similar to: how to define the bound between parameters in nls()

The multiple exponential problem you are attempting has a well-known and long history. Lanczos 1956 book showed that changing the 4th decimal in a data set changes the parameters hugely. Nevertheless, if you just need a "fit" and not reliable paramters, you could reparameterize to k1 and k2diff=k2-k1, so k2=k1+kdiff. Then kdiff has a lower bound of 0, though putting 0 will almost

Thanks to Dieter Menne and Spencer Graves I started to get my way through lsoda() Now I need to use it in with nls() to assess parameters I have a go with a basic example dy/dt = K1*conc I try to assess the value of K1 from a simulated data set with a K1 close to 2. Here is (I think) the best code that I've done so far even though it crashes when I call nls()

Error in model.frame When I run the following nls model an error message appears and I dont know how to solve that. Could you help me?? mat = c(1,2,3,4,5,6,7,8,9,12,16,24,36,48,60) for (i in 1:length(j30)) { bliss = nls(c(j[i,1:length(mat)]) ~ b0 + b1*((1-exp(-k1*mat))/(k1*mat)) + b2*(((1-exp(-k2*mat))/(k2*mat))-exp(-k2*mat)), start = list(k1=0.1993, k2=0.1993, b0= 22.0046,

Hi, I have a density that I need to get MLEs from, which includes definite integrals both in the denominator and in the numerator of the density function. It looks like the outcome depends on the initial values given. My program is shown below: library(circular) ######################################## 4 parameters ######################################## z<-rvonmises(100,0,1)

Hello, I am trying to find the analytical solution to this differential equation dR/dt = k1*R (1-(R/Rmax)^k2); R(0) = Ro k1, k2, and Rmax are parameters that need to fitted, while Ro is the baseline value (which can be fitted or fixed). The response (R) increases initially at an exponential rate governed by the rate constant k1. Response has a S-shaped curve as a function of time and it

Dear Michael, One key is adjustment of nls optimizer tolerance. I notice it has to be higher than usual, but, I recovered your noisy "known" parameter values with an error of K1 (-7%) and k1 (-6%): #### Miller problem with Dalgaard modifications ## Linares 1/22/2004 ## Solution 1 nls(noisy ~ lsoda(xstart, time, one.compartment.model, c(K1=K1, k2=k2))[,2], data=C1.lsoda,

Hi.. i have an expression of the form: model1<-nls(y~beta1*(x1+(k1*x2)+(k1*k1*x3)+(k2*x4)+(k2*k1*x5)+(k2*k2*x6)+(k3*x7)+(k3*k4*x8)+(k3*k2*x9)+(k3*k3*x10)+ (k4*x11)+(k4*k1*x12)+(k4*k2*x13)+(k4*k3*x14)+(k4*k4*x15)+(k5*x16)+(k5*k1*x17)+(k5*k2*x18)+(k5*k3*x19)+

Dear Colleagues, Our group is also working on implementing the use of R for pharmacokinetic compartmental analysis. Perhaps I have missed something, but > fit <- nls(noisy ~ lsoda(xstart, time, one.compartment.model, c(K1=0.5, k2=0.5)), + data=C1.lsoda, + start=list(K1=0.3, k2=0.7), + trace=T + ) Error in eval(as.name(varName), data) : Object

Hello...i want to find the empirical rate for type 1 error using fixed trimmed mean. To make it easy, i'm referring to journal given by this website http://www.academicjournals.org/ajmcsr/PDF/pdf2011/Yusof%20et%20al.pdf. I already run the programme and there is no error in it but i got zero for the empirical rate of type 1 error. The empirical rate for the type 1 error given in the journal

Hi, The code below is giving me this error message: Error in while (err > eps) { : missing value where TRUE/FALSE needed In addition: Warning messages: 1: In dnbinom(x, size, prob, log) : NaNs produced 2: In dnbinom(x, size, prob, log) : NaNs produced I know from the help files that for dnbinom "Invalid size or prob will result in return value NaN, with a warning", but I am not able

