similar to: help computing a covariance

Hello R-help, I need to compute matrices of first derivatives of a covariance matrix C with entries given by c_ij=theta*exp(-0.5* sum(eta*(x[i,]-x[j,])^2)), wrt to elements of eta, a m-dimensional vector of parameters, given a n*m data matrix x. So far, I have been computing matrices for each parameter (given by par[index]) analytically, using the following kmatder<- function(x, par, index) {

Hi! I would like to conduct a Cochran`s Q Test in R, but have not found any suitable function. My first idea was: J <- as.table(matrix(c(6,16,2,4),ncol=2, dimnames = list("C" = c("Favorable","Unfavorable"),"Drug A Favorable"=c("B Favorable","B Unfavorable")))) L <- as.table(matrix(c(2,4,6,6),ncol=2, dimnames =

Does anyone know which package include the computation of Cochran’s Q statistic in R? jlfmssm [[alternative HTML version deleted]]

I want to understand ANOVA better. But a few textbook that I have do not describe Cochran's Theorem in details. Could somebody recommend a book for me?

Hello, I've a question concerning the display of interval data. A sample dataset where X is an interval between Xa and Xb which should be displayed: Y=c(15,14,23,18,19,9,19,13) Xa=c(17,22,21,18,19,25,8,19) Xb=c(22,22,29,34,19,26,17,22) X = (Xa+Xb)/2 It's easily possible to plot the mean of the interval like: plot(X,Y) afterwards I can create lines for the interval with:

Dear all This may be obvious, but I cannot get it working. I'm trying to subset & sort a data frame in one go. x <- iris x$Species1 <- as.character(x$Species) ##subsetting alone works fine with(x, x[Sepal.Length==6.7,]) ##sorting alone works fine with(x, x[order(Sepal.Length, rev(sort(Species1))),]) ##gets subsetted, but not sorted as expected with(x, x[(Sepal.Length==6.7) &

Sir, I have a problem here while applying chisquare test to the following Data ( below the subject of this mail) ...when I wanted to test the significance using three different free statistical packages, here R, EpiInfo and OpenEpi. *Only OpenEpi accepts the test based on Cochran's Recommendations. * R says " chi squared approximation may be incorrect." Does it mean the same as

Hi everyone, ? I am trying to get a point of intersection between a polyline and a straight line ?.. and get the x and y coordinates of this point. For exemplification consider this: ? ? set.seed(123) ? k1 <-rnorm(100, mean=1.77, sd=3.33) ?k1 <- sort(k1) q1 <- rnorm(100, mean=2.37, sd=0.74) q1 <- sort(q1, decreasing = TRUE) plot(k1, q1, xlim <- c((min(k1)-5),

Dear all, I have a function to optimize for a set of parameters and want to set a constraint on only one parameter. Here is my function. What I want to do is estimate the parameters of a bivariate normal distribution where the correlation has to be between -1 and 1. Would you please advise how to revise it? ex=function(s,prob,theta1,theta,xa,xb,xc,xd,t,delta) { expo1=

Dear All, could someone please shed some light on the use of the .C or .Fortran function: I am trying load and running on R the following function // psi.cpp -- psi function for real arguments. // Algorithms and coefficient values from "Computation of Special // Functions", Zhang and Jin, John Wiley and Sons, 1996. // // (C) 2003, C. Bond. All rights reserved. // //

Dear all, I am trying to look for a built in function that performs the cochran Q test. that is, cochranq.test(X) where X is a contingency table (maybe a matrix or data.frame). The output will naturally be the test statisitcs, p-value, etc. A quick search on Google gives me the cochran.test in the 'outlier' package, but I had a look at the description of the test and it doesn't look

Hi all I have a 3d array containing missing values. > Xa , , 1 [,1] [,2] [1,] 1 3 [2,] NA 4 , , 2 [,1] [,2] [1,] 5 7 [2,] NA NA , , 3 [,1] [,2] [1,] 9 11 [2,] 10 12 I want to replace the missing values with the mean, but the mean of each 'page' in the array (wrong terminology I'm sure). So - for the array above - [2,1,2] and

Hello Gurus I have two correlation matrices 'xa' and 'xb' set.seed(100) d=cbind(x=rnorm(20)+1, x1=rnorm(20)+1, x2=rnorm(20)+1) d1=cbind(x=rnorm(20)+2, x1=rnorm(20)+2, x2=rnorm(20)+2) xa=cor(d,use='complete') xb=cor(d1,use='complete') I want to combine these two to get a third matrix which should have half values from 'xa' and half

Hi any, Can some please detail me the createX command in bayesm package? To make things easy for you to help me, let me put forward my problem Suppose I have 3 covariates (say X matrix) and my Y has 3 categories say (1,2,3). Now from the CreateX I understand that the data matrix say 'Xa' must be of dimension n* (naxp), where 'na' is the number of variables and 'p' is

Dear all Is there a simpler method to achieve the following: When I obtain an empty data.frame after subsetting, I need for it to contain one line of NAs. Here's a dummy example: > (.xb <- iris[ iris$Species=='zz', ]) [1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species <0 rows> (or 0-length row.names) > dim(.xb) [1] 0 5 > (.xa <-

Dear all I have a 'character' vector containing mixed formats (thanks Excel!) and I'd like to translate it into a default "%Y-%m-%d" Date vector. x <- c("1/3/2005", "13/04/2004", "2/5/2005", "2/5/2005", "7/5/2007", "22/04/2004", "21/04/2005", "20080430", "13/05/2003",

Dear All,   is there a function in R that would help me convert a covariance matrix built based on arithmetic returns to a covariance matrix from log-returns?   As an example of the means and covariance from arithmetic:   mu <-c(0.094,0.006,1.337,1.046,0.263) sigma

I have two lists: xa <- list( X=c(1,2,3), Y=c(4,5,6), Z=c(7,8,9) ) xb <- with( barley, tapply( X=seq(1:nrow(barley)), INDEX=site , FUN=function(z)yield[z])) I can convert xa to a dataframe easily with: as.data.frame(xa) But if i try the same with xb I get: as.data.frame(xb) Error in as.data.frame.default(xb) : can't coerce array into a data.frame What

I have been told by the CRAN administrators that the following code generated an error on 64-bit Fedora Linux (gcc, clang) and on Solaris machines (sparc, x86), but runs well on all other systems): > fn <- function(x, y) ifelse(x^2 + y^2 <= 1, 1 - x^2 - y^2, 0) > tol <- 1.5e-8 > fy <- function(x) integrate(function(y) fn(x, y), 0, 1,

Hi, This looks OK: > x <- c("_1_", "1_9", "2_9") > rank(x) [1] 1 2 3 But this does not: > xa <- paste(x, "a", sep="") > xa [1] "_1_a" "1_9a" "2_9a" > rank(xa) [1] 2 1 3 Cheers, H. > sessionInfo() R version 2.14.0 (2011-10-31) Platform: x86_64-unknown-linux-gnu (64-bit) locale: [1]