similar to: uniroot

Dear R-user, I am trying to use the R "nlme" function to fit a non linear mixed effects model. The model I wand to fit is an individual somatic growth model with 4 parameters. For all parameters both fixed and random effects have to be estimated, as well as their covariance matrix (see the formula bellow). The data are simulated with the same growth model as in the nlme, with know

Hi, I have a strange situation where when I get different timestamps depending on if a use ls -l or ls -ln: --------------- sftp> ls -l -rw-rw-rw- 1 0 0 39649 Sep 20 11:30 C20060920.1115-20060920.1130 sftp> ls -ln -rw------- 0 0 0 39649 Sep 20 13:30 C20060920.1115-20060920.1130 -------------- The machine is in timezone UTC+2. According to the

Hi, I am having some problems using the if statement correctly. I have used it many times previously so I dona't know what is different with this case. Here is my problem: I have a 1X10 matrix of values as follows: > H.MC [,1] [1,] 4.257669 [2,] 7.023242 [3,] 4.949857 [4,] 5.107000 [5,] 4.257669 [6,] 4.257669 [7,] 4.257669 [8,] 4.257669 [9,] 4.257669 [10,] 4.257669

# Hi All, # # I need to solve a somewhat complex equation at many parameter values for # a number of different parameters. # A simplified version of the equation is: 0= (d1/(h1^2))-(h2*(d2^2)) # I'd like to solve it across a parameter space of d1 and d2, holding # h1 and h2 constant. # It seems that uniroot() can do it, but I don't see how to vectorize it. #

If I have a vector of n elements, e.g. a vector of length 4 with elements 10, 20, 30, 40 and want to find the different values of x such that x^2=10, x^2=20, x^30 and x^2=40, how could I do this in R? I'm thinking of using the uniroot function, but am finding difficult applying it to a vector. Thanks -- View this message in context:

Nurza, Try running a while loop steping out until you have a start and finish thats the function is opposite in sign. You need a "start" and "finish" where F is + and - on either side of the loop. Graphing F might help. step<-10 checkme<-F(start)*F(finish+step) while(checkme>0){ initialstep<-initialstep*2 checkme<-F(start)*F(finish+step) }

fisher.test(matrix(c(1,20,246,6873),2),hybrid=F) Error in if (f(lower, ...) * f(upper, ...) >= 0) stop("f() values at end points not of opposite sign") : missing value where logical needed Thomas Lumley Assistant Professor, Biostatistics University of Washington, Seattle -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list --

Full_Name: Chris Andrews Version: 2.2.1 OS: Windows Submission from: (NULL) (128.205.94.95) The function page for uniroot indicates If the algorithm does not converge in 'maxiter' steps, a warning is printed and the current approximation is returned. I have not been able to get a warning message even when I think I should get one (see code below). Perhaps the bug is in the

Hi, I wonder if it would make sense to make uniroot detect zeros at the endpoints, eg if f(lower)==0, return lower as the root, else if f(upper)==0, return upper as the root, else stop if f(upper)*f(lower) > 0 (currently it stops if >=), else proceed to the algorithm proper. Currently I am using a wrapper function to implement this, and I found it useful. But I didn't want to send a

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

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 all, I would like to fit growth data to a smooth curve (preferably using a Stineman interpolation). Then I would like to find the intersection of the curve with a horizontal line (e.g. 0). See below for a coding example. The root finding using 'uniroot' works fine if I fit the data with 'smooth.spline', but not if I fit the data with 'stinterp' (stinepack) or

Hello all, I was going to solve (n-m)! * (n-k)! = 0.5 *n! * (n-m-k)! for m when values of n and k are provided n1<-c(10,13,18,30,60,100,500)         # values of n kx<-seq(1,7,1);                               # values of k slv2<-lapply(n1,function(n){    slv1<-lapply(kx,function(k){              lhs<-function(m)              {                

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:

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)) provides a root at -6.00e-05, which is outside of the specified bounds. The default value of the "extendInt" argument to uniroot() is "no", as far as I can see ... $root [1] -6.003516e-05 $f.root [1] -74453981 $iter [1] 1 $init.it [1] NA

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.

Hi R-users,I'm trying to solve a non linear equation, to find the degrees of freedom of a mixture of t student. I'm sure i wrote the minimization equation in the right way, but when i try to run the EM algorithm to estimate the parameters of the mixture, the following error will appear: Error in uniroot(function(z) log(z/2) - digamma(z/2) + 1 - log((z + d)/2) + : f() values at end

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

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: >