Maximize result of bitwise AND Unix Exit Command How to get the last monday of every month Scroll a quarter (25%) of the screen up or down Why don't cameras offer I am sure the problem is with nls,because the external fitting algorithm perfectly fits it in less than asecond. If that is the case remove it from the start list and set it to the known value T <- ... Thanks in advance.Alternatively, what do you suggest I should do?

Is powered by WordPress using a bavotasan.com design. What nls() means when it gives you the message "singular gradient matrix" is that the Jacobian matrix J(x) is nearly rank deficient. (I don't know why they call it a "gradient" Thanks in advance.Alternatively, what do you suggest I should do? Many thanks indeed.

This could occur merely due to thechoice of convergence criteria (step size, choice of objective function,etc.). Example taken from here: x <- 0:140 y <- 200 / (1 + exp(17 - x)/2) * exp(-0.02*x) yeps <- y + rnorm(length(y), sd = 2) nls(yeps ~ p1 / (1 I have the results of a natural experiment and I want to show the number of consecutive occurrences of an event fit an exponential distribution. Look: # indentifiability No <- 100; a <- 1; b <- -1; T <- 2 Ne <- seq(1, 10, l=8) curve(No*(1-exp(a*(b*x-T))), 0, 10) abline(h=No*(1-exp(a*(b*0-T)))) # intercept C <- a*b; D <-

up vote 5 down vote favorite 3 I have some basic data on emission reductions and cost per car: q24 <- read.table(text = "reductions cost.per.car 50 45 55 55 60 62 There are many ways to follow us - By e-mail: On Facebook: If you are an R blogger yourself you are invited to add your own R content feed to this bsnrh Threaded Open this post in threaded view ♦ ♦ | Report Content as Inappropriate ♦ ♦ Re: NLS "Singular Gradient" Error In reply to this post by Gabor Grothendieck Ravi. -----Original Message----- From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Corrado Sent: Wednesday, March 31, 2010 9:13 AM Cc: r-help at r-project.org Subject: Re: [R] Error "singular gradient

Can anyone please advise as to what I may be doing wrong? I am using a modification of Holling's (1959) disc equation to account for non-replacement of prey; Ne=No{1-exp[a(bNe-T)]} where a is the attack rate, b is the handling time, and T is Why aren't they right? I have tried > modifying the parameter start points with no success. > > Many thanks, > -- > View this message in context: http://r.789695.n4.nabble.com/NLS-Singular-Gradient-Error-tp2069029p2069029.html> Sent from the R help mailing

reply | permalink Bert Gunter Corrado: Over parameterization/non-identifiability is not determined by the ratio of the number of data values to the number of parameters: if you try to fit a asked 5 years ago viewed 16372 times active 4 years ago Blog Stack Overflow Podcast #92 - The Guerilla Guide to Interviewing Linked 4 How to fit an exponential equation of John C Nash wrote:If you have a perfect fit, you have zero residuals. Thank you for your suggestions; The first returns this error; "Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : NA/NaN/Inf in foreign function call (arg 4)" I

Gabor Grothendieck at Mar 30, 2010 at 11:24 am ⇧ You could try method="brute-force" in the nls2 package to find starting values.On Tue, Mar 30, 2010 at 7:03 AM, Corrado wrote:I The Michaelis-Menten function has this form: where r is the growth rate (the dependent variable), C is the concentration (the independent variable), Vm is the maximum rate in the system, and x <- seq(min(time), max(time), length=100) y <- predict(logisticModel, list(time=x)) points(x, y, type='l', col='blue') The added step of finding good starting parameters can be bypassed by using the self-starting nls logistic model Thanks in advance.Alternatively, what do you suggest I should do?

How to make sure that my operating system is not affected by CVE-2016-5195? It is from Juliano's (2001) Nonlinear Curve Fitting chapter in Scheiner and Gurevitch. In that case it's often convenient (and fast) to use the previous solutions as initial estimates for the next ones. The easiest way to evaluate the fitted function is with predict().

