Displaying 20 results from an estimated 12000 matches similar to: "How to store the parameter estimated by nls( ) to a variable?"
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
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
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
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
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:
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
2007 Aug 03
2
Saving an expression to a file
Dear Friends.
I have a very long expression and I use function D to find its derivative,
which will be even longer. I save the resulting expression in a variavle,
say bbb. But when I tried to display bbb on the screen, the R screen is not
long enough for me to view it. Is there a way to save the expression to a
file?
Best Wishes
Yuchen Luo
[[alternative HTML version deleted]]
2010 Apr 15
2
using nls for gamma distribution (a,b,d)
Dear all
i want to estimated the parameter of the gamma density(a,b,d)
f(x) = (1/gamma(b)*(a^b)) * ((x-d)^(b-1)) * exp{-(x-d)/a)} for x>d
f(x) = Age specific fertility rate
x = age
when i run this in R by usling nls()
gamma.asfr <- formula(asfr ~ (((age-d)^(b-1))/((gamma(b))*(a^b)))*
exp(-((age-d)/a)))
gamma.asfr1 <- nls(gamma.asfr, data= asfr.aus, start = list(b = 28, a = 1,
d= 0.5),
2007 Aug 13
2
How to write to a table column by column?
Dear friends.
Every loop of my program will result in a list that is very long, with a
structure similar to the one below:
Lst <- list(name="Fred", wife="Mary", daily.incomes=c(1:850))
Please notice the large size of "daily.incomes".
I need to store all such lists in a csv file so that I can easily view them
in Excel. Excel cannot display a row of more than 300
2010 Apr 19
3
nls for piecewise linear regression not converging to least square
Hi R experts,
I'm trying to use nls() for a piecewise linear regression with the first
slope constrained to 0. There are 10 data points and when it does converge
the second slope is almost always over estimated for some reason. I have
many sets of these 10-point datasets that I need to do. The following
segment of code is an example, and sorry for the overly precise numbers,
they are just
2008 Aug 29
1
nls() fails on a simple exponential fit, when lm() gets it right?
Dear R-help,
Here's a simple example of nonlinear curve fitting where nls seems to get
the answer wrong on a very simple exponential fit (my R version 2.7.2).
Look at this code below for a very basic curve fit using nls to fit to (a)
a logarithmic and (b) an exponential curve. I did the fits using
self-start functions and I compared the results with a more simple fit
using a straight lm()
2010 Apr 19
2
nls minimum factor error
Hi,
I have a small dataset that I'm fitting a segmented regression using nls on.
I get a step below minimum factor error, which I presume is because residual
sum of square is still "not small enough" when steps in the parameter space
is already below specified/default value. However, when I look at the trace,
the convergence seems to have been reached. I initially thought I might
2005 Jan 06
1
nls - convergence problem
Dear list,
I do have a problem with nls. I use the following data:
>test
time conc dose
0.50 5.40 1
0.75 11.10 1
1.00 8.40 1
1.25 13.80 1
1.50 15.50 1
1.75 18.00 1
2.00 17.00 1
2.50 13.90 1
3.00 11.20 1
3.50 9.90 1
4.00 4.70 1
5.00 5.00 1
6.00 1.90 1
7.00 1.90 1
9.00 1.10 1
12.00 0.95 1
14.00
2004 Nov 05
1
Problems running a 4-parameter Weibull function with nls
Hi,
I am rather new to R, but both myself and another much more experience user cannot figure this out. I have a collegue who has a 800+ nonlinear regressions to run for seed germination (different species, treatments, etc.) over time. I created a looping structure to extract the parameters from each regression; I will then use the parameters themselves for further analysis. I would like to
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
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
2009 Jul 09
1
nls, reach limit bounds
Hi,
I am trying to fit a 4p logistic to this data, using nls function. The function didn't freely converge; however, it converged if I put a lower and an upper bound (in algorithm port). Also, the b1.A parameter always takes value of the upper bound, which is very strange. Has anyone experienced about non-convergent of nls and how to deal with this kind of problem?
Thank you very much.
2009 Jul 20
1
Argument problem in function wrapper
Dear all,
we are writing a wrapper for the nls function in library stats. We are
having a problem with one of the arguments (weightsArgument) which seems not
to reach nls even if we explicitly assign it in the function call. We are
attaching the simplest code reproducing the error and the output calling the
wrapper and calling nls directly. We are using R 2.9.0
library(stats)
wrappernls <-
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())
2007 Jun 23
3
Setting up a blank table with column names in the hard drive
Dear Friends.
Greetings!
This should be a very common operation and I believe there should be a nice
way in R to handle it. I couldn't find it in the manual or by searching on
line. I am wondering if I could ask for some help in this community.
I am trying to record the results of my program to a csv file in the hard
drive so as to save memory space and also to read the results in excel