Displaying 20 results from an estimated 3000 matches similar to: "(no subject)"
2012 Apr 17
3
error using nls with logistic derivative
Hi
I?m trying to fit a nonlinear model to a derivative of the logistic function
y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function with nls)
The derivative calculated with D function is:
> logis<- expression(a/(1+exp((b-x)/c)))
> D(logis, "x")
a * (exp((b - x)/c) * (1/c))/(1 + exp((b - x)/c))^2
So I enter this expression in the nls function:
2005 Feb 22
3
problems with nonlinear fits using nls
Hello colleagues,
I am attempting to determine the nonlinear least-squares estimates of
the nonlinear model parameters using nls. I have come across a common
problem that R users have reported when I attempt to fit a particular
3-parameter nonlinear function to my dataset:
Error in nls(r ~ tlm(a, N.fix, k, theta), data = tlm.data, start =
list(a = a.st, :
step factor 0.000488281
2007 Apr 15
1
nls.control( ) has no influence on nls( ) !
Dear Friends.
I tried to use nls.control() to change the 'minFactor' in nls( ), but it
does not seem to work.
I used nls( ) function and encountered error message "step factor
0.000488281 reduced below 'minFactor' of 0.000976563". I then tried the
following:
1) Put "nls.control(minFactor = 1/(4096*128))" inside the brackets of nls,
but the same error message
2008 Apr 10
1
(no subject)
Subject: nls, step factor 0.000488281 reduced below 'minFactor' of
0.000976563
Hi there,
I'm trying to conduct nls regression using roughly the below code:
nls1 <- nls(y ~ a*(1-exp(-b*x^c)), start=list(a=a1,b=b1,c=c1))
I checked my start values by plotting the relationship etc. but I kept
getting an error message saying maximum iterations exceeded. I have
tried changing these
2006 Aug 04
1
gnlsControl
When I run gnls I get the error:
Error in nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xy, :
step factor 0.000488281 reduced below 'minFactor' of 0.000976563
My first thought was to decrease minFactor but gnlsControl does not contain
minFactor nor nlsMinFactor (see below). It does however contain nlsMaxIter
and nlsTol which I assume are the analogs of
2011 Mar 28
1
error in nls, step factor reduced below minFactor
Hello,
I've seen various threads on people reporting:
step factor 0.000488281 reduced below `minFactor' of 0.000976563
While I know how to set the minFactor, what I'd like to have happen is for nls to return to me, the last or closest fitted parameters before it errors out. In other words, so I don't get convergence, I'd still like to acquire the values of the parameters
2002 Apr 23
1
Use of nls command
Hello.
I am trying to do a non-linear fit using the 'nls' command.
The data that I'm using is as follows
pH k
1 3.79 34.21
2 4.14 25.85
3 4.38 20.45
4 4.57 15.61
5 4.74 12.42
6 4.92 9.64
7 5.11 7.30
8 5.35 5.15
9 5.67 3.24
with a transformation of pH to H <- 10^-pH
When using the nls command for a set of parameters - a, b and c, I receive
two sets of errors:
>
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
2012 Jan 25
1
solving nls
Hi,
I have some data I want to fit with a non-linear function using nls, but it
won't solve.
> regresjon<-nls(lcfu~lN0+log10(1-(1-10^(k*t))^m), data=cfu_data,
> start=(list(lN0 = 7.6, k = -0.08, m = 2)))
Error in nls(lcfu ~ lN0 + log10(1 - (1 - 10^(k * t))^m), data = cfu_data, :
step factor 0.000488281 reduced below 'minFactor' of 0.000976562
Tried to increase minFactor
2008 Apr 14
3
Logistic regression
Dear all,
I am trying to fit a non linear regression model to time series data.
If I do this:
reg.logis = nls(myVar~SSlogis(myTime,Asym,xmid,scal))
I get this error message (translated to English from French):
Erreur in nls(y ~ 1/(1 + exp((xmid - x)/scal)), data = xy, start =
list(xmid = aux[1], :
le pas 0.000488281 became inferior to 'minFactor' of 0.000976562
I then tried to set
2008 Sep 02
2
nls.control()
All -
I have data:
TL age
388 4
418 4
438 4
428 5
539 10
432 4
444 7
421 4
438 4
419 4
463 6
423 4
...
