similar to: Error messages using nonlinear regression function (nls)

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