similar to: The nls2 function automatically prints the object!

Displaying 20 results from an estimated 4000 matches similar to: "The nls2 function automatically prints the object!"

2010 Sep 02
1
How using the weights argument in nls2?
Good morning gentlemen! How using a weighted model in nls2? Values with the nls are logical since values with nls2 are not. I believe that this discrepancy is due to I did not include the weights argument in nls2. Here's an example: MOISTURE <- c(28.41640, 28.47340, 29.05821, 28.52201, 30.92055, 31.07901, 31.35840, 31.69617, 32.07168, 31.87296, 31.35525, 32.66118, 33.23385,
2010 May 11
1
nls() and nls2() behavior?
first, apologies for so many posts yesterday and today. I am wrestling with nls() and nls2(). I have tried to whittle it down to a simple example that still has my problem, yet can be cut-and-pasted into R. here it is: library(nls2) options(digits=12); y= c(0.4334,0.3200,0.5848,0.6214,0.3890,0.5233,0.4753,0.2104,0.3240,0.2827,0.3847,0.5571,0.5432,0.1326,0.3481) x=
2004 Jul 22
1
package nls2 for windows
Dear Madam or sir, Does anyone know if there is a pre-compiled version of package nls2 for windows, please? Thank you. Souleymane
2010 Jan 18
2
Problem extracting from mer objects
I am having a problem extracting from "mer" objects.    I have constructed my problem using existing datasets.   Using the following commands:   require(lme4) fm1 <- lmer(Yield ~ 1 + (1 | Batch), Dyestuff) fixef(fm1) I get the following error message: "Error in UseMethod("fixef") : no applicable method for "fixef""   I know that "fixef" is in
2013 Jan 02
1
Need help with self-defined function to perform nonlinear regression and get prediction interval
Dear All, I was trying to call a self-defined function that performs nonlinear regression and gets the corresponding prediction upper limit using nls2 package. However, weird thing happened. When I called the function in the main program, an error message "fitted(nlsmodel): object 'nlsmodel' not found" came up. But when I directly ran the codes inside the function, no error came
2010 May 18
1
Using the zero-inflated binomial in experimental designs
I'm trying to use the inflated binomial distribution of zeros (since 75% of the values are zeros) in a randomized block experiment with four quantitative treatments (0, 0.5, 1, 1.5), but I'm finding it difficult, since the examples available in VGAM packages like for example, leave us unsure of how it should be the data.frame for such analysis. Unfortunately the function glm does not have
2009 Feb 10
3
summary of a list
Hello, I'm using the following for loop to find regression curves using a list of functions (formList), a list of starting values (startList), uppervalues (upperList) and lower values (lowerList). A sample of the list of function I use in the loop is the following: FormList <- list(PTG.P ~ fz1(Portata, a, b), PTG.P ~ fz2(Portata, a, b), PTG.P ~ fz3(Portata,a, b, d, e), PTG.P ~
2009 Jun 11
1
Error in 1:p : NA/NaN argument when running model comparisons
Hi there, I am trying to compare nonlinear least squares regression with AIC and anova. The simplest model is one nonlinear curve, and in the more complex model I have a categorical variable (producing parameter estimates for four curves). Both models run fine, but when I try to produce an AIC value for the second model I get the error: > AIC(pow.nls1) [1] 114408.3 > AIC(pow.nls2) Error in
2010 Apr 30
2
Curve Fitting
I am having troubles in fitting functions of the form y~a*x^b+c to data, for example x<-c(0.1,0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23) y<-c(8.09,9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86) I tried for example with nls, which did only work with really good initial guessed values. Any suggestion, what I should use? Thanks a lot Thomas [[alternative HTML version deleted]]
2012 Jun 04
2
Non-linear curve fitting (nls): starting point and quality of fit
Hi all, Like a lot of people I noticed that I get different results when I use nls in R compared to the exponential fit in excel. A bit annoying because often the R^2 is higher in excel but when I'm reading the different topics on this forum I kind of understand that using R is better than excel? (I don't really understand how the difference occurs, but I understand that there is a
2008 May 23
3
nls diagnostics?
