similar to: error using boxcox.nls during non linear estimation

Hi all I?m trying to get confidence intervals for parameters from nls modeling. I fitted a nls model to the following variables: > x [1] 2 1 1 5 4 6 13 11 13 101 101 101 > y [1] 1.281055090 1.563609934 0.001570796 2.291579783 0.841891853 [6] 6.553951324 14.243274230 14.519899320 15.066473610 21.728809880 [11] 18.553054450 23.722637370 The model fitted was:

Hi all I have fitted a model usinf nls function to these data: > x [1] 1 0 0 4 3 5 12 10 12 100 100 100 > y [1] 1.281055090 1.563609934 0.001570796 2.291579783 0.841891853 [6] 6.553951324 14.243274230 14.519899320 15.066473610 21.728809880 [11] 18.553054450 23.722637370 The model fitted is: modellogis<-nls(y~SSlogis(x,a,b,c)) It runs OK. Then I calculate

I'm using nls to fit a variety of different models. Here I use SSgompertz as an example. I want the ability to fix one (or more) of the coefficients that would normally be optimised (e.g. fix b3=0.8). Examples; based on and using data from example(SSgompertz) #--------------------- # vanilla call to nls, no coefficients fixed, works fine nls(density ~ SSgompertz(log(conc), Asym, b2, b3),

I'm trying to fit a Gompertz sigmoid as follows: x <- c(15, 16, 17, 18, 19) # arbitrary example data here; y <- c(0.1, 1.8, 2.2, 2.6, 2.9) # actual data is similar gm <- nls(y ~ a+b*exp(-exp(-c*(x-d))), start=c(a=?, b=?, c=?, d=?)) I have been unable to properly set the starting value '?'s. All of my guesses yield either a "singular gradient" error if they

Hi, I am using the nls function and it stops because the number of iterations exceeded 50, but i used the nls.control argument to allow for 500 iterations. Do you have any idea why it's not working? fm1 <- nls(npe ~ SSgompertz(npo, Asym, b2, b3), data=f,control=nls.control(maxiter=500)) Error in nls(y ~ exp(-b2 * b3^x), data = xy, algorithm = "plinear", start = c(b2 =

Dear all i want to fit the self start model in nls. i have two question. i have a function, (asfr ~ I(((a*b)/c))+ ((c/age)^3/2)+ exp((-b^2)*(c/age)+(age/c)-2) i am wondering how to build the selfstart model. there is lost of example, (i.e. SSgompertz, SSmicman, SSweibull, etc). my question is, how to derive the function of parameters. and also which model to use for get the initials values. In the

Hi, I am using the nls function and it stops because the number of iterations exceeded 50, but i used the nls.control argument to allow for 500 iterations. Do you have any idea why it's not working? fm1 <- nls(npe ~ SSgompertz(npo, Asym, b2, b3), data=f,control=nls.control(maxiter=500)) Thanks for your help, Cheers, Karine.

Hi there are some predefined self-start functions, like SSmicmen, SSbiexp, SSasymp, SSasympOff, SSasympOrig, SSgompertz, SSflp, SSlogis, SSweibull, Quadratic, Qubic, SSexp (nlrwr) Btw, do you know graphic examples for this functions? The SSexpDecay (exponential decay) for y = (y0 - plateau)*exp(-k*x) + plateau from

Hi everyone, I'm working on a script trying to use foreach %dopar% but without success, so I manage to run the code with foreach %do% and looks like this: The code is part of a MCMC model for projects valuation, returning the most important results (VPN, TIR, EVA, etc.) of the simulation. foreach (simx = NsimT, .combine=cbind, .inorder=FALSE, .verbose=TRUE) %do% { MCPVMPA = MCVAMPA[simx]

Dear r-helpers, Prior to analysis of variance, I ran the Boxcox function (MASS library) to find the best power transformation of my data. However, reading the Boxcox help file, I cannot figure out if this function (through its associated log-likelihood function) corrects for * normality only * or if it also induces * homogeneity of variances *. I found in Biometry (Sokal and Rohlf, p. 419)

Hi, Thakns all for your help I am doing the next in my dataframe tabla, column pend1, because the Lilliefors (Kolmogorov-Smirnov) test give me a pvalue < alfa. (data no normal distribution). I need do a transformation with box-cox or other: > bc <- boxcox.fit(tabla$pend1) R send to me: Error in boxcox.fit(tabla$pend1) : Transformation requires positive data The summary for my data

Hi, Hope this does not sound too ignorant . I am trying to detrend and transform variables to achieve normality and stationarity (for time series use, namely spectral analysis). I am using the boxcox transformations. As my dataset contains zeros, I found I need to add a constant to it in order to run "boxcox". I have ran tests adding several types of constants, from .0001

Hello, I am implementing the BoxCox transformation for multiple regression using the function boxcox() from library MASS (this seems to be the best one). To avoid negative values in the variable to be transformed however, one must specify a shift parameter. However different choices for the value of the shift parameter result in different estimates for the exponent (lambda) and in different

Hi, I'm doing some work with linear models, and I've scaled my data using the scale(dataset) function. This was great at removing the skew, but I now can't perform the Box Cox transformation on the data set (using the boxcox(dataset) function), as the scaling has returned negative values. So my question is: how can I get the scale function to return a positive set of data (so I can

I've got a vector of data (hours to drive from a to b) y. After a qqplot I know, that they don't fit the normal probability. I would like to transform these data with the boxcox transformation (MASS), that they fit the model. When I try ybx<-boxcox(y~1,0) qqnorm(ybx) the plot is different from library (TeachingDemos) ybct<-bct(y,0) // qqnorm(ybct) How can I transform

Hi, I am new to R. I need help with regards to box cox transformation. I have phenotypic data for e.g. plant height. data is non-normal. Skewness is 0.34. Could you please help me? Regards, Yogi -- View this message in context: http://r.789695.n4.nabble.com/boxcox-transformations-tp4682077.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]]

Hmm, this looks like a buglet/infelicity in update.lm rather than MASS::boxcox per se. Moving to R-devel. I think the story is that update.lm eventually does eval(call, parent.frame()) where the call is extracted from the lm object, but call$formula is unevaluated, and does not contain environment information like formula(obj) would do. Then when the call is evaluated and parent.frame()

Help page for boxcox function in MASS library says that the transformation is y^lambda, which is different from the Y' = log(Y) if lambda = 0 , Y' = ((Y ^ lambda) - 1)/lambda otherwise I'm used to. Is this just a help page typo ? Thanks. ---------------------------------------------------------------- This message was sent using IMP, the Internet Messaging Program.

Hi, I can't seem to find a function in R that will reverse a BoxCox transformation. Can somebody help me locate one please? Thanks in advance. Best wishes, Des [[alternative HTML version deleted]]

Hi!! Can anyone help me, i have problems to converge the following data with 5 nonlinears models that i evaluated. Firtly, i send my data (totalsinatipicos) that i just try to fit with the nonlinear models. Next, i have the following script where i called the data as totalsinatipicos. I made selfstarting each nonlinear model. ###Library library(NRAIA) ###Data d<-totalsinatipicos