similar to: R_using non linear regression with constraints

https://cran.r-project.org/web/views/Optimization.html (Cran's optimization task view -- as always, you should search before posting) In general, nonlinear optimization with nonlinear constraints is hard, and the strategy used here (multiplying by a*b < 1000) may not work -- it introduces a discontinuity into the objective function, so gradient based methods may in particular be

> On Jun 18, 2017, at 6:24 AM, Manoranjan Muthusamy <ranjanmano167 at gmail.com> wrote: > > I am using nlsLM {minpack.lm} to find the values of parameters a and b of > function myfun which give the best fit for the data set, mydata. > > mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) > > myfun=function(a,b,r,t){ > prd=a*b*(1-exp(-b*r*t)) >

I ran the following script. I satisfied the constraint by making a*b a single parameter, which isn't always possible. I also ran nlxb() from nlsr package, and this gives singular values of the Jacobian. In the unconstrained case, the svs are pretty awful, and I wouldn't trust the results as a model, though the minimum is probably OK. The constrained result has a much larger sum of squares.

I am not as expert as John, but I thought it worth pointing out that the variable substitution technique gives up one set of constraints for another (b=0 in this case). I also find that plots help me see what is going on, so here is my reproducible example (note inclusion of library calls for completeness). Note that NONE of the optimizers mentioned so far appear to be finding the true best

I've seen a number of problems like this over the years. The fact that the singular values of the Jacobian have a ration larger than the usual convergence tolerances can mean the codes stop well before the best fit. That is the "numerical analyst" view. David and Jeff have given geometric and statistical arguments. All views are useful, but it takes some time to sort them all out and

Is there a way to call a function, and explicitly set an argument to 'not specified'? My situation is the following. I have a function which passes on most of its arguments to another function. The second function, myfun2, serializes all arguments and is out of my control. myfun <- function(...){ return(myfun2(...)); } now, the value for arguments of myfun are stored in variables.

Hello all, I'm trying to use the apply function on a data frame, by applying a function that takes a one row data.frame as argument . Here's the example : myfun = function(x) paste(x$f1 , x$f2) df = data.frame(f1 = c(1,4,10),f2 = "hello") apply(df,1,myfun) ==> Does not work (I get "character(0)" ) Though : myfun(df[1,]) works, and myfun(df) works as well. So if

Hi All I have a dataframe: myframe<-data.frame(ID=c("first","second"),x=c(1,2),y=c(3,4)) And I have a function myfun: myfun<-function(x,y) x+y I would like to write a function myfun2 that takes myframe and myfun as parameters and returns a list as below: mylist $first [1] 4 $second [2] 6 Could you please help me with this? Doesn't seem like the

Dear R experts. I might be missing something obvious. I have been trying to fix this problem for some weeks. Please help. #data ped <- c(rep(1, 4), rep(2, 3), rep(3, 3)) y <- rnorm(10, 8, 2) # variable set 1 M1a <- sample (c(1, 2,3), 10, replace= T) M1b <- sample (c(1, 2,3), 10, replace= T) M1aP1 <- sample (c(1, 2,3), 10, replace= T) M1bP2 <- sample (c(1, 2,3), 10, replace= T)

Dear all, I am using the nlsLM function to fit a Lorentzian function to my experimental data. The LM algorithm should allow to specify limits, but the upper limit appears not to work as expected in my code. The parameter 'w', which is peak width at half maximuim always hits the upper limit if the limit is specified. I would expect the value to be in-between the upper and lower limit with

Dear all, I'd appreciate any help to rectify what must be a misconception of mine how environments work: ########################## myEnv <- new.env() myEnv$a.env <- 1 myEnv$symbols.env <- "a.env" a.global <- 2 symbols.global <- "a.global" myFun <- function(symbols){do.call("print", lapply(symbols, FUN=as.name))} do.call("myFun",

I am trying to write a function, which would allow to call various methods and would pass to them extra arbitrary parameters. My first attempt was to use call() as illustrated below, but apparently '...' cannot be used in such context. How can this be achieved ? Best regards, Ryszard > myfun <- function(method, x, ...) { + v <- eval(call(method, x, ...)) + } > method =

Dear R-users; I'm working with a a dataset that was previously used to fit a nonlinear model of the form: Y ~ a * (1 + b * log(1 - c * X^d)) The parameters published elsewhere are: a = 1.758863, b = .217217, c = .99031, and d = .054589 However, there is no way I can replicate this result. I've tried several options (including SAS) w/o success. The data is: X <-

Hi, suppose that I have the following data.frame: cnae4 cnpj 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 Y 24996 10020470 1 1 2 12 16 21 17 51 43 19 183 24996 10020470 69 91 79 92 91 77 90 96 98 108 891 36145 10020470 0 0 0 0 2 83 112 97 91 144 529 44444 10023333 5 20 60 0 0 0 0 5 20 1000 1110 I would like to create a new variable X that indicates which

? Mon, 6 Nov 2023 17:53:49 +0100 Troels Ring <tring at gvdnet.dk> ?????: > Hence I wonder if I could somehow have non linear regression to find > the 3 pK values. Below is HEPESFUNC which delivers charge in the > fluid for known pKs, HEPTOT and SID. Is it possible to have > root-finding in the formula with nls? Sure. Just reformulate the problem in terms of a function that

Dear R useRs, I have a problem with nls.lm function of minpackl.lm package. I need to fit the Van Genuchten Model to a set of data of Theta and hydraulic conductivity with nls.lm function of minpack.lm package. For the first fit, the parameter estimates keep changing even after 1000 iterations (Th) and I have a following error message for fit of hydraulic conductivity (k); Reason for

Thanks a lot! This was amazing. I'm not sure I see how the conditiion pK1 < pK2 < pK3 is enforced? - it comes from the derivation via generalized Henderson-Hasselbalch but perhaps it is not really necessary. Anyway, the use of Vectorize did the trick! Best wishes Troels Den 06-11-2023 kl. 19:19 skrev Ivan Krylov: > ? Mon, 6 Nov 2023 17:53:49 +0100 > Troels Ring <tring at

I won't send to list, but just to the two of you, as I don't have anything to add at this time. However, I'm wondering if this approach is worth writing up, at least as a vignette or blog post. It does need a shorter example and some explanation of the "why" and some testing perhaps. If there's interest, I'll be happy to join in. And my own posting suggests how the

Dear All, I'm having trouble on 2 laptops Lenovo B580 since upgrading to Centos6.5. ( Because it's a Lenovo I cannot switch the network card for a better supported network card. ) There on the latest kernel : root at jac network-scripts]# uname -a Linux jac.cawdekempen 2.6.32-431.1.2.0.1.el6.x86_64 #1 SMP Fri Dec 13 13:06:13 UTC 2013 x86_64 x86_64 x86_64 GNU/Linux With the help of

I'm drawing a fitted normal distribution over a histogram. The use case is trivial (fitting normal distributions on densities) but I want to extend it to other fitting scenarios. What has stumped me so far is how to take the list that is returned by nls() and use it for curve(). I realize that I can easily do all of this with a few intermediate steps for any specific case. But I had expected