Hi, All: What tools exist for diagnosing singular gradient problems with 'nls'? Consider the following toy example: DF1 <- data.frame(y=1:9, one=rep(1,9)) nlsToyProblem <- nls(y~(a+2*b)*one, DF1, start=list(a=1, b=1), control=nls.control(warnOnly=TRUE)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial
2008 May 09
2
Regarding anova result
  Hi, I fitted tree growth data with Chapman-Richards growth function using nls. summary(fit.nls) Formula: Parameters: Estimate Std. Error t value Pr Signif. codes: 0 ''***'' 0.001 ''**'' 0.01 ''*'' 0.05 ''.'' 0.1 '' '' 1 Residual standard error: 1.879 on 713 degrees of freedom Algorithm
2005 Mar 08
4
Non-linear minimization
hello, I have got some trouble with R functions nlm(), nls() or optim() : I would like to fit 3 parameters which must stay in a precise interval. For exemple with nlm() : fn<-function(p) sum((dN-estdata(p[1],p[2],p[3]))^2) out<-nlm(fn, p=c(4, 17, 5), hessian=TRUE,print.level=2) with estdata() a function which returns value to fit with dN (observed data vactor) My problem is that only
2009 Aug 25
3
Covariates in NLS (Multiple nonlinear regression)
Dear R-users, I am trying to create a model using the NLS function, such that: Y = f(X) + q + e Where f is a nonlinear (Weibull: a*(1-exp(-b*X^c)) function of X and q is a covariate (continous variable) and e is an error term. I know that you can create multiple nonlinear regressions where x is polynomial for example, but is it possible to do this kind of thing when x is a function with unknown
2011 Dec 11
1
nls start values
I'm using nls to fit periodic gene-expression data to sine waves. I need to set the upper and lower boundaries, because I do not want any negative phase and amplitude solutions. This means that I have to use the "port" algorithm. The problem is, that depending on what start value I choose for phase, the fit works for some cases, but not for others. In the example below, the fit works
2010 Jul 06
1
nls + quasi-poisson distribution
Hello R-helpers, I would like to fit a non-linear function to data (Discrete X axis, over-dispersed Poisson values on the Y axis). I found the functions gnlr in the gnlm package from Jim Lindsey: this can handle nonlinear regression equations for the parameters of Poisson and negative binomial distributions, among others. I also found the function nls2 in the software package
2008 Apr 10
1
(no subject)
Subject: nls, step factor 0.000488281 reduced below 'minFactor' of 0.000976563 Hi there, I'm trying to conduct nls regression using roughly the below code: nls1 <- nls(y ~ a*(1-exp(-b*x^c)), start=list(a=a1,b=b1,c=c1)) I checked my start values by plotting the relationship etc. but I kept getting an error message saying maximum iterations exceeded. I have tried changing these
2012 May 16
2
confidence intervals for nls or nls2 model
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
2011 Oct 21
0
nls making R "not responding"
Here is the code I am running: library(nls2) modeltest<- function(A,mu,l,b,thour){ out<-vector(length=length(thour)) for (i in 1:length(thour)) { out[i]<-b+A/(1+exp(4*mu/A*(l-thour[i])+2)) } return(out) } A=1.3 mu=.22 l = 15 b = .07 thour = 1:25 Yvals<-modeltest(A,mu,l,b,thour)-.125+runif(25)/4 st2 <- expand.grid(A = seq(0.1, 1.6,.5), mu = seq(0.01, .41,.1), l=1, b =seq(0,.6,.3))
2009 Mar 12
3
avoiding termination of nls given convergence failure
Hello. I have a script in which I repeatedly fit a nonlinear regression to a series of data sets using nls and the port algorithm from within a loop. The general structure of the loop is: for(i in 1:n){ … extract relevant vectors of dependent and independent variables … … estimate starting values for Amax and Q.LCP…