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# R Error In Nls Singular Gradient

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 The problem still stays.Gabor Grothendieck wrote:Sorry, its algorithm="brute-force"On Tue, Mar 30, 2010 at 10:29 AM, Corrado wrote:Hi Gabor,same problem even using nls2 with method=brute-force to calculate theinitial parameters.Best,Gabor Grothendieck wrote:You could I've read that when using SSasymp: b is 'the horizontal asymptote (a) -the response when x is 0' while c is the rate constant. H(x) has the form J(x)^T J(x) + B(x), where B(x) vanishes at points where r(x) = 0, so what Gauss-Newton is doing is using J(x)^T J(x) to approximate the Hessian matrix

r self-study exponential share|improve this question edited Jul 9 '15 at 13:33 whuber♦ 146k18285546 asked Jul 8 '15 at 20:00 Amanda 13416 I would suggest starting your deciphering by kn).Can anyone help me with suggestions? more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed I think I mis-read this as an exponential function.

Copyright © 2016 R-bloggers. In a saturating curve, the rate constant is determined from the curvature. Here you will find daily news and tutorials about R, contributed by over 573 bloggers. The help page specifically states: Warning Do not use nls on artificial "zero-residual" data.

Gabor Grothendieck Threaded Open this post in threaded view ♦ ♦ | Report Content as Inappropriate ♦ ♦ Re: NLS "Singular Gradient" Error In reply to this post by bsnrh Functions of the form y = b * x / (c+x) are concave up when b < 0 and c > 0; they are concave down when b > 0 and Initializing ‘a’ to '1.'. Free forum by Nabble Edit this page R › R help Search everywhere only in this topic Advanced Search Non linear Regression: "singular gradient matrix at initial parameter estimates" ‹ Previous

Or the convergence could be to a local minimum, not a global one.Comparison of performance of optimization algorithms is, again, a cottageindustry for which serious expertise is required. (This should not Both $\log(a)$ and $b$ can be estimated with least squares. singular gradientmatrix at initial parameter estimatesI have read all the previous postings and the documentation, but to noavail: the error is there to stay. Grothendieck Aug 21 '13 at 19:08 OK, got it.

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 Because your data is "nearly" linear, and there is substantial scatter, the best fit (e.g., the set of parameters a, b, and c which minimize the residual sum-of-squares), is concave down I saidthat an external algorithms fits the model without any problems: with ~500,000 data points and 19 paramters (ki in the original equation), itfits the model in less than 1 second. I am sure the problem is with nls,because the external fitting algorithm perfectly fits it in less than asecond.

• Gabor Grothendieck at Mar 30, 2010 at 3:02 pm ⇧ What do you mean the problem still stays?
• It had no effect on the model.
• But in the nlsmanual page we have:Warning:*Do not use ?nls?
• r nls share|improve this question edited Aug 20 '15 at 22:17 DJJ 685823 asked Aug 19 '15 at 11:56 Rachael 83 This error usually happens when you have chosen

Error t value Pr(>|t|) # a -179.17 22.86 -7.837 5.06e-05 *** # b 1009.36 2556.44 0.395 0.703 # c -5651.20 11542.41 -0.490 0.638 # --- # Signif. Is it safe for a CR2032 coin cell to be in an oven? install.packages(c("bbmle","emdbook")) library(bbmle) library(emdbook) ?mle2 ?lambertW (You can address further questions to me off-list ...) ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland An extended example of a (moderately difficult) nonlinear fit whose initial values can be determined in this way is described in my answer at http://stats.stackexchange.com/a/15769.

In that case it's often convenient (and fast) to use the previous solutions as initial estimates for the next ones. If $b$ is much outside a fairly narrow interval around those two values, you'll run into some problems. [Alternatively try a different algorithm] –Glen_b♦ Jul 8 '15 at 23:09 The function is: $$y=a+b\cdot r^{(x-m)}+c\cdot x$$ It is effectively an exponential curve with a linear section, as well as an additional horizontal shift parameter (m). reply | permalink Corrado Hi Gabor, same problem even using nls2 with method=brute-force to calculate the initial parameters.

The presence of the exponential encourages us to use logarithms--but the addition of $c$ makes it difficult to do that. Are you claiming that every single point on the grid fails? How large a grid are you using? A short example: #parameters used to generate the data reala=-3 realb=5 realc=0.5 realr=0.7 realm=1 x=1:11 #x values - I have 11 timepoint data #linear+exponential function y=reala + realb*realr^(x-realm) + realc*x #add

For the model to be estimable in a region of the current estimates, this matrix must have full column rank. The solutionis unique and the rapidity of convergence is practically independentfrom the selection of start conditions (with a reasonable selection ofstart conditions at least). Visual analysis of a scatterplot (to determine initial parameter estimates) is described and illustrated at http://stats.stackexchange.com/a/32832.

## Also, if my n is 4, then the nls works perfectly (but thatexcludesall the k5 ....

House, 6 - 8 Stuart Street, LU1 2SJ. kn).Can anyone help me with suggestions? How do I translate "hate speech"? Um=nls(N ~ Ymin + ((A-Ymin)/(1+(exp(((B - Distance) + (C*(Distance)^2)*D))))), start=list(Ymin=1,A=30,B=0.1,C=0.01, D=-0.1), nls.control( maxiter = 500, tol = 2e-05)) But, I obtained: Error in nlsModel(formula, mf, start, wts) : singular gradient

The next iterate is then x + a d, for some positive scalar a. Why aren't they right? How do I install the latest OpenOffice? Here is the revised code: c.0 <- min(q24\$cost.per.car) * 0.5 model.0 <- lm(log(cost.per.car - c.0) ~ reductions, data=q24) start <- list(a=exp(coef(model.0)[1]), b=coef(model.0)[2], c=c.0) model <- nls(cost.per.car ~ a * exp(b *

Ben Bolker Threaded Open this post in threaded view ♦ ♦ | Report Content as Inappropriate ♦ ♦ Re: NLS &quot;Singular Gradient&quot; Error Ben Bolker ufl.edu> writes: > mario On 12-Apr-11 18:01, Felix Nensa wrote: > fit = nls(yeps ~ p1 / (1 + exp(p2 - x)) * exp(p4 * x)) > -- Ing. Cantonale Galleria 2, 6928 Manno, Switzerland | Fax: +41 (91) 610.82.82 ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code. The modifications were made so that the formula is transformed into a function that returns a vector of (weighted) residuals whose sum square is minimized by nls.lm.

The main reason is the one given by @whuber and @marco. Unix Exit Command Word for making your life circumstances seem much worse than they are Animate a circle "rolling" along a complicated 3D curve What's a Shady Word™? After giving decent starting values as suggested by a different post here I get the singular gradient error. I think this is because I've confused R about a, b and c (?).

The issue I initially got was infinity, which I don't get since none of the values are 0. All we can do is guess unless you provide a reproducible example which means that if we paste it in from your post it will give the same errors you see.