Displaying 20 results from an estimated 1000 matches similar to: "Error messages using nonlinear regression function (nls)"
2017 Oct 20
1
Error messages using nonlinear regression function (nls)
Hi
Keep your messages in the list, you increase your chance to get some answer.
I changed your data to groupedData object (see below), but I did not find any problem in it.
plot(wlg)
gives reasonable picture and I am not such expert to see any problem with data. Seems to me, that something has to be wrong with nlsList function.
> wheat.list <- nlsList(Prop ~ SSlogis(end,Asym, xmid,
2003 Aug 14
1
gnls - Step halving....
Hi all,
I'm working with a dataset from 10 treatments, each
treatment with 30 subjects, each subject measured 5
times. The plot of the dataset suggests that a
3-parameter logistic could be a reasonable function to
describe the data. When I try to fit the model using
gnls I got the message 'Step halving factor reduced
below minimum in NLS step'. I´m using as the initial
values of the
2017 Oct 20
1
Error messages using nonlinear regression function (nls)
Thank you Martin.
If I understand correctly, OP could do
wheat.list <- nlsList(Prop ~ SSfpl(end, A, B, xmid, scal), data=wlg)
or add some small value to all zeroes
wlg$prop < -wlg$Prop+1e-7
wheat.list <- nlsList(prop ~ SSlogis(end,Asym, xmid, scal), data=wlg)
which gives fairly reasonable results.
plot(augPred(wheat.list))
Am I correct?
Cheers
Petr
> -----Original Message-----
2017 Oct 20
0
Error messages using nonlinear regression function (nls)
>>>>> PIKAL Petr <petr.pikal at precheza.cz>
>>>>> on Fri, 20 Oct 2017 06:33:36 +0000 writes:
> Hi
> Keep your messages in the list, you increase your chance to get some answer.
> I changed your data to groupedData object (see below), but I did not find any problem in it.
> plot(wlg)
> gives reasonable picture and I am
2011 Apr 10
1
survival object
Hi All,
I am trying to do a survivorship analysis with library(survival)from a data
set that looks like this:
I followed a bunch of naturally germinated seedlings of an annual plant from
germination to death (none made it to reproduce, and died in a period of ~60
days after germination.)
I also know the size of the seed of every individual censused. So I am
trying to analyze seedling survival as
2009 Mar 06
1
fitting a gompertz model through the origin using nls
Dear all!
I tried to fit Gompertz growth models to describe cummulative germination rates
using nls. I used the following code:
germ.model<-nls(percent.germ~a*exp(-b*exp(-k*day)),data=tab,start=list(a=100,b=10,k=0.5))
My problem is that I want that the fitted model goes through the origin, since
germination cannot start before the experiment was started, and y-max should be
100.
Does anyone
2011 Feb 06
1
anova() interpretation and error message
Hi there,
I have a data frame as listed below:
> Ca.P.Biomass.A
P Biomass
1 334.5567 0.2870000
2 737.5400 0.5713333
3 894.5300 0.6393333
4 782.3800 0.5836667
5 857.5900 0.6003333
6 829.2700 0.5883333
I have fit the data using logistic, Michaelis?Menten, and linear model,
they all give significance.
> fm1 <- nls(Biomass~SSlogis(P, phi1, phi2, phi3), data=Ca.P.Biomass.A)
2008 Apr 04
1
Problems with Unit Root testing using ur.df function
Hi All,
I'm new to R and am trying to run a unit root test on the vector "y" (a time
series of inflation (i.e. changes in the Consumer Price Index quarter on
quarter)).
I've run the Augmented-Dickey-Fuller Test below (R's URCA package). It gives
me an error that it cannot find the function ur.df unless I comment out the
third last line of code (see below).
I try to call
2010 Apr 14
0
ur.df ADF Unit Root Test: what is the meaning of phi1 and phi2 test statistic?
Hello,
I am using the ur.df function from the {arca} package to run the augmented
Dickey-Fuller unit root test on several time series. However; I do not
understand the econometric interpretation of the the "phi1" and "phi2"
test-statisitc which are output if you choose a "trend" or "drift" model. I
looked at the source code for the function but I do not
2012 Dec 01
1
reading json tables
I'm trying to read two data sets in json format from a single .js file.
I've tried fromJSON()
in both RJSONIOIO and RJSON packages, but they require that the lines be
pre-parsed somehow in ways I don't understand. Can someone help?
> wheat <- readLines("http://mbostock.github.com/protovis/ex/wheat.js")
> str(wheat)
chr [1:70] "var wheat = [" "
2004 Nov 05
1
Problems running a 4-parameter Weibull function with nls
Hi,
I am rather new to R, but both myself and another much more experience user cannot figure this out. I have a collegue who has a 800+ nonlinear regressions to run for seed germination (different species, treatments, etc.) over time. I created a looping structure to extract the parameters from each regression; I will then use the parameters themselves for further analysis. I would like to
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 =
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
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
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
Dear Bert,
Thank you for your extremely valuable feedback. Now, I just want to
understand why the signs for those starting values, given the following:
> #Fiting intermediate model to get starting values
> intermediatemod <- lm(log(y - .37) ~ x, data=mod14data2_random)
> summary(intermediatemod)
Call:
lm(formula = log(y - 0.37) ~ x, data = mod14data2_random)
Residuals:
Min
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
Basic algebra and exponentials/logs. I leave those details to you or
another HelpeR.
-- Bert
On Sun, Aug 20, 2023 at 12:17?PM Paul Bernal <paulbernal07 at gmail.com> wrote:
> Dear Bert,
>
> Thank you for your extremely valuable feedback. Now, I just want to
> understand why the signs for those starting values, given the following:
> > #Fiting intermediate model to get
2004 May 15
2
questions about optim
Hi,
I am trying to do parameter estimation with optim, but I can't get it to
work quite right-- I have an equation X = Y where X is a gaussian, Y is a
multinomial distribution, and I am trying to estimate the probabilities of
Y( the mean and sd of X are known ), Theta1, Theta2, Theta3, and Theta4; I
do not know how I can specify the constraint that Theta1 + Theta2 + Theta3 +
Theta4 = 1 in
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
2008 Apr 22
2
optimization setup
Hi, here comes my problem, say I have the following functions (example case)
#------------------------------------------------------------
function1 <- function (x, theta)
{a <- theta[1] ( 1 - exp(-theta[2]) ) * theta[3] )
b <- x * theta[1] / theta[3]^2
return( list( a = a, b = b )) }
#-----------------------------------------------------------
function2<-function (x, theta)
{P
2007 Aug 23
1
degrees of freedom question
R2.3, WinXP
Dear all,
I am using the following functions:
f1 = Phi1+(Phi2-Phi1)/(1+exp((log(Phi3)-log(x))/exp(log(Phi4)))
f2 = Phi1+(Phi2-Phi1)/(1+exp((log(Phi3)-log(r)-log(x))/exp(log(Phi4)))
subject to the residual weighting
Var(e[i]) = sigma^2 * abs( E(y) )^(2*Delta)
Here is my question, in steps:
1. Function f1 is separately fitted to two different datasets
corresponding to