similar to: "firstthat"?

Dear All, I have a function f(x,y,z)=exp(x^3+y^4+x^2*y+x*z^2+y/z) over D, where is D={ (x,y,z)| 0 <z<Inf, 0<y<c1*z, 0<x<c2*/y}. x,y,z are all vectors and c1 and c2 are constants. I tried the "adapt" package and I get some error. This is the error message: "Error in function (z, y, x) : argument "x" is missing, with no default" I included my R

R version: 2.7.0 Running on: WinXP I am trying to model damage from fire losses (given that the loss occurred). Since I have the individual insured amounts, rather than sampling dollar damage from a continuous distribution ranging from 0 to infinity, I want to sample from a percent damage distribution from 0-100%. One obvious solution is to use runif(n, min=0, max=1), but this does not seem to be

Hello all how can sample from f(x)~x^(a-1)*ind(0,min(b,-log(u)) in R? where a and b is positive constand and 0<u<1 thanks khazaei

Hello everyone, I have a function here that I wrote but doesn't seem to work quite right. Attached is the code. In the calib funcion under the for loop Bt[i+2]<-(1-m)*Bt[i+1]+Rt[i]*Rerr-Ct[i+1] returns NA's for everything after years 1983 and 1984. However the code works when it reads Bt[i+2]<-(1-m)*Bt[i+1]+Rt[i]*Rerr-Ct[i]. I don't quite understand why since it should be

Hi, I'm looking the best way to fill irregular polygons with patterns, Something like the function grid.pattern do, but my case is with irregular polygons. Whit this script I can get it, but I'm looking for an "elegant" solution.. library(grid) grid.polygon(x=c(0.2, 0.8, 0.6, 0.6, 0.8, 0.2), y=c(0.2, 0.2, 0.3, 0.5, 0.7,0.7), gp=gpar(fill="grey",

Hello everyone, I have a function here that I wrote but doesn't seem to work quite right. Attached is the code. In the calib funcion under the for loop Bt[i+2]<-(1-m)*Bt[i+1]+Rt[i]*Rerr-Ct[i+1] returns NA's for everything after years 1983 and 1984. However the code works when it reads Bt[i+2]<-(1-m)*Bt[i+1]+Rt[i]*Rerr-Ct[i]. I don't quite understand why since it should be

The loop is correct, you just need to make sure that your result is computed and stored as the n-th element that is returned by the loop. Pick up any manual of R, and looping will be explained there. Also, I would recommend that you draw a random number for every iteration of the loop. Defining the random vectors outside the loop make sense to me only if they are the same length as n.

Hi All, I am attempting to simulation an inventory model on R and I am having some problems. I believe the problem is when I define my demand rate is stays constant throughout so when I do need to reorder the model does not recognise it as I have the initial supply arrival time set to infinity at the start as no order has been played but I have an if statement saying, if the level falls below a

I've redirected this reply from r-help to the bugs list. On 11/3/2006 8:25 AM, vito muggeo wrote: > Dear all, > I am dealing with the following (apparently simple problem): > For some reasons I am interested in passing variables from a dataframe > to a specific environment, and in fitting a standard glm: > > dati<-data.frame(y=rnorm(10),x1=runif(10),x2=runif(10)) >

In package fOptions, there are functions that generate Halton sequences. The van der Corput sequence for base 2 is a particular case of the Halton sequence generated by: n <- 8 # anything here... x <- runif.halton(n, 1) In fact, x <- runif.halton(n, dim) will generate the van der Corput sequences for the base b as the i-th prime number in x[,i]. (in other words, if I want the van der

As a followup to pi-day, I attempted to make a .gif of a simulation based estimation of pi by plotting points inside a single quadrant of a circle (a la?http://www.drewconway.com/zia/?p=2667 ). ?When rendering the individual x,y pairs with points() I intermittently see points crop up around (2,0.5) but the input values for x and y are bounded between 0 and 1. square<-structure(c(0, 0, 1, 1, 0,

hello R: we have a two-parameter IRT simulation code. The goal is to generate a response matrix.But the "for" part doesn't run. we don't know what is wrong with it. Thanks so much~~~ I <- 10 J <- 5 response <- matrix(0, 10, 5) pij <- function(a,b,theta) { a <- rnorm(J, 0.8, 0.04) a b <- rnorm(J, 0, 1) b theta <- rnorm(I, 0,1) theta for( i in 1:I ) { for(

Hi this is my first post. I am trying to run a simulation for a computer playing Von Neumann poker and adjusting it's expectation of an opponent's behavior according to how the opponent plays. This program involves random generation of "hands" and shifting of parameters. However, when I run the code, no errors come up, but the program doesn't do anything. Could someone

Hi, I fit a Cox PH model to estimate the cause-specific hazards (in a competing risks setting). Then , I compute the survival estimates for all the individuals in my data set using the `survfit' function. I am currently playing with a data set that has about 6000 observations and 12 covariates. I am finding that the survfit function is very slow. Here is a simple simulation example

Hi guy, I have recently encountered a problem while I was just trying to generate some random numbers with the function "rnorm", the problem is shown below: ########case 1############ > rnorm(20*0.2) [1] -1.2765922 -0.5732654 -1.2246126 -0.4734006 ########case 2########### *> rnorm(20*(1-0.8)) [1] -0.62036668 0.04211587 -0.91092165* #########case 3############ > a<-0.2 >

Hello! I am performing coupling of chains in MCMC and I need the same value of seed for two chains. I will show demo of what I want: R code, which might show my example is: niter <- 3 nchain <- 2 tmpSeed <- 123 for (i in 1:niter) { # iterations for (j in 1:nchain) { # chains set.seed(tmpSeed) a <- runif(1) cat("iter:", i, "chain:", j,

Hellow everybody, I have a dataframe with 468 individuals (rows) that I captured at least once during 28 visits (columns), it looks like: mortality[1:10,] X18.10.2004 X20.10.2004 X22.10.2004 X24.10.2004 X26.10.2004 X28.10.2004 X30.10.2004 X01.11.2004 X03.11.2004 X07.11.2004 1 1 0 0 0 1 1 1 0 0

Dear All, I am new to R, I have one question which might be easy. I have a large data with more than 250 variable, i am reducing number of variables by redun function as in the example below, n <- 100 x1 <- runif(n) x2 <- runif(n) x3 <- x1 + x2 + runif(n)/10 x4 <- x1 + x2 + x3 + runif(n)/10 x5 <- factor(sample(c('a','b','c'),n,replace=TRUE)) x6 <-

Hi Bert, On Tue, Nov 14, 2017 at 8:11 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > Could someone please explain the following? I did check bug reports, but > did not recognize the issue there. I am reluctant to call it a bug, as it > is much more likely my misunderstanding. Ergo my request for clarification: > > ## As expected: > >> lapply(1:3, rnorm, n = 3)

Thanks, I tried someting like this, but computation takes times for large matrices btransf <- function(y,X=length(y)^4) { N <- length(y) bm <- matrix(rep(1/N,N^2),N,N) for(j in 1:X){ coord <- sample(1:N,4,replace=T) d <- runif(1,0,min(bm[coord[1],coord[2]],bm[coord[3],coord[4]]))