Displaying 20 results from an estimated 10000 matches similar to: ""Error : singular gradient matrix at initial parameters estimates""
2011 Apr 11
1
Non linear Regression: "singular gradient matrix at initial parameter estimates"
Hi,
I am using nls to fit a non linear function to some data but R keeps giving
me "singular gradient matrix at initial parameter estimates" errors.
For testing purposes I am doing this:
### R code ###
x <- 0:140
y <- 200 / (1 + exp(17 - x)/2) * exp(-0.02*x) # creating 'perfect' samples
with fitting model
yeps <- y + rnorm(length(y), sd = 2) # adding noise
# results
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
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
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
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:
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
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
2008 Feb 15
3
Error 'singular gradient' in nonlinear model fitting
w.age.female.2004 <- nls(WEIGHT ~ (alpha*TOTAL^beta)/454,
start=list(alpha=1, beta=3),
data=spottedseatrout2004.female.data)
I am trying to fit above model to length-weight data of a fish species
(spotted seatrout) by year (1999-2006). The convergence occurred for all
the years except 2002 and 2004. In these two year, R shows the error
called
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
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 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
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
2007 Feb 07
1
Singular Gradient
I tried to fit data with the following function:
fit<-nls(y~ Is*(1-exp(-l*x))+Iph,start=list(Is=-2e-5,l=2.3,Iph=-0.3
),control=list(maxiter=500,minFactor=1/10000,tol=10e-05),trace=TRUE)
But I get only a singular Gradient warning...
the data can by found attached(there are two sampels of data col 1/2 and
3/4).
I tried to fix it by chanching the start parameters but that didn't solve
the
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
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
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
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())
2006 Sep 29
1
linear gradient in nls
Hello,
I hope this doesn't turn into a statistics question but here I
go. I am using the nls function with a Gaussian distribution, see coding
below. When I run the nls I get an error back saying that I have a linear
gradient. I then, of course am unable to do anything else. The data that I
am using are intensity values from some mass spectrometry data. Is there
something I can
2012 Oct 15
2
fit a "threshold" function with nls
I am trying to model a dependent variable as a threshold function of
my independent variable.
What I mean is that I want to fit different intercepts to y following 2
breakpoints, but with fixed slopes.
I am trying to do this with using ifelse statements in the nls function.
Perhaps, this is not an appropriate approach.
I have created a very simple example to illustrate what I am trying to do.
2004 Jun 29
1
nls fitting problems (singularity)
Hallo!
I have a problem with fitting data with nls. The first
example with y1 (data frame df1) shows an error, the
second works fine.
Is there a possibility to get a fit (e.g. JMP can fit
also data I can not manage to fit with R). Sometimes I
also got an error singularity with starting
parameters.
# x-values
x<-c(-1,5,8,11,13,15,16,17,18,19,21,22)
# y1-values (first data set)