similar to: Avoiding Loops When Iterating Over Statement That Updates Its Input

I'm interested in using a data frame as if it were a hash table. For instance if I had the following, > (d <- data.frame(key=seq(0.5, 3, 0.5), value=rnorm(6))) key value 1 0.5 -1.118665122 2 1.0 0.465122921 3 1.5 -0.529239211 4 2.0 -0.147324638 5 2.5 -1.531503795 6 3.0 -0.002720434 Then I'd like to be able to quickly retrieve the "value" of "key" 1.5

Hi, Is there a way to specify the last element of a vector, similar to "end" in MATLAB? v[end] would be MATLAB for v(length(v)) in R. While `v(length(v))' does yield the last element, that approach fails in the following, rep(v, each=2)[-c(1,length(v))] which is meant to duplicate all elements of `v' except for the first and last. (I.e., if `v <- 1:4', then

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

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

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

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:

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

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 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

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.

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 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

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 all, This is probably a blindingly obvious question: Why does it matter in the uniroot function whether the f() values at the end points that you supply are of the same sign? For example: f <- function(x,y) {y-x^2+1} #this gives a warning uniroot(f,interval=c(-5,5),y=0) Error in uniroot(f, interval=c(-5, 5), y = 0) : f() values at end points not of opposite sign #this doesn't give a

Hi, Could somebody help me with the following. I want to calculate the integral over an implicit function. I thought to integrate over a function depending on uniroot. In previous topics I found a thread about finding the root of an integral. And that works. But the other way around, does not work. Does R support this? I included the following example. The function in the example is very easy

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

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

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

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

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