Displaying 20 results from an estimated 4000 matches similar to: "Covariates in NLS (Multiple nonlinear regression)"
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
2011 Apr 27
3
MASS fitdistr with plyr or data.table?
I am trying to extract the shape and scale parameters of a wind speed
distribution for different sites. I can do this in a clunky way, but
I was hoping to find a way using data.table or plyr. However, when I
try I am met with the following:
set.seed(144)
weib.dist<-rweibull(10000,shape=3,scale=8)
weib.test<-data.table(cbind(1:10,weib.dist))
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
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
2009 Jun 12
2
Comparing model fits for NLME when models are not nested
Hi there,
I am looking to compare nonlinear mixed effects models that have different nonlinear functions (different types of growth curve)embedded. Most of the literature I can find focuses on comparing nested models with likelihood ratios and AIC. Is there a way to compare model fits when models are not nested, i.e. when the nonlinear functions are not the same?
Many thanks in advance!
Lindsay
2010 Jan 13
1
Problem fitting a non-linear regression model with nls
Hi,
I'm trying to make a regression of the form :
formula <- y ~ Asym_inf + Asym_sup * ( (1 / (1 + (n1 * (exp( (tmid1-x)
/ scal1) )^(1/n1) ) ) ) - (1 / (1 + (n2 * (exp( (tmid2-x) / scal2)
)^(1/n2) ) ) ) )
which is a sum of the generalized logistic model proposed by richards.
with data such as these:
x <- c(88,113,128,143,157,172,184,198,210,226,240,249,263,284,302,340)
y <-
2005 Jun 09
2
Weibull survival modeling with covariate
I was wondering if someone familiar
with survival analysis can help me with
the following.
I would like to fit a Weibull curve,
that may be dependent on a covariate,
my dataframe "labdata" that has the
fields "cov", "time", and "censor". Do
I do the following?
wieb<-survreg(Surv(labdata$time,
labadata$censor)~labdata$cov,
2009 Dec 18
2
NLS-Weibull-ERROR
Hello
I was trying to estimate the weibull model using nls after putting OLS
values as the initial inputs to NLS.
I tried multiple times but still i m getting the same error of Error in
nlsModel(formula, mf, start, wts) :
singular gradient matrix at initial parameter estimates.
The Program is as below
> vel <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14)
> df <- data.frame(conc, vel)
>
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
2005 Feb 24
2
survreg with gamma distribution: re-post
Dear r-help subscribers,
A couple of weeks ago I sent the following message to the r-help mail
list. It hasn't generated any response, and I could really use some help
on this. Anyone able to help?
Thanks again,
Roger Dungan
>>
I am working on some survival analysis of some interval censored failure
time data in R. I have done similar analysis before using PROC LIFEREG
in SAS. In
2007 Feb 13
1
nls: "missing value or an infinity" (Error in numericDeriv) and "singular gradient matrix"Error in nlsModel
Hi,
I am a non-expert user of R. I am essaying the fit of two different functions to my data, but I receive two different error messages. I suppose I have two different problems here... But, of which nature? In the first instance I did try with some different starting values for the parameters, but without success.
If anyone could suggest a sensible way to proceed to solve these I would be
2009 May 11
1
Error in NLME (nonlinear mixed effects model)
Hi there,
I have been trying to fit an NLME to my data. My dataset has two category levels - one is a fixed effect (level1) and one is a random effect (level2), though so far I have only experimented with the highest level grouping (fixed, level1), with the following code:
mod1 <- nlme(H ~ a*(1-exp(-b*D^c)),
data=sizes,
fixed=a+b+c~factor(Loc),
start=c(a=75,b=0.05,c=0.7))
This returns the
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),
2023 Aug 20
3
Issues when trying to fit a nonlinear regression model
Dear friends,
This is the dataset I am currently working with:
>dput(mod14data2_random)
structure(list(index = c(14L, 27L, 37L, 33L, 34L, 16L, 7L, 1L,
39L, 36L, 40L, 19L, 28L, 38L, 32L), y = c(0.44, 0.4, 0.4, 0.4,
0.4, 0.43, 0.46, 0.49, 0.41, 0.41, 0.38, 0.42, 0.41, 0.4, 0.4
), x = c(16, 24, 32, 30, 30, 16, 12, 8, 36, 32, 36, 20, 26, 34,
28)), row.names = c(NA, -15L), class =
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
I got starting values as follows:
Noting that the minimum data value is .38, I fit the linear model log(y -
.37) ~ x to get intercept = -1.8 and slope = -.055. So I used .37,
exp(-1.8) and -.055 as the starting values for theta0, theta1, and theta2
in the nonlinear model. This converged without problems.
Cheers,
Bert
On Sun, Aug 20, 2023 at 10:15?AM Paul Bernal <paulbernal07 at
2023 Aug 20
2
Issues when trying to fit a nonlinear regression model
Dear Bert,
Thank you so much for your kind and valuable feedback. I tried finding the
starting values using the approach you mentioned, then did the following to
fit the nonlinear regression model:
nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x),
start =
list(theta1 = 0.37,
theta2 = exp(-1.8),
theta3 =
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=
2013 Nov 18
1
Ajuste curva
Hola,
quiero ajustar una curva sinoidal de la forma "f(x)=k/( 1+(c/log(x))^n)"
mediante la función 'nls' pero me da error el siguiente código:
>datos<-read.table(file="datos.csv", header=TRUE,sep=";",dec=",")
>library(nls)
>fit <- nls(y ~ k/(1+(c/log(x))^n), datos, start = list (k=100 , c
=5*10^(-6), n=1))
Error en
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
Oh, sorry; I changed signs in the model, fitting
theta0 + theta1*exp(theta2*x)
So for theta0 - theta1*exp(-theta2*x) use theta1= -.exp(-1.8) and theta2 =
+.055 as starting values.
-- Bert
On Sun, Aug 20, 2023 at 11:50?AM Paul Bernal <paulbernal07 at gmail.com> wrote:
> Dear Bert,
>
> Thank you so much for your kind and valuable feedback. I tried finding the
> starting