Displaying 20 results from an estimated 1000 matches similar to: "How can I extract information from list which class is nls"
2009 Nov 09
1
Parameter info from nls object
Hi!
When checking validity of a model for a large number
of experimental data I thought it to be interesting
to check the information provided by
the summary method programmatically.
Still I could not find out which method to
use to get to those data.
Example (not my real world data, but to show the point):
[BEGIN]
> DNase1 <- subset(DNase, Run == 1)
> fm1DNase1 <- nls(density ~
2017 Apr 01
6
Intervalos de confianza de la varianza de los residuos en un modelo no lineal.-
Hola amigos,
Supongamos que se quiere ejecutar un modelo no lineal con nls. Pensemos en
el ejemplo de la ayuda:
DNase1 <- subset(DNase, Run == 1)
fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1)
summary(fm1DNase1)
Aquí se está modelando la densidad óptica de un ensayo relacionada de forma
no lineal (logística) con (el logaritmo) de la concentración de una
proteína.
2003 May 21
2
Access Object's Objects HELP
Dear WizaRds,
A run of nls produces the following concise summary:
> summary(cs.wt)
Formula: 0 ~ wt.MM(conc, time, A1, a1, A2, a2)
Parameters:
Estimate Std. Error t value Pr(>|t|)
A1 4.814e+02 2.240e+01 21.495 0.0296 *
a1 7.401e-01 7.435e-02 9.956 0.0637 .
A2 1.613e+02 1.738e+01 9.280 0.0683 .
a2 1.770e-02 7.324e-03 2.417 0.2497
2000 Oct 14
2
Access to calculations in nls
Hi,
I would like to be able to access the calculated results from the nls package.
Using the example in R, fm3DNase1 we can reurn certain parts of the
calculations:
> coef(fm3DNase1)
Asym xmid scal
2.345179 1.483089 1.041454
> resid(fm3DNase1)
[1] -0.0136806237 -0.0126806237 0.0089488569 0.0119488569 -0.0025803222
[6] 0.0064196778 0.0026723396 -0.0003276604
2006 Apr 18
1
Nonlinear Regression model: Diagnostics
Hi,
I am trying to run the following nonlinear regression model.
> nreg <- nls(y ~ exp(-b*x), data = mydf, start = list(b = 0), alg = "default", trace = TRUE)
OUTPUT:
24619327 : 0
24593178 : 0.0001166910
24555219 : 0.0005019005
24521810 : 0.001341571
24500774 : 0.002705402
24490713 : 0.004401078
24486658 : 0.00607728
24485115 : 0.007484372
2004 Jul 16
1
Does AIC() applied to a nls() object use the correct number of estimated parameters?
I'm wondering whether AIC scores extracted from nls() objects using
AIC() are based on the correct number of estimated parameters.
Using the example under nls() documentation:
> data( DNase )
> DNase1 <- DNase[ DNase$Run == 1, ]
> ## using a selfStart model
> fm1DNase1 <- nls( density ~ SSlogis( log(conc), Asym, xmid, scal ),
DNase1 )
Using AIC() function:
>
2011 Nov 17
3
Obtaining a derivative of nls() SSlogis function
Hello, I am wondering if someone can help me. I have the following function
that I derived using nls() SSlogis. I would like to find its derivative. I
thought I had done this using deriv(), but for some reason this isn't
working out for me.
Here is the function:
asym <- 84.951
xmid <- 66.90742
scal <- -6.3
x.seq <- seq(1, 153,, 153)
nls.fn <- asym/((1+exp((xmid-x.seq)/scal)))
2008 Jan 04
3
nls (with SSlogis model and upper limit) never returns (PR#10544)
Full_Name: Hendrik Weisser
Version: 2.6.1
OS: Linux
Submission from: (NULL) (139.19.102.218)
The following computation never finishes and locks R up:
> values <- list(x=10:30, y=c(23.85, 28.805, 28.195, 26.23, 25.005, 20.475,
17.33, 14.97, 11.765, 8.857, 5.3725, 5.16, 4.2105, 2.929, 2.174, 1.25, 1.0255,
0.612, 0.556, 0.4025, 0.173))
> y.max <- max(values$y)
> model <- nls(y ~
2001 Oct 07
1
Bug in Deriv? (PR#1119)
deriv seems to have problems with a minus-sign before a bracket.
Below are four examples of the same function, the top one
is wrong, all others are correct (hopefully).
Rest of expression not shown, it is the same for all versions.
_
platform i386-pc-mingw32
arch x86
os Win32
system x86, Win32
status
major 1
minor 3.0
year 2001
month 06
day 22
language R
2011 Aug 09
1
nls, how to determine function?