[truncated]
and I'm trying to fit a simple Von Bertalanffy growth curve with program:
#Creates a Von Bertalanffy growth model
VonB=nls(TL~Linf*(1-exp(-k*(age-t0))), data=box5.4,
start=list(Linf=1000, k=0.1, t0=0.1), trace=TRUE)
#Scatterplot of the data
plot(TL~age, data=box5.4,
2009 Oct 02
1
nls not accepting control parameter?
Hi
I want to change a control parameter for an nls () as I am getting an error
message "step factor 0.000488281 reduced below 'minFactor' of 0.000976562".
Despite all tries, it seems that the control parameter of the nls, does not
seem to get handed down to the function itself, or the error message is
using a different one.
Below system info and an example highlighting the
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:
2004 Mar 18
1
profile error on an nls object
Hello all,
This is the error message that I get.
> hyp.res <- nls(log(y)~log(pdf.hyperb(theta,X)), data=dataModel,
+ start=list(theta=thetaE0),
+ trace=TRUE)
45.54325 : 0.1000000 1.3862944 -4.5577142 0.0005503
3.728302 : 0.0583857346 0.4757772859 -4.9156128701 0.0005563154
1.584317 : 0.0194149477 0.3444648833 -4.9365149150 0.0004105426
1.569333 :
2015 Mar 18
1
Help
Hi to All,
I am fitting some models to a data using non linear least square, and
whenever i run the command, parameters value have good convergence but I
get the error in red as shown below. Kindly how can I fix this problem.
Convergence of parameter values
0.2390121 : 0.1952981 0.9999975 1.0000000
0.03716107 : 0.1553976 0.9999910 1.0000000
0.009478433 : 0.2011017 0.9999798 1.0000000
2004 Feb 04
1
Fitting nonlinear (quantile) models to linear data.
Hello.
I am trying to fit an asymptotic relationship (nonlinear) to some
ecological data, and am having problems. I am interested in the upper
bound on the data (i.e. if there is an upper limit to 'y' across a range
of 'x'). As such, I am using the nonlinear quantile regression package
(nlrq) to fit a michaelis mention type model.
The errors I get (which are dependant on
2007 Oct 10
11
please help me
dear list
I am student M.S. statistics in department statistics . I am working in the function "nls" in the [R 2.3.1] with 246 data and want to fit the "exp" model to vectors( v and u ) but I have
a problem to use it
u
5.000000e-13 2.179057e+03 6.537171e+03 1.089529e+04 1.525340e+04
1.961151e+04 2.396963e+04 2.832774e+04 3.268586e+04 3.704397e+04
4.140209e+04
2023 Aug 20
2
Issues when trying to fit a nonlinear regression model
Dear Bert,
Thank you so much for your kind and valuable feedback. I tried finding the
starting values using the approach you mentioned, then did the following to
fit the nonlinear regression model:
nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x),
start =
list(theta1 = 0.37,
theta2 = exp(-1.8),
theta3 =
2011 Jun 12
1
Error in NLS example in the documentation
Hello there,
I am trying to use R function NLS to analyze my data and one of the examples
in the documentation is -
## the nls() internal cheap guess for starting values can be sufficient:
x <- -(1:100)/10
y <- 100 + 10 * exp(x / 2) + rnorm(x)/10
nlmod <- nls(y ~ Const + A * exp(B * x), trace=TRUE)
plot(x,y, main = "nls(*), data, true function and fit, n=100")
curve(100 +
2013 Oct 27
1
nls function error
data(Boston, package='MASS')
y <- Boston$nox
x <- Boston$dis
nls(y~ A + B * exp(C * x), start=list(A=1, B=1, C=1))
Error in nls(y ~ A + B * exp(C * x), start = list(A = 1, B = 1, C = 1), :
step factor 0.000488281 reduced below 'minFactor' of 0.000976562
I don't know how to fix this error. I think my problem is that I set the
wrong start. Could somebody help please?