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Singular Gradient Nls

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model1<-nls(Flux~b*Par/(c+Par)-a, data = curve1, start=list(a=180, b=-200, c=-2000)) plot(Flux~Par,curve1) curve(predict(model1,newdata=data.frame(Par=x)),add=TRUE) summary(model1) # Formula: Flux ~ b * Par/(c + Par) - a # # Parameters: # Estimate Std. Ben Bolker ______________________________________________ [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. Comment every line of code? Browse other questions tagged r regression curve-fitting nls or ask your own question.

Replace "/U+[0-9A-F]{4}/" with proper unicode character in shell pipeline At age 25, is it still okay to wear dental braces to work? My script is as follows; model<-nls(Ne~No*(1-exp(a*(b*Ne-T))),start=list(a=1.2,b=0.015,T=24)) This returns the following error; "Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates". Sometimes, though, I have data that doesn't go well with those parameters and I get errors from nls (e.g. "singular gradient" or things like that). Try our newsletter Sign up for our newsletter and get our top new questions delivered to your inbox (see an example).

Singular Gradient Nls

You don't need the line: curve1<-do.call("cbind", curve1) share|improve this answer answered Aug 20 '15 at 20:18 jlhoward 36.9k32865 Ah Ok I understand. asked 5 years ago viewed 17009 times active 4 years ago Get the weekly newsletter! If you set c=0 and take log of y (making a linear relationship), you can use regression to get initial estimates for log($a$) and $b$ that will suffice for your data 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

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. I think I mis-read this as an exponential function. 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 Nls Step Factor Reduced Below Minfactor I have tried calling T<-24 and rerunning with no success sadly.

What's a word/phrase like "compulsion" or "addiction" that is more about guilt than physiology or anxiety? If the latter is the case, is there a way to force R to do its best to fit a non-linear model to it anyway? 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). May I hold the DSLR in part by the hotshoe flash?

It had no effect on the model. Nlslm 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) current community Authoritative source that <> and != are identical in performance in SQL Server Shortest code to throw SIGILL Why use a kettle to heat water? 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.

  • So, what's similar and what's, well, better… library(minpack.lm) ### Examples from 'nls' doc ### DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nlsLM(density ~ Asym/(1 + exp((xmid - log(conc))/scal)), data =
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Error In Nlsmodel(formula, Mf, Start, Wts) : Singular Gradient Matrix At Initial Parameter Estimates

Mathematically, for x << c b * x / (c + x) ~ (b/c) * x in your case the slope is about -0.25, so b/c ~ -0.25. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the Singular Gradient Nls How to compose flowering plants? R Nls Missing Value Or An Infinity Produced When Evaluating The Model dat2 cuts off at 14 since the 0 count bins would throw off the exponential regression, and I really only need to fit those first 14.

It solved this problem. All I needed was poly() model <- lm(cost.per.car ~ poly(reductions, 3), data=q24) new.data <- data.frame(reductions = c(91,92,93,94)) predict(model, new.data) plot(q24) lines(q24$reductions, predict(model, list(reductions = q24$reductions))) Or, using lattice: xyplot(cost.per.car ~ reductions, I just added a silent="TRUE" parameter to try, so I don't get errors printed out. –nico Jun 3 '10 at 8:47 @kohske The logic of your answer makes sense. Your data is somewhat pathological. Nls Starting Values

Terms and Conditions for this website Never miss an update! 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 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 Another method to estimate initial values relies on understanding what they mean, which can be based on experience, physical theory, etc.

Where were sacrifices offered if not in the Temple? Number Of Iterations Exceeded Maximum Of 50 I was wondering if it's possible to find that package on the internet? When automating this, you might perform some quick analyses of the residuals, such as comparing their extremes to the spread in the ($y$) data.

Have I just fundamentally misunderstood how to use SSasymp?

Related 7Curve fitting in R using nls0How do I fit a multiple regression model in R using gam8Fitting a function in R0“minimum count is not zero” error for zero inflated model0Matplot The nls.lm function must be supplied with an objective function that returns a vector of residuals to be minimized. Both $\log(a)$ and $b$ can be estimated with least squares. `par' That You Provided Is Non-list And Non-numeric! I was > wondering if it's possible to find that package on the internet? > > Many thanks, > Neal If the (draft) PDF says that, it's an error.

Any ideas? Join them; it only takes a minute: Sign up R nls singular gradient up vote 2 down vote favorite 2 I've tried searching the other threads on this topic but none 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 How to change ip address of host at runtime?

The issue I initially got was infinity, which I don't get since none of the values are 0. 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 <-