Hi R help,
I am trying to determine how nls() generates a function based on the
self-starting SSlogis and what the formula for the function would be.
I've scoured the help site, and other literature to try and figure
this out but I still am unsure if I am correct in what I am coming up
with.
**************************************************************************
dat <-
2006 May 17
1
nlme model specification
Hi folks,
I am tearing my hair out on this one.
I am using an example from Pinheiro and Bates.
### this works
data(Orange)
mod.lis <- nlsList(circumference ~ SSlogis(age, Asymp, xmid, scal),
data=Orange )
### This works
mod <- nlme(circumference ~ SSlogis(age, Asymp, xmid, scal),
data=Orange,
fixed = Asymp + xmid + scal ~ 1,
start =
2008 Sep 10
3
writing simple function through script
Hi all,
I try to write a simple function in a script. The script is as follows
yo<-function(Xdata)
{
n<-length(Xdata[,1])
Lgm<-nls(formula=LgmFormula,
data=Xdata,
start=list(a=1500,b=0.1),weights=Xdata$Qe)
return(Lgm)
}
After the execution of the script, when I call the function yo on data
called NC60.DATA I get an error.
#yo(NC60.DATA)
Erreur dans eval(expr, envir, enclos)
2004 Feb 20
1
nlme and multiple comparisons
This is only partly a question about R, as I am not quite sure about the
underlying statistical theory either.
I have fitted a non-linear mixed-effects model with nlme. In the fixed
part of the model I have a factor with three levels as explanatory
variable. I would like to use Tukey HSD or a similar test to test for
differences between these three levels.
I have two grouping factors:
2004 Oct 01
4
gnls or nlme : how to obtain confidence intervals of fitted values
Hi
I use gnls to fit non linear models of the form y = alpha * x**beta
(alpha and beta being linear functions of a 2nd regressor z i.e.
alpha=a1+a2*z and beta=b1+b2*z) with variance function
varPower(fitted(.)) which sounds correct for the data set I use.
My purpose is to use the fitted models for predictions with other sets
of regressors x, z than those used in fitting. I therefore need to
2006 Sep 11
4
syntax of nlme
Hello,
How do I specify the formula and random effects without a startup object
? I thought it would be a mixture of nls and lme.
after trying very hard, I ask for help on using nlme.
Can someone hint me to some examples?
I constructed a try using the example from nls:
#variables are density, conc and Run
#all works fine with nls
DNase1 <- subset(DNase, Run == 1 )
fm2DNase1 <- nls(
2009 May 04
1
how to change nlme() contrast parametrization?
How to set the nlme() function to return the answer without the intercept parametrization?
#=========================================================================================
library(nlme)
Soybean[1:3, ]
(fm1Soy.lis <- nlsList(weight ~ SSlogis(Time, Asym, xmid, scal),
data = Soybean))
(fm1Soy.nlme <- nlme(fm1Soy.lis))
fm2Soy.nlme <- update(fm1Soy.nlme,
2001 Aug 08
1
NLME augPred error
Could someone explain the meaming of this error message from augPred:
> augPred(area3.pen.nlme, primary=~day)
Error in predict.nlme(object, value[1:(nrow(value)/nL), , drop =
FALSE], :
Levels 1,2,3 not allowed for block
>
predict.nlme(area3.pen.nlme) does not produce an error.
area3.pen.nlme was created with:
> area3.pen.nlme <- nlme(area ~ SSlogis(day, Asym, xmid, scal),
2009 Oct 02
1
nls not accepting control parameter?
Hi
I want to change a control parameter for an nls () as I am getting an error
message "step factor 0.000488281 reduced below 'minFactor' of 0.000976562".
Despite all tries, it seems that the control parameter of the nls, does not
seem to get handed down to the function itself, or the error message is
using a different one.
Below system info and an example highlighting the
2008 Sep 27
1
seg.fault from nlme::gnls() {was "[R-sig-ME] GNLS Crash"}
>>>>> "VW" == Viechtbauer Wolfgang (STAT) <Wolfgang.Viechtbauer at STAT.unimaas.nl>
>>>>> on Fri, 26 Sep 2008 18:00:19 +0200 writes:
VW> Hi all, I'm trying to fit a marginal (longitudinal)
VW> model with an exponential serial correlation function to
VW> the Orange tree data set. However, R crashes frequently
VW>
2007 Jun 07
2
Nonlinear Regression
Hello
I followed the example in page 59, chapter 11 of the 'Introduction to R'
manual. I entered my own x,y data. I used the least squares. My function has
5 parameters: p[1], p[2], p[3], p[4], p[5]. I plotted the x-y data. Then I
used lines(spline(xfit,yfit)) to overlay best curves on the data while
changing the parameters. My question is how do I calculate the residual sum
of squares.