I took the liberty of formatting your answer, I hope you don't mind. In S-PLUS, one such method is used by the function nlregb(). summary(logisticModel) coef(logisticModel) As in the Michaelis-Menton example, adding the fitted curve requires evaluating the function over the range of the data, and adding it with the points() function. Also, if my n is 4, then the nls works perfectly (but that excludesall the k5 ....

SSH makes all typed passwords visible when command is provided as an argument to the SSH command more hot questions question feed default about us tour help blog chat data legal Also, if my n is 4, then the nls works perfectly (but that excludesall the k5 .... Another method to estimate initial values relies on understanding what they mean, which can be based on experience, physical theory, etc. Are youclaiming that every single point on the grid fails?

Let's plot it: plot(q24) p <- coef(model) curve(p["a"] * exp(p["b"] * x) + p["c"], lwd=2, col="Red", add=TRUE) It worked well! logisticModel <- nls(population~K/(1+exp(Po+r*time)), start=list(Po=5, r=2, K=5)) Non-linear fits can be sensitive to the initial guesses. r nonlinear-regression nls share|improve this question edited Jul 15 '11 at 7:51 mpiktas 24.8k449104 asked Jul 14 '11 at 17:10 steiny 78116 2 Hint: look at the coefficient of $r^x$ Bert Gunter Genentech Nonclinical Biostatistics -----Original Message----- From: r-help-bounces at r-project.org Bert Gunter at Mar 30, 2010 at 5:45 pm ⇧ Your model is almost certainly over-parameterized (given the data that

Related 4How to get p value and confidence intervals for nls functions?1Finding starting values for nls for critical exponential function2How do I incorporate 2 peaks in a liquidity model using curve-fitting It is also not hard to see why the gradient is singular. I do believe that the code couldat least give a better diagnostic message. Cantonale Galleria 2, 6928 Manno, Switzerland | Fax: +41 (91) 610.82.82 > > -- Felix Nensa Luisenstr. 15-17 44787 Bochum Germany mail: [hidden email] mobile: +49 171 958 51 40

Comments are closed. If r(x) is the vector of residuals (response - fitted values), then at each iteration the Gauss-Newton method moves from the current parameter values, x, along a direction d that is Chapter 20 of The R Book by Michael J. Cheers, Andrej Filed under: R Internals Related To leave a comment for the author, please follow the link and comment on their blog: Rmazing.

For example, a neural network? –osazuwa Jul 20 '15 at 19:47 add a comment| up vote 13 down vote The answers above are, of course, correct. It > had no effect on the model. > > Thank you for your suggestions; > > The first returns this error; > "Error in lm.fit(x, y, offset = offset, singular.ok Subscribe to R-bloggers to receive e-mails with the latest R posts. (You will not see this message again.) Submit Click here to close (This popup will not appear again) R › current community blog chat Cross Validated Cross Validated Meta your communities Sign up or log in to customize your list.

Shall I abandon nls infavour of optim?Regards--Corrado TopiPhD ResearcherGlobal Climate Change and BiodiversityArea 18,Department of BiologyUniversity of York, York, YO10 5YW, UKPhone: + 44 (0) 1904 328645, E-mail: ct529 at york.ac.uk______________________________________________R-help I think requiring the $r \in (0,1)$ would do the job. –Macro Jul 14 '11 at 19:02 add a comment| 2 Answers 2 active oldest votes up vote 11 down vote asked 3 years ago viewed 9387 times active 3 years ago Blog Stack Overflow Podcast #92 - The Guerilla Guide to Interviewing Linked 1 Exponential fitting with R Related 5Modifying a DDoS: Why not block originating IP addresses?

Are you claiming that every single point on the grid fails? Messages sorted by: [ date ] [ thread ] [ subject ] [ author ] More information about the R-help mailing list current community chat Stack Overflow Meta Stack Overflow your I am using a modification of Holling's (1959) disc equation to > > account for non-replacement of prey; > > > > Ne=No{1-exp[a(bNe-T)]} > > > > where a is the The resulting residuals are approximatelynormally distributed with mean 0 and sd ~ 4.23.2) I agree with the comment of Bert on over-parametrization, but againthe model is not overparamterised, and it is