similar to: Non-linear curve fitting (nls): starting point and quality of fit

Displaying 20 results from an estimated 10000 matches similar to: "Non-linear curve fitting (nls): starting point and quality of fit"

2012 Jul 11
2
nls problem: singular gradient
Why fails nls with "singular gradient" here? I post a minimal example on the bottom and would be very happy if someone could help me. Kind regards, ########### # define some constants smallc <- 0.0001 t <- seq(0,1,0.001) t0 <- 0.5 tau1 <- 0.02 # generate yy(t) yy <- 1/2 * ( 1- tanh((t - t0)/smallc) * exp(-t / tau1) ) + rnorm(length(t))*0.01 # show the curve
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
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
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 <-
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
2009 Jul 12
2
Nonlinear Least Squares nls() programming help
Hi, I am trying to use the nls() function to closely approximate a vector of values, colC and I'm running into trouble. I am not sure how if I am asking the program to do what I think its doing, because the same minimization in Excel's Solver does not run into problems. If anyone can tell me what is going wrong, and why I'm getting a singular convergence(7) error, please tell me. I
2013 Mar 14
2
question about nls
Hi,all: I met a problem of nls. My data: x y 60 0.8 80 6.5 100 20.5 120 45.9 I want to fit exp curve of data. My code: > nls(y ~ exp(a + b*x)+d,start=list(a=0,b=0,d=1)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates I can't find out the reason for the error. Any suggesions are welcome. Many thanks. [[alternative HTML
2013 Jul 09
3
fitting log function: errors using nls and nlxb
Hi- I am trying to fit a log function to my data, with the ultimate goal of finding the second derivative of the function. However, I am stalled on the first step of fitting a curve. When I use the following code: FG2.model<-(nls((CO2~log(a*Time)+b), start=setNames(coef(lm(CO2 ~ log(Time), data=FG2)), c("a", "b")),data=FG2)) I get the following error: Error in
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]]
2013 Oct 03
2
SSweibull() : problems with step factor and singular gradient
SSweibull() :  problems with step factor and singular gradient Hello I am working with growth data of ~4000 tree seedlings and trying to fit non-linear Weibull growth curves through the data of each plant. Since they differ a lot in their shape, initial parameters cannot be set for all plants. That’s why I use the self-starting function SSweibull(). However, I often got two error messages:
2012 Apr 10
3
nls function
Hi, I've got the following data: x<-c(1,3,5,7) y<-c(37.98,11.68,3.65,3.93) penetrationks28<-dataframe(x=x,y=y) now I need to fit a non linear function so I did: fit <- nls(y ~ I(a+b*exp(1)^(-c * x)), data = penetrationks28, start = list(a=0,b = 1,c=1), trace = T) The error message I get is: Error in nls(y ~ I(a + b * exp(1)^(-c * x)), data = penetrationks28, start = list(a =
2012 Apr 17
3
error using nls with logistic derivative
Hi I?m trying to fit a nonlinear model to a derivative of the logistic function y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function with nls) The derivative calculated with D function is: > logis<- expression(a/(1+exp((b-x)/c))) > D(logis, "x") a * (exp((b - x)/c) * (1/c))/(1 + exp((b - x)/c))^2 So I enter this expression in the nls function:
2009 Aug 25
1
Help with nls and error messages singular gradient
Hi All, I'm trying to run nls on the data from the study by Marske (Biochemical Oxygen Demand Interpretation Using Sum of Squares Surface. M.S. thesis, University of Wisconsin, Madison, 1967) and was reported in Bates and Watts (1988). Data is as follows, (stored as mydata) time bod 1 1 0.47 2 2 0.74 3 3 1.17 4 4 1.42 5 5 1.60 6 7 1.84 7 9 2.19 8 11 2.17 I then
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=
2005 Jun 21
2
nls(): Levenberg-Marquardt, Gauss-Newton, plinear - PI curve fitting
Hello, i have a problem with the function nls(). This are my data in "k": V1 V2 [1,] 0 0.367 [2,] 85 0.296 [3,] 122 0.260 [4,] 192 0.244 [5,] 275 0.175 [6,] 421 0.140 [7,] 603 0.093 [8,] 831 0.068 [9,] 1140 0.043 With the nls()-function i want to fit following formula whereas a,b, and c are variables: y~1/(a*x^2+b*x+c) With the standardalgorithm
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),
2012 May 17
3
nls and if statements
Hi All, I have a situation where I want an 'if' variable to be parameterized. It's entirely possible that the way I'm trying to do this is wrong, especially given the error message I get that indicates I can't do this using an 'if' statement. Essentially, I have data where I think a relationship enters when a variable (here Pwd) is below some value (z). I don't
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 =
2012 Jun 05
1
nls: how do you know if the model is significant?
Hi all, I'm struggling with nls. How do you know if your model is significant? For a lm, you get a p-value, but you don't get it for a nls. Is there a way to calculate it? For a lm I use this: a<-summary(lm(model ~obs)) f.stat<-a$fstatistic p.value<-1-pf(f.stat["value"],f.stat["numdf"],f.stat["dendf"]) Is there something similar for a nls?
2017 Jun 21
1
fitting cosine curve
Using a more stable nonlinear modeling tool will also help, but key is to get the periodicity right. y=c(16.82, 16.72, 16.63, 16.47, 16.84, 16.25, 16.15, 16.83, 17.41, 17.67, 17.62, 17.81, 17.91, 17.85, 17.70, 17.67, 17.45, 17.58, 16.99, 17.10) t=c(7, 37, 58, 79, 96, 110, 114, 127, 146, 156, 161, 169, 176, 182, 190, 197, 209, 218, 232, 240) lidata <- data.frame(y=y, t=t) #I use the