similar to: How would you avoid loops in this simple example?

Warning: I am a complete newbie to R. I have read ISwR, but I am still finding myself completely stuck on some simple concepts. I have tried everything I can think of to solve this one, and finally decided that enough was enough and I need a pointer to a solution. I have the following summary from lm(): ---- > summary(lm(nu1~nu4)) Call: lm(formula = nu1 ~ nu4) Residuals: Min 1Q

Hi all; I recently have been used 'maxLik' function for maximizing G2StNV178 function with gradient function gradlik; for receiving this goal, I write the following program; but I have been seen an error  in calling gradient  function; The maxLik function can't enter gradlik function (definition of gradient function); I guess my mistake is in line ******** ,that the vector  ‘h’ is

Hi all, I wondered whether anyone has some advice on a stats-related 'sanity check', as I ran a nonparametric multivariate test (mulrank function as decribed by R. Wilcox, 2005) on both systems, but got different results (please see below for the system-specific outputs)! The functions I used are attached as well. Any advice would be much appreciated! Thanks in advance for getting back to

Dear R users, i would like to transform the following function from R-code to C-code and call it from R in order to speed up the computation because in my other functions this function is called many times. `dgcpois` <- function(z, lambda1, lambda2) { `f1` <- function(alpha, lambda1, lambda2) return(exp(log(lambda1) * (alpha - 1) - lambda2 * lgamma(alpha))) `f2` <-

Goal: Suppose you have a vector that is a discrete variable with values ranging from 1 to 3, and length of 10. We'll use this as the example: y <- c(1,2,3,1,2,3,1,2,3,1) ...and suppose you want your new vector (y.new) to be equal in length to the possible discrete values (3) times the length (10), and formatted in such a way that if y[1] == 1, then y.new[1:3] == c(1,0,0), and if y[2] ==

Hi, My name is Laura. I'm a PhD student at Queen's University Belfast and have just started learning R. I was wondering if somebody could help me to see where I am going wrong in my code for estimating the parameters [mu1, mu2, lambda1] of a 2-phase Coxian Distribution. cox2.lik<-function(theta, y){ mu1<-theta[1] mu2<-theta[2] lambda1<-theta[3]

Hi, I'm very new to R and absolutely love it. Does anyone know how to use something in R that functions like a BY command in SAS? For example, let's say you have a variable x, and you want to see the mean. Easy... > mean(x) But what if you want to see the mean of x conditional on another discrete variable? My best attempts so far are something like... > mean(x, y_cat=1)

I use "while" loop but it produces an errro. I have no idea about this. Error in "[<-"(`*tmp*`, i, value = numeric(0)) : nothing to replace with The problem description is The likelihood includes two parameters to be estimated: lambda (=beta0+beta1*x) and alpha. The algorithm for the estimation is as following: 1) with alpha=0, estimate lambda (estimate beta0

How to convert the following ms() in Splus to Optim in R? The "Calc" function is also attached. ms(~ Calc(a.init, B, v, off, d, P.a, lambda.a, P.y, lambda.y, 10^(-8), FALSE, 20, TRUE)$Bic, start = list(lambda.a = 0.5, lambda.y = 240), control = list(maxiter = 10, tol = 0.1)) Calc <- function(A.INIT., X., V., OFF., D., P1., LAMBDA1., P2., LAMBDA2., TOL., MONITOR.,

Hi, I'm sure that a large fixed width file, such as 300 million rows and 1,000 columns, is too large for R to handle on a PC, but are there ways to deal with it? For example, is there a way to combine some sampling method with read.fwf so that you can read in a sample of 100,000 records, for example? Something like this may make analysis possible. Once analyzed, is there a way to, say, read

I want to apply a more complicated function than what I use in my example, but the idea is the same: Suppose you have a data frame named x and you want to a function applied to each variable, we'll just use the quantile function for this example. I'm trying all sorts of apply functions, but not having luck. My best guess would be: sapply(x, FUN=quantile) -- View this message in

I have an error for a simple optimization problem. Is there anyone knowing about this error? lambda1=-9 lambda2=-6 L<-function(a){ s2i2f<-(exp(-lambda1*(250^a)-lambda2*(275^a-250^a)) -exp(-lambda1*(250^a)-lambda2*(300^a-250^a))) logl<-log(s2i2f) return(-logl)} optim(1,L) Error in optim(1, L) : function cannot be evaluated at initial parameters Thank you in advance -- View this

Dear all I want to simulate a stochastic jump variance process where N is Bernoulli with intensity lambda0 + lambda1*Vt. lambda0 is constant and lambda1 can be interpreted as a regression coefficient on the current variance level Vt. J is a scaling factor How can I rewrite this avoiding the loop structure which is very time-consuming for long simulations? for (i in 1:N){ ... N <- rbinom(n=1,

Dear All I new to using R and am struggling with some matrix multiplication. I have two matrices, one containing random numbers, these are multiplied together to get another matrix which is different each time. When I put in another for loop to repeat this process a multiple times the matrices are all the same. I?m sure there is a way to keep the randomness of the different matrices but I think

Thanks everyone for the replies. After some experimentation, I found that the order in which the passes are specified matters: opt -O3 -profile-loader matmult.bc -o matmult.opt.bc (works) opt -profile-loader -O3 matmult.bc -o matmult.opt.bc (does not work) Also, I am able to avoid the inconsistency warning only for optimization levels -O3 and -O2. I get that warning when using -O1 and

Dear Friends. I found something very puzzling with constOptim(). When I change the parameters for ConstrOptim, the error messages do not seem to be consistent with each other: > constrOptim(c(0.5,0.3,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in constrOptim(c(0.5, 0.3, 0.5), f = fit.error, gr = fit.error.grr, : initial value not feasible > constrOptim(c(0.5,0.9,0.5),

Memdisk 2.08 cannot boot an XP boot floppy image (XP appears to use WinME boot files for its bootable floppy). For example: http://members.iinet.net.au/~bdriver/bootdisk/download.htm This disk will not boot with Memdisk (Memdisk starts to load the image and then hangs). I'm using Bart's tools to build a boot CD: http://www.nu2.nu/corpmodboot/ Which uses Isolinux and Memdisk.

Any one know is there any package or function to generate bivariate exponential distribution? I gusee there should be three parameters, two rate parameters and one correlation parameter. I just did not find any function available on R. Any suggestion is appreciated. -- View this message in context:

Hi, I use R at home, and am interested in using it at my work company (which is in the Fortune 100). I began the request, and our legal team has given some gruff about the open source license. Not boring you with the details here, but I used some info on gnu.org as a rebuttal, and someone at the company replied that the generalities of GNU GPL may differ from R's specific GNU GPL license,

Dear r-help, I have an array a1 with dimensions [1:660,1:65,1:25] I would like the first dimension to be the last one. That is I want and array [1:65,1:25,1:660] The only way to do this, I know, is tmp.a<-array(dim=dim(a1)[c(2,3,1)]) for(i in 1:dim(a1)[1]) tmp.a[,,i]<-a1[i,,] a1<-tmp.a rm(tmp.a) Is it possible to avoid 'for' loop here? Thank you! ---