Hi I want to sum over one of the dimensions of a n x k1 x k2 array in which each column is the product of the corresponding columns from two matrices with dimensions n x k1 and n x k2. I can see two approaches: a loop on k1 and a loop on k2. But I cannot figure a solution that avoids the loop? Is it possible? (I don't refer to apply or lapply etc either as they are just hidden loops so,

Hello, I haven't found errors in your code. I implemented the test in the paper (the first, fixed symetric mean) and it also gives me zero Type I errors, when alpha = 0.05. Try to see the value of min(pv) or to plot the histogram of 'pv', hist(pv) and you'll see that there are no significant p-values, at that level. Anyway I'll continue to look at it, but my first

Hi everyone! I'm matching two samples to create one sample that have pairs of observations equal for the k1 variable. Merge() doesn't work because I dont't want to recycle the values. x <- data.frame(k1=c(1,1,2,3,3,5), k2=c(20,21,22,23,24,25)) x y <- data.frame(k1=c(1,1,2,2,3,4,5,5), k2=c(10,11,12,13,14,15,16,17)) y merge(x,y,by="k1") k1 k2.x k2.y 1 1 20

comp_keys is duplicating what is done in btrfs_comp_cpu_keys, so just call it. Signed-off-by: Diego Calleja <diegocg@gmail.com> --- fs/btrfs/ctree.c | 14 +------------- 1 file changed, 1 insertion(+), 13 deletions(-) Index: linux/fs/btrfs/ctree.c =================================================================== --- linux.orig/fs/btrfs/ctree.c 2009-07-24 00:47:20.936410297 +0200 +++

Dear R Gang, I'm interested in using R and the nls package for fitting kinetic models. I'm having some difficulty getting a model specified for nls though. The math for the model that I want to fit is dg(t)/dt = K1 f(t) - k2 g(t) where g(t) and f(t) are measured data at a sequence of times t. K1 and k2 are the parameters of the model. If I solve this, the solution is g(t) = K1

Jim has been exceedingly patient (and may well continue to be so), but this smells like "failure to launch". At what point will you start showing your (failed) attempts at solving your own problems so we can help you work on your specific weaknesses and become self-sufficient? -- Sent from my phone. Please excuse my brevity. On February 25, 2018 7:55:55 AM PST, Val <valkremk at

On Tue, 23 Jun 2015, Jakub Jelen wrote: > > On 05/29/2015 09:12 AM, Damien Miller wrote: > > Hi, > > > > OpenSSH 6.9 is almost ready for release, so we would appreciate testing > > on as many platforms and systems as possible. This release contains > > some substantial new features and a number of bugfixes. > Tested basic configuration on Fedora 22. With

I would like to simulate a stochastic process by specifying an intensity function, say lambda(t)=0.05 + 3 exp(鈭抰) I have the following code, but it has a number of problems. One issue I have is I do not know where the time variable is. I would appreciate your help with this code. library("survival") n <- 5000 lamc<-0.08 k1<-0.05 k2<-3 ld<-function(t){k1+k2}

Sorry I had typo in previous email, this typo corrected version: Dear R experts I have simple question, please execuse me: #example data, the real data consists of 20000 pairs of variables K1 <- c(1,2,1, 1, 1,1); K2 <- c(1, 1,2,2, 1,2); K3 <- c(3, 1, 3, 3, 1, 3) M1a <- rep( K1, 100); M1b <- rep(K2, 100) M2a <- rep(K1, 100); M2b <- rep(K1, 100) M3a <- rep(K1, 100); M3b

HI Jim and all, I want to put one more condition. Include col2 and col3 if they are not in col1. Here is the data mydat <- read.table(textConnection("Col1 Col2 col3 K2 X1 NA Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) The desired out put would be Col1 Col2 col3 1 X1 0 0 2 K1 0 0 3 Y1 0 0 4 W1 0 0 6 K2 X1