similar to: Algo. for matrix inverse

Dear R-devel: I could use some advice about matrix calculations and steps that might make for faster computation of generalized inverses. It appears in some projects there is a bottleneck at the use of svd in calculation of generalized inverses. Here's some Rprof output I need to understand. > summaryRprof("Amelia.out") $by.self self.time self.pct

I have 2 vecros : x<-c(100,96,88,100,100,96,80,68,92,96,88,92,68,84,84,88,72,88,72,88) x1 = sample(x, 5, replace=FALSE) Now i want to get remaining values of vector "x" those are not member of vector "x1". Can anyone please tell me how to do that?

Hi, In function chol2inv with the option LINPACK set to false (default), it raises an error when the matrix is 1x1 matrix (i.e. just a real) saying 'a' must be a numeric matrix This error is raised by the underlying C function (modLa_chol2inv in function Lapack.c). Everything is normal, but I wonder if we could have another behavior when we pass a 1x1 matrix. I spent time this morning

Thanks David, It is working. Holtman's also gave me a solution but, I wanted to have a color pallet for description of colors, that was not in his solution. However I need one small modification. If I want to plot only lower diagonal elements of 'dat' then how should I proceed? What I want is, to visualize only lower diagonal elements and having the color pallet on them only. Also

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

Let me take an artifical matrix : dat = matrix(rnorm(200*200), 200, 200) My goal is to visualize this matrix according to the procedure, described in previous mails. I took Mendelssohn's advice and got following advice : ?plot.im Z <- setcov(owin()) plot(Z) .................... etc However I can not reproduce this example in my problem. How I can change my data

Dear all R users, I have a small doubt about panel data analysis. My basic understanding on Panel data is a type of data that is collected over time and subjects. Vector Autoregressive Model (VAR) model used on this type of data. Therefore can I say that, one of statistical tools used for analysis of panel data is VAR model? If you clarify my doubt I will be very grateful. Thanks and regards,

Hi, I have created a list object like that : x = vector("list") for (i in 1:5) x[[i]] = rnorm(2) x Now I want to do two things : 1. for each i, I want to do following matrix calculation : t(x[[i]]) %*% x[[i]] i.e. for each i, I want to get a 2x2 matrix 2. Next I want to get x[[1]] + x[[2]] +.... I did following : res=vector("list"); res = sapply(x, function(i) t(x[[i]]) %*%

hi, hi, good morning everyone. I have a time series with binary outputs like : 0001011110100.................etc. Now I want to forecast the future values of that. Can anyone please tell me whether there is any tools exist in literature for dealing with this kind of binary observation? If possible please provide me some good references in net as well. rgd, Megh

[My previous message rejected, therefore I am sending same one with some modification] I have 3 vectors with object name : dat1, dat2, dat3 Now I want to create a loop, like : for (i in 1:3) { cat(sd(dati)) } How I can do this in R? Regards,

Dear all, I have following 2 zoo objects. However when I try to merge those 2 objects into one, nothing is coming as intended. Please see below the objects as well as the merged object: > dat11 V2 V3 V4 V5 2010-10-15 13:43:54 73.8 73.8 73.8 73.8 2010-10-15 13:44:15 73.8 73.8 73.8 73.8 2010-10-15 13:45:51 73.8 73.8 73.8 73.8 2010-10-15 13:46:21 73.8 73.8 73.8 73.8

Hi, I am trying to create a vector of length 10 (say), wherein each element will be average of random sample of size 100, from a distribution, say Normal. Can anyone please tell me without creating a "for" loop, how I can do that? Regards, -- View this message in context: http://www.nabble.com/Alternate-to-for-loop-tp22035954p22035954.html Sent from the R help mailing list archive at

Dear BioConductor and R fellow users I apologize in advance for double posting, but I am not sure which list would actually be best fit for this message. I am experiencing a weird error with my R installation on Ubuntu 10.04.4 (LTS) 64bit: When I run R on the terminal everything goes smoothly: $R R version 2.15.2 (2012-10-26) -- "Trick or Treat" Copyright (C) 2012 The R Foundation

Hi, I have following kind of dataset (all are dates) in my Excel sheet. 09/08/08 09/05/08 09/04/08 09/02/08 09/01/08 29/08/2008 28/08/2008 27/08/2008 26/08/2008 25/08/2008 22/08/2008 21/08/2008 20/08/2008 18/08/2008 14/08/2008 13/08/2008 08/12/08 08/11/08 08/08/08 08/07/08 However I want to use R to compile those data to make all dates in same format. Can anyone please tell me any automated way

I have following " 1975 01 7711.16" Here I need to identify where the <space> is there and then concatenate rest of the digits without <space>, i.e. I want to have "1975017711.16". Is there any R function? Regards,

I used ?image function to do that, like below : require(grDevices) # for colours x <- y <- seq(-4*pi, 4*pi, len=27) r <- sqrt(outer(x^2, y^2, "+")) image(x, y, r, col=gray((0:32)/32)) However my next problem to add a color pallet for color description [as shown in following link]. If anyone here tell me how to do that, it will be good for me. Regards, Megh Dal

I have following dataset: > res [,1] [,2] [,3] [1,] 1946 4 1.27 [2,] 1946 5 1.27 [3,] 1946 6 1.27 [4,] 1946 7 1.27 [5,] 1946 8 1.52 [6,] 1946 9 1.52 [7,] 1946 10 1.52 [8,] 1946 11 1.52 [9,] 1946 12 1.62 [10,] 1947 1 1.62 [11,] 1947 2 1.62 [12,] 1947 3 1.62 [13,] 1947 4 1.87 [14,] 1947 5 1.87 [15,] 1947 6 1.87 Now I write following code

Is there any R function to calculate automatically the last day of a particular month? For example "sep2009" should be converted to last day of September of 2009? Thanks -- View this message in context: http://www.nabble.com/How-to-get-last-day-of-a-month--tp25425645p25425645.html Sent from the R help mailing list archive at Nabble.com.

Dear all, Please forgive me if there is a duplicate post; my previous mail perhaps didnt reach the list....... Let say I have following time series library(zoo) > dat1 <- zooreg(rnorm(10), start=as.Date("2010-01-01"), frequency=1) > dat1[c(3, 7,8)] = NA > dat1 2010-01-01 2010-01-02 2010-01-03 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 2010-01-09

Dear all R users, Please forgive me if my question is too trivial. Suppose I have two variables, (x,y) which is log-normally distributed with expected value (mu1, mu2) and some variance-covariance matrix. Now I want to draw a random sample of size 1000 from this distribution. Is there any function available to do this? Thanks and regards, Megh