y, data = y.na, correlation=corARMA(p=5,

Hi! This is probably a very trivial question. Is there an easy way to invert the y-axis (low values on top) when using the function barplot()? Thanks, Jörg [[alternative HTML version deleted]]