similar to: Is there a bisection method in R?

Im trying to ascertain whether or not the facilities of R are sufficient for solving an optimization problem I've come accross. Because of my limited experience with R, I would greatly appreciate some feedback from more frequent users. The problem can be delineated as such: A utility function, we shall call g is a function of x, n ... g(x,n). g has the properties: n > 0, x lies on the

Hi R-users, Is it possible to make a bisection in R? [[alternate HTML version deleted]]

Hi R users, Simply question: On the command line, how do I list the datasets contained within a package, e.g. MASS? I scanned the mailing list history but was unable to find the answer. Thanks in advance. Gregory --------------------------------- [[alternative HTML version deleted]]

I am looking for an efficient way to solve the following problem. I have a large matrix of continuous values with a small proportion of missing values. Columns correspond to variables where each variable has two measurements, call them A and B. The matrix is such that the columns are in sequence with respect to the variables. I would like to sum up the two measurements for each variable and each

Hi fellow R-users, Are there any text editors that recognize sweave (.rnw) files? I am running Windows Vista and in the past I used Tinn-R for R files but it (surprisingly) doesn't recognize rnw files and does not do any syntax highlighting for them. Thanks in advance, Greg __________________________________________________________________ Make your browsing faster, safer, and

dear R experts--- I have (many) unidimensional root problems. think loc.of.root <- uniroot( f= function(x,a) log( exp(a) + a) + a, c(.,9e10), a=rnorm(1) ) $root (for some coefficients a, there won't be a solution; for others, it may exceed the domain. implied volatilities in various Black-Scholes formulas and variant formulas are like this, too.) except I don't need 1 root, but a

Hi all, I have the following two function f1 and f2. f1 <- function(lambda,z,p1){ lambda*(p1*exp(-3*z-9/2)+(0.2-p1)*exp(4*z-8))-(1-lambda)*0.8} f2 <- function(p1,cl, cu){ 0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl+3)+(1-pnorm(cu+3)))+(0.2-p1)*(pnorm(cl-4)+(1-pnorm(cu-4))))}-0.05 First fix p1 to be 0.15. (i) choose a lambda value, say lamda=0.6, (ii)

Greetings, I have a seemingly simple task which I have not been able to solve today and I checked all of the help archives on this and have been unable to find anything useful. I want to construct a symmetric matrix of arbtriray size w/o using loops. The following I thought would do it: p <- 6 Rmat <- diag(p) dat.cor <- rnorm(p*(p-1)/2) Rmat[outer(1:p, 1:p, "<")] <-

Greetings, I have a seemingly simple task which I have not been able to solve today. I want to construct a symmetric matrix of arbtriray size w/o using loops. The following I thought would do it: p <- 6 Rmat <- diag(p) dat.cor <- rnorm(p*(p-1)/2) Rmat[outer(1:p, 1:p, "<")] <- Rmat[outer(1:p, 1:p, ">")] <- dat.cor However, the problem is that the matrix

Hi fellow R-users, I have run into a problem when trying to identify the number of rows in a matrix. Say we have an arbitrary 5 by 5 matrix called temp: temp <- matrix(norm(25), nrow=5) The problem is that nrow(temp[1,]) returns NULL. I would like it to return 1 because in my larger program I am indexing the rows of large matrices according to another variable and I need to test when the

Hi, I am new to R, and as a first exercise, I decided to try to implement an XIRR function using the secant method. I did a quick search and saw another posting that used the Bisection method but wanted to see if it was possible using the secant method. I would input a Cash Flow and Date vector as well as an initial guess. I hardcoded today's initial date so I could do checks in Excel.

Dear R-users, Say that I have a sequence of zeroes and ones: x <- c(1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0) The sequences of ones represent segments and I want to report the starting and endpoints of these segments. For example, in 'x', the first segment starts at location 1 and ends at 3, and the second segment starts at location 8 and ends at location 10. Is there an efficient

Hi guys im having a problem getting R to numerically integrate for some function, say f(bhat)*optimize(G(bhat)), over bhat. Where id like to integrate this over some finite range, so that here as we integrate over bhat optimize would return a different optimum. For instance consider this simple example for which I cannot get R to return the desired result: f <- function(bhat) exp(bhat) g

Hello fellow R users, I have a matrix computation that I imagine should be relatively easy to do, however I cannot figure out a nice way to do it. I have two matrices, for example mat1 <- matrix(c(1:5,rep(NA,5), 6:10), nrow=3, byrow=T) mat2 <- matrix(c(2:6, 6:10, rep(NA,5)), nrow=3, byrow=T) I'd like to compute the element-wise average for non-NA entries. Of course (mat1+mat2)/2

Hi everyone, My goal is to sample from a two-dimensional grid. Consider the following example of code: n.grid <- 500 muA.grid <- seq(-4,4, length=n.grid) muB.grid <- seq(-4,4, length=n.grid) mu.p <- matrix(NA, nrow=n.grid, ncol=n.grid) for(i in 1:n.grid){ for(j in 1:n.grid){ mu.p[i,j] <- dnorm(muA.grid[i], 0, 1)*dnorm(muB.grid[j], 0, 0.5) } } mu.p <-

curiosity---given that vector operations are so much faster than scalar operations, would it make sense to make uniroot vectorized? if I read the uniroot docs correctly, uniroot() calls an external C routine which seems to be a scalar function. that must be slow. I am thinking a vectorized version would be useful for an example such as of <- function(x,a) ( log(x)+x+a ) uniroot( of, c(

Can anyone tell me the trick for obtaining the smoother matrix from smooth.spline when there are non-unique values for x. I have the following code but, of course, it only works when all values of x are unique. ## get the smoother matrix (x having unique values smooth.matrix = function(x, df){ n = length(x); A = matrix(0, n, n); for(i in 1:n){ y = rep(0, n); y[i]=1; yi =

I have the following function for which I need to find the root of a: f <- function(R,a,c,q) sum((1 - (1-R)^a)^(1/a)) - c * q To give context for the problem, this is a psychometric issue where R is a vector denoting the percentage of students scoring correct on test item i in class j, c is the proportion correct on the test by student k, and q is the number of items on the test in total. I

Dear R users, I'd like to evaluate a vector of characters. For example, say I have a data frame called Data including the field names x1, x2, x3, and I'd like to a list out of paste("Data$x", 1:3, sep=""). How can I get list to evaluate paste("Data$x", 1:3, sep="") as an R object rather than a string? Thanks in advance for you assistance. Gregory

Fellow R-users, I have a longitudinal data set with missing values in it. I would like to produce a residual plot for each time using panel.xyplot function but I get an error message. Here's a simple example, library(nlme) set.seed(1544) longdata <- data.frame(ID=gl(10,1,50), y=rnorm(50), time = as.numeric(gl(5,10,50)), x = rnorm(50)) longdata$y[5] <- NA longdata$x[35] <- NA