similar to: nls() and "singular grandient"

Displaying 20 results from an estimated 10000 matches similar to: "nls() and "singular grandient""

2005 Mar 23
1
nl regression with 8 parameters, help!
I'm doing a non linear regression with 8 parameters to be fitted: J.Tl.nls<-nls(Gw~(a1/(1+exp(-a2*Tl+a3))+a4)*(b1/(1+exp(b2*Tl-b3))+b4),data=Enveloppe, start=list(a1=0.88957,a2=0.36298,a3=10.59241,a4=0.26308, b1=0.391268,b2=1.041856,b3=0.391268,b4=0.03439)) First, I fitted my curve on my data by guessing the parameters'
2010 Dec 13
2
Complicated nls formula giving singular gradient message
I'm attempting to calculate a regression in R that I normally use Prism for, because the formula isn't pretty by any means. Prism presents the formula (which is in the Prism equation library as Heterologous competition with depletion, if anyone is curious) in these segments: KdCPM = KdnM*SpAct*Vol*1000 R=NS+1 S=(1+10^(X-LogKi))*KdCPM+Hot a=-1*R b=R*S+NS*Hot+BMax c = -1*Hot*(S*MS+BMax) Y
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 Mar 28
1
Singular Gradient in nls
//Referring to the response posted many years ago, copied below, what is the specific criterium used for singularity of the gradient matrix? Is a Singular Value Decomposition used to determine the singular values? Is it the gradient matrix condition number or some other criterion for determining singularity? // //Glenn // / / /> What does the error 'singular gradient' mean
2005 Apr 23
1
start values for nls() that don't yield singular gradients?
I'm trying to fit a Gompertz sigmoid as follows: x <- c(15, 16, 17, 18, 19) # arbitrary example data here; y <- c(0.1, 1.8, 2.2, 2.6, 2.9) # actual data is similar gm <- nls(y ~ a+b*exp(-exp(-c*(x-d))), start=c(a=?, b=?, c=?, d=?)) I have been unable to properly set the starting value '?'s. All of my guesses yield either a "singular gradient" error if they
2013 Jun 19
1
nls singular gradient ..as always..
Hi all. Sorry for posting again such a topic but I went through previous posts but couldn't find a solution. I use the following code to fit an exponential model to my data. I have 4 different datasets. For 3 datasets nls seems to work fine and I have no error messages. But for 1 dataset I am getting the "world known" singular gradient error. xfit.dNEE <-
2010 Apr 28
1
NLS "Singular Gradient" Error
Hello, I am trying to model a type II functional response of number of prey eaten (Ne) against number supplied (No) with a non-linear least squares regression (nls). 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 the experimental period. My script is as
2007 Sep 05
3
'singular gradient matrix’ when using nls() and how to make the program skip nls( ) and run on
Dear friends. I use nls() and encounter the following puzzling problem: I have a function f(a,b,c,x), I have a data vector of x and a vectory y of realized value of f. Case1 I tried to estimate c with (a=0.3, b=0.5) fixed: nls(y~f(a,b,c,x), control=list(maxiter = 100000, minFactor=0.5 ^2048),start=list(c=0.5)). The error message is: "number of iterations exceeded maximum of
2002 Jul 11
1
nls() singular graident matrix error
R-helpers; I used Proc Model in SAS to fit the following model to data: proc model data = dbsmv; a = a1*F**2; b = b1*F + b2*T + b3*F*T; tph2 = tph1 *((1 - exp(-a*age2)) / (1 - exp(-a*age)))**-b; fit tph2; and yielded the following estimated parameters after iterations: a1 = -0.15943, a2 = -1.8177, b1 = -0.01911, b2
2007 Feb 13
1
nls: "missing value or an infinity" (Error in numericDeriv) and "singular gradient matrix"Error in nlsModel
Hi, I am a non-expert user of R. I am essaying the fit of two different functions to my data, but I receive two different error messages. I suppose I have two different problems here... But, of which nature? In the first instance I did try with some different starting values for the parameters, but without success. If anyone could suggest a sensible way to proceed to solve these I would be
2011 Jun 30
1
Error "singular gradient matrix at initial parameter estimates" in nls
Greetings, I am struggling a bit with a non-linear regression. The problem is described below with the known values r and D inidcated. I tried to alter the start values but get always following error message: Error in nlsModel(formula, mf, start, wts): singular gradient matrix at initial parameter estimates Calls: nls -> switch -> nlsModel I might be missing something with regard to the
2011 Oct 11
1
singular gradient error in nls
I am trying to fit a nonlinear regression to infiltration data in order to determine saturated hydraulic conductivity and matric pressure. The original equation can be found in Bagarello et al. 2004 SSSAJ (green-ampt equation for falling head including gravity). I am also VERY new to R and to nonlinear regressions. I have searched the posts, but am still unable to determine why my data come up
2005 Mar 18
3
Non linear modeling
AFAIK most model fitting techniques will only deal with additive errors, not multiplicative ones. You might want to try fitting: log(y-x) = a*x + e which is linear. Andy > From: Angelo Secchi > > Hi, > is there a way in R to fit a non linear model like > > y=x+exp(a*x)*eps > > where a is the parameter and eps is the error term? > Thanks > Angelo > >
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 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
2011 Jun 09
0
nls Error Message - Singular Gradient Matrix
I've been trying to run some analysis using the nls function in R but keep coming up with an error message which I don't understand how to fix. The message follows here: Error in nls(formula = f.p, data = n.data, start = list(S = 1, a = -0.1, : singular gradient In addition: Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing
2007 Oct 01
1
[nls] singular gradient
Hi, I am new to R. I don't have strong background of statistics. I am a student of Geotechnical Engineering. I tried to run a nonlinear regression for a three-variable function, that is N = f(CSR, ev) # N is a function of CSR and ev, and N = CSR/(A +B*CSR), wherer (A,B) are function of ev. N, CSR and ev are observed in the experiments. Following is my R script. rm(list=ls())
2001 Oct 05
1
nls() fit to a lorentzian - can I specify partials?
First, thanks to all who helped me with my question about rescaling axes on the fly. Using unlist() and range() to set the axis ranges in advance worked well. I've since plotted about 300 datasets with relative ease. Now I'm trying to fit a lossy oscillator resonance to (the square root of) a lorentzian (testframe$y is oscillator amplitude, testframe$x is drive frequency): lorentz
2005 Mar 21
1
menu() and commands assigned to choices ->break out a repeat loop so that the next code lines to be read
first:I'd like to have the choice between breaking out a repeat loop or continue it then: i'd would like my code after the end of my repeat loop not to be read unless I type 2 (to continue executing the content of my repeat loop) THAT'S THE PROBLEM Is there any way to go straight out of the loop (like a "goto" associated to a "label") or a pause like command to
2005 Mar 21
1
classes of data (with a variable size)
How is it possible to "split" a data.frame in order to get classes with variable size. actually I'd like to get classes of data with classe size (cs) so that cs becomes bigger with an other increasing value. Guillaume Storchi