Displaying 20 results from an estimated 100 matches similar to: "nls() and nls2() behavior?"
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 Nov 24
1
The nls2 function automatically prints the object!
Good morning gentlemen!
When I use the function nls2, and store it in an object, that object is
automatically printed, without the summary or to draw the object. For
example.
model <- nls2 (...)
Number of iterations to convergence: ...
Achieved convergence tolerance: ...
Nonlinear regression model
model: ... ~ ...
Date: NULL
The B k
... ... ...
residual sum-of-squares: ...
Number
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
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
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
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 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 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
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
2007 Dec 05
1
confidence intervals for y predicted in non linearregression
Hi Thanks for your suggestion, I'm trying to install this package in Ubuntu
(7.10) but unsuccessfully. Also tried in MacOSX, and no success too.
_____
De: Ndoye Souleymane [mailto:ndoye_p@hotmail.com]
Enviado el: miércoles, 05 de diciembre de 2007 13:38
Para: bady@univ-lyon1.fr; Florencio González
CC: r-help@stat.math.ethz.ch
Asunto: RE: [R] confidence intervals for y predicted in
2016 Apr 28
0
Linear Regressions with constraint coefficients
The nls2 package can be used to get starting values.
On Thu, Apr 28, 2016 at 8:42 AM, Aleksandrovic, Aljosa (Pfaeffikon)
<Aljosa.Aleksandrovic at man.com> wrote:
> Hi Gabor,
>
> Thanks a lot for your help!
>
> I tried to implement your nonlinear least squares solver on my data set. I was just wondering about the argument start. If I would like to force all my coefficients to
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]]
2010 Mar 30
6
Error "singular gradient matrix at initial parameter estimates" in nls
I am using nls to fit a non linear function to some data.
The non linear function is:
y= 1- exp(-(k0+k1*p1+ .... + kn*pn))
I have chosen algorithm "port", with lower boundary is 0 for all of the
ki parameters, and I have tried many start values for the parameters ki
(including generating them at random).
If I fit the non linear function to the same data using an external
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
2011 Mar 02
1
power regression: which package?
Dear R users and R friends,
I have a little problem... I don't know anymore which package to use if
I want to perform a power regression analysis.
To be clear, I want to fit a regression model like this:
fit <- ....(y ~ a * x ^ b + c)
where a, b and c are coefficients of the model.
The R Site does not have the answer I want...
Thanks in advance and with kind regards,
David
2004 Aug 25
3
Samba as NT Domain Member via Winbind - After Upgrade users prompted for password for any shares
Hello,
Apparently, I did a stupid thing today. I used apt-get on my
Debian Woody system to upgrade my Samba packages from 3.0.2 to 3.0.6.
Since doing so, all my users are prompted for a password when trying to
access shares. Even just listing the IPC$, Windows XP systems prompt for
user name and password. Windows 98 machines prompt for password. None
are successful. I believe
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
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
2012 Nov 03
6
Parámetros iniciales para ajustes no lineales
Hola a todos
estoy aplicando la función polinómica de Hossfeld [1], y algunos otros modelos no lineales para tratar de ajustarlos a un grupo de datos forestales,
[1] Y= b*t*exp(c)/(t*exp(c)+a)
Al colocar la función en R con parámetros estimados, me devuelve los siguiente:
## model1 <- nls(ho ~ (b*edad*exp(c)/(edad*exp(c)+a)), data=nigra,
start=list(a=0.005,b=0.08,c=-0.00006),