similar to: how to convert a vector to an upper matrix

I am trying to teach my models how to render themselves, i.e. <%= my_model_object.render() %> Let me explain my reasoning and proposed method before this gets shot down as anti-MVC. Let''s say I am writing a contact-management application. I have a class Contact. I will need to display this class all over the application. My first choice is to use a partial. Now I can

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

This below is not solvable with uniroot to find "a": fn=function(a){ b=(0.7/a)-a (1/(a+b+1))-0.0025 } uniroot(fn,c(-500,500)) gives "Error in uniroot(fn, c(-500, 500)) : f() values at end points not of opposite sign" I read R-help posts and someone wrote a function: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/92407.html but it is not very precise. Is there any

Dear all I am trying to calculate the value of n using uniroot. Here is the message I am having: <<< Error in uniroot(integ, lower = 0, upper = 1000, n) : 'interval' must be a vector of length 2 >>> Please would you be able to give me an indication on why I am having this error message. Many thanks EXAMPLE BELOW: ## t = statistics test from t -distribution

With main R releases only happening yearly in spring, now is good time to consider *and* discuss new features for what we often call "R-devel" and more officially is R Under development (unstable) (.....) -- "Unsuffered Consequences" Here is one such example I hereby expose to public scrutiny: A few minutes ago, I've committed the following to R-devel (the

In looking at rootfinding for the histoRicalg project (see gitlab.com/nashjc/histoRicalg), I thought I would check how uniroot() solves some problems. The following short example ff <- function(x){ exp(0.5*x) - 2 } ff(2) ff(1) uniroot(ff, 0, 10) uniroot(ff, c(0, 10), trace=1) uniroot(ff, c(0, 10), trace=TRUE) shows that the trace parameter, as described in the Rd file, does not seem to be

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda value

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:

Hi, I'm trying to calculate the value of the variable, dp, below, in the argument to the integral of dnorm(x-dp) * pnorm(x)^(m-1). This corresponds to the estimate of the sensitivity of an observer in an m-alternative forced choice experiment, given the probability of a correct response, Pc, a Gaussian assumption for the noise and no bias. The function that I wrote below gives me an error:

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

dear R experts--- I am experimenting with multicore processing, so far with pretty disappointing results. Here is my simple example: A <- 100000 randvalues <- abs(rnorm(A)) minfn <- function( x, i ) { log(abs(x))+x^3+i/A+randvalues[i] } ?## an arbitrary function ARGV <- commandArgs(trailingOnly=TRUE) if (ARGV[1] == "do-onecore") { ?library(foreach) ?discard <-

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

Dear all, I have the following problem with the uniroot function. I want to find roots for the fucntion "Fp2" which is defined as below. Fz <- function(z){0.8*pnorm(z)+p1*pnorm(z-u1)+(0.2-p1)*pnorm(z-u2)} Fp <- function(t){(1-Fz(abs(qnorm(1-(t/2)))))+(Fz(-abs(qnorm(1-(t/2)))))} Fp2 <- function(t) {Fp(t)-0.8*t/alpha} th <- uniroot(Fp2, lower =0, upper =1,

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(

I have a strange problem with uniroot() function. Here is the result : > uniroot(th, c(-20, 20)) $root [1] 4.216521e-05 $f.root [1] 16.66423 $iter [1] 27 $estim.prec [1] 6.103516e-05 Pls forgive for not reproducing whole code, here my question is how "f.root" can be 16.66423? As it is finding root of a function, it must be near Zero. Am I missing something? -- View this message

Sir, I want to solve the equation Q(u)=mean, where Q(u) represents the quantile function. Here my Q(u)=(c*u^lamda1)/((1-u)^lamda2), which is the quantile function of Davies (Power-pareto) distribution. Hence I want to solve , *(c*u^lamda1)/((1-u)^lamda2)=28353.7....(Eq.1)* where lamda1=0.03399381, lamda2=0.1074444 and c=26104.50. When I used the package 'Davies' and solved Eq 1, I got the

On Fri, 25 Aug 2023 22:17:05 +0530 ASHLIN VARKEY <ashlinvarkey at gmail.com> wrote: > Sir, Please note that r-help is a mailing list, not a knight! ?? > I want to solve the equation Q(u)=mean, where Q(u) represents the > quantile function. Here my Q(u)=(c*u^lamda1)/((1-u)^lamda2), which is > the quantile function of Davies (Power-pareto) distribution. Hence I > want to

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

Despite my years with R, I didn't know about trace(). Thanks. However, my decades in the minimization and root finding game make me like having a trace that gives some info on the operation, the argument and the current function value. I've usually found glitches are a result of things like >= rather than > in tests etc., and knowing what was done is the quickest way to get there.

I'd like to plot a point at the intersection of these two curves. Thanks x <- seq(.2, .3, by = .01) f <- function(x){ x*cos(x)-2*x**2+3*x-1 } plot(x,f(x), type = "l") abline(h = 0)