similar to: vecortizing uniroot() for numerical solutions

Hi All, I need to sample a vector ("old"), with replacement, up to the point where my vector of samples ("new") sums to a predefined value ("target"), shortening the last sample if necessary so that the total sum ("newsum") of the samples matches the predefined value. While I can easily do this with a "while" loop (see below for example

Dear All, I am hoping to implement a barchart using trellis graphics where all elements have a line width of 2. Using trellis.par.set(), I am able to make most elements lwd=2, but not all. In particular, the top of the box (above the upper most strip) and the left y-axis remain one point. Code with a barchart() example is below. Any advice would be greatly appreciated. Thanks for your

Dear R-Helpers, Can anyone please help me to interpret warning messages from zeroinfl (package pscl) while fitting a zero inflated negative binomial model? The console reports convergence and the parameters seam reasonable, but these <<Warning messages: 1: algorithm did not converge in: glm.fit(X, Y, family = poisson()) 2: fitted rates numerically 0 occurred in: glm.fit(X, Y, family =

Dear All, I need to produce a multi-panel plot where: 1) groups within panels receive distinct symbols; 2) a linear regression is fit to the entire panel, not the individual groups; 3) a common abline is plotted in each panel. Essentially I would like to merge the two plots below, but I can't seem to figure out how to call both lmline and abline, and also retain the grouping symbols.

Hi All, convhulln {geometry} computes the convex hull of a set of points in n- dimensions via .call, returning the hull itself, and also "unavoidably generates a diagnostic report on an Rterm console." See the example below. I need to access the results of the diagnostic report (specifically the computed hull volume) from within R, but I cannot see how. I've looked at

On 06/10/2020 11:00 a.m., Sorkin, John wrote: > Colleagues, > I am trying to learn to use uniroot to solve a simple linear equation. I define the function, prove the function and a call to the function works. When I try to use uniroot to solve the equation I get an error message, > Error in yfu n(x,10,20) : object 'x' not found. > > I hope someone can tell we how I can fix

Colleagues, I am trying to learn to use uniroot to solve a simple linear equation. I define the function, prove the function and a call to the function works. When I try to use uniroot to solve the equation I get an error message, Error in yfu n(x,10,20) : object 'x' not found. I hope someone can tell we how I can fix the problem

Hi , I would like to know if a R function have the same behaviour than the matlab solve function. I tried something with uniroot but I have some problems: The equation I need to solve is : exp(c0-r0)*(bb0+x)*(bb1-x)=(bb0+x+1)(bb1-x-1) So I tried this: STEP 1: my function test test <- function(x,bb0=-3,bb1=5,c0=2,r0=0) { + ((exp(c0-r0)*(bb0+x)*(bb1-x))/((bb0+x+1)(bb1-x-1))-1)} STEP 2: >

Hi All, I am writing a simulation that examines the effects of species extinctions on ecological communties by sequentially removing individuals of a given species (sometimes using weighted probabilities) and replacing the lost individuals with species identities randomly sampled from the remaining individuals. Thus I use two dataframes. One contains all the individuals and their species

I am calling the uniroot function from inside another function using these lines (last two lines of the function) : d <- uniroot(k, c(.001, 250), tol=.05) return(d$root) The problem is that on occasion there's a problem with the values I'm passing to uniroot. In those instances uniroot stops and sends a message that it can't calculate the root because f.upper * f.lower is greater

I tend to avoid the the trace/verbose arguments for the various root finders and optimizers and instead use the trace function or otherwise modify the function handed to the operator. You can print or plot the arguments or save them. E.g., > trace(ff, print=FALSE, quote(cat("x=", deparse(x), "\n", sep=""))) [1] "ff" > ff0 <- uniroot(ff, c(0, 10))

Hi, I'm using the uniroot function, and would like to detect an error which occurs, for instance, when the values at endpoints are not of opposite signs. For example: uniroot( function(x) x^2+1, lower=1, upper=2 ). I want to say something like: if "error in uniroot(...)" return NA else return uniroot$root Thanks a lot! Asaf -- View this message in context:

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(

Dear Prof Azzalini, You have an interesting example of recursive use of uniroot(). [re-cited at the end] However, note that R currently has the same problem as S-plus: Uniroot() doesn't work reliably, recursively. When you found it to be better, you were just lucky. The relevant file in R's source, src/main/optimize.c says /* WARNING : As things stand, these routines should

Dear All I am trying to find a uniroot of a function within another function (see example) but I am getting an error message (f()values at end points not of opposite sign). I was wondering if you would be able to advise how redefine my function so that I can find the solution. In short my first function calculates the intergrale which is function of "t" , I need to find the uniroot of

Hi, I have tried to use uniroot to solve a value (value a in my function) that gives f=0, and I repeat this process for 10000 times(stimulations). However error occures from the 4625th stimulation - Error in uniroot(f, c(0, 2), maxiter = 1000, tol = 0.001) : f() values at end points not of opposite sign I have also tried interval of (lower=min(U), upper=max(U)) and it won't work as well.

I have one equation, two unknowns, so I am trying to build the solution set by running through possible values for one unknown, and then using uniroot to solve for the accompanying second solution, then graphing the two vectors. p0 = .36 f = function(x) 0.29 * exp(5.66*(x - p0)) f.integral = integrate(f, p0, 1) p1 = p0 + .01 i = 1 n = (1 - p0)/.01 p1.vector = rep(0,n) p2.vector = rep(0,n) for (i

A recent message on ansari.test() prompted me to play with the examples. This doesn't work for me in R version 2.4.1 R> ansari.test(rnorm(100), rnorm(100, 0, 2), conf.int = TRUE) Error in uniroot(ab, srange, tol = 1e-04, zq = qnorm(alpha/2, lower = FALSE)) : object "ab" not found It looks like there's a small typo in ccia() inside ansari.test.default() in which

Hello, I am struggling to find the root of a exponent function. "uniroot" is complaining about a values at end points not of opposite sign? s<- sapply(1:length(w),function(i) + { + + + + + uniroot(saeqn,lower=-5000,upper=0.01036597923,l=list(t=w[i],gp=gp))$root + }) Error in uniroot(saeqn, lower = -5000, upper = 0.01036597923, l = list(t = w[i], : f() values at

I wrote first cut at unirootR for Martin M and he revised and put in Rmpfr. The following extends Ben's example, but adds the unirootR with trace output. c1 <- 4469.822 c2 <- 572.3413 f <- function(x) { c1/x - c2/(1-x) }; uniroot(f, c(1e-6, 1)) uniroot(f, c(1e-6, 1)) library(Rmpfr) unirootR(f, c(1e-6, 1), extendInt="no", trace=1) This gives more detail on the iterations,