Displaying 20 results from an estimated 3000 matches similar to: "nls.regression syntax problem"
2005 Apr 01
1
optim problem, nls regression
Hi,
I try to fit a non linear regression by minimising the sum of the sum of
squares.
The model is number[2]-(x/number[1])^number[3]
Number [2] and number [1] change as the data changes but for all the set of
data number[3] must be identical.
I have 3 set of data (x1,y1), (x2,y2), (x3,y3).
x_a<-c(0,0.5,1,1.5,2,3,4,6)
y_a<-c(5.4,5,4.84,4.3,4,2,1.56,1.3)
2005 Apr 08
0
TR: The results of your email commands
Hi,
I try to minimize the sum of the sum of sce. The following program has
been
created but it only takes in consideration the last kinetic and not the
first ones. I think that I have forget a subscrib but I don't know
where. so
if you can help me, it will be great....
I have try an other program where y and x are directly calculate in sce
function but the time of
2008 Feb 26
0
NLS -- multiplicative errors and group comparison
Hello,
I am attempting to fit a non-linear model (Von Bertalanffy growth model)
to fish length-at-age data with the purpose of determining if any of the
three parameters differ between male and female fish. I believe that I
can successfully accomplish this goal assuming an additive error
structure (illustrated in section 1 below). My trouble begins when I
attempt this analysis using a model
2003 Aug 14
0
Bug in numericDeriv (was: [R] nls confidence intervals) (PR#3746)
Moved from r-help:
On Thu, 14 Aug 2003 09:08:26 -0700, Spencer Graves
<spencer.graves@pdf.com> wrote :
>p.s. The following command in S-Plus 6.1 seems to work fine but
>produces an error in R 1.7.1:
>
>nls(y~a, data=tstDf, start=list(a=1))
>Error in nlsModel(formula, mf, start) : singular gradient matrix at
>initial parameter estimates
This looks like a bug in
2010 Oct 13
2
Using NLS with a Kappa function
Hi Everyone,
I am trying to use NLS to fit a dataset using a Kappa function, but I am
having problems. Depending on the start values that I provide, I get
either:
Error in numericDeriv(form[[3L]], names(ind), env) :
Missing value or an infinity produced when evaluating the model
Or
Error in nls(FldFatRate ~ funct3(MeanDepth_m, h, k, z, a), data = data1, :
singular gradient
I think these
2003 Mar 26
1
nls
Hi,
df <- read.table("data.txt", header=T);
library(nls);
fm <- nls(y ~ a*(x+d)^(-b), df, start=list(a=max(df->y,na.rm=T)/2,b=1,d=0));
I was using the following routine which was giving Singular Gradient, Error in
numericDeriv(form[[3]], names(ind), env) :
Missing value or an Infinity produced when evaluating the model errors.
I also tried the
2004 Mar 12
0
Basic questions on nls and bootstrap
Dear R community,
I have currently some problems with non linear regression analysis in R.
My data correspond to the degradation kinetic of a pollutant in two
different soil A and B, x data are time in day and y data are pollutant
concentration in soil.
In a first time, I want to fit the data for the soil A by using the Ct =
C0*exp(-k*Tpst) with Ct the concentration of pollutant at time t, C0
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
2003 Jun 27
2
nls question
I'm running into problems trying to use the nls function to fit the some
data. I'm invoking nls using
nls(s~k/(a+r)^b, start=list(k=1, a=13, b=0.59))
but I get errors indicating that the step has been reduced below the
minimum step size or an inifinity is generated in numericDeriv. I've
tried to use a variety of starting values for a, b, k but get similar
errors.
Is there
2005 Aug 18
1
How to put factor variables in an nls formula ?
Hello,
I want to fit a Gompertz model for tree diameter growth that depends on a 4
levels edaphic factor (?Drain?) and I don?t manage to introduce the factor
variable in the formula.
Dinc is the annual diameter increment and D is the Diameter.
>treestab
> Dinc D Drain
[1,] 0.03 26.10 2
[2,] 0.04 13.05 1
[3,] 0.00 24.83 1
[4,] 0.00 15.92 4
2004 Feb 19
1
controlling nls errors
Hello. I am using the nonlinear least squares function (nls). The
function that I am trying to fit seems to be very sensitive to the
starting values and, if these are not chosen properly, the nls function
stops and gives an error message:
Error in numericDeriv(form[[3]], names(ind), env) :
Missing value or an Infinity produced when evaluating the model
In addition: Warning
2006 Nov 08
1
nls
> y
[1] 1 11 42 64 108 173 214
> t
[1] 1 2 3 4 5 6 7
> nls(1/y ~ c*exp(-a*b*t)+1/b, start=list(a=0.001,b=250,c=5), trace=TRUE)
29.93322 : 0.001 250.000 5.000
Error in numericDeriv(form[[3]], names(ind), env) :
Missing value or an infinity produced when evaluating the model
# the start value for b is almost close to final estimates,
# a is usually
2011 Oct 01
1
Problem with logarithmic nonlinear model using nls() from the `stats' package
Example:
> f <- function(x) { 1 + 2 * log(1 + 3 * x) + rnorm(1, sd = 0.5) }
> y <- f(x <- c(1 : 10)); y
[1] 4.503841 5.623073 6.336423 6.861151 7.276430 7.620131 7.913338 8.169004
[9] 8.395662 8.599227
> nls(x ~ a + b * log(1 + c * x), start = list(a = 1, b = 2, c = 3), trace = TRUE)
37.22954 : 1 2 3
Error in numericDeriv(form[[3L]], names(ind), env) :
Missing value or an
2011 Sep 20
1
NLS error
Hello there,
I am using NLS for fitting a complex model to some data to estimate a couple
of the missing parameters. The model is -
y ~ (C+((log(1-r))*exp(-A*d)*(-1+r+exp(d*(A-B)))/(r*(-A*d+d*B+log(1-r)))))
where A, B and C are unknown.
In order to test the model, I generate data by setting values for all
parameters and add some noise (C).
A <- 20
B <- 500
r <- 0.6881
d <-
2012 Dec 16
1
nls for sum of exponentials
Hi there,
I am trying to fit the following model with a sum of exponentials -
y ~ Ae^(-md) + B e^(-nd) + c
the model has 5 parameters A, b, m, n, c
I am using nls to fit the data and I am using DEoptim package to pick the
most optimal start values -
fm4 <- function(x) x[1] + x[2]*exp(x[3] * -dist) + x[4]*exp(x[5] * -dist)
fm5 <- function(x) sum((wcorr-fm4(x))^2)
fm6 <- DEoptim(fm5,
2012 Jan 30
1
Problem in Fitting model equation in "nls" function
Dear R users,
I am struggling to fit expo-linear equation to my data using "nls" function. I am always getting error message as i highlighted below in yellow color:
### Theexpo-linear equation which i am interested to fit my data:
response_variable = (c/r)*log(1+exp(r*(Day-tt))), where "Day" is time-variable
## my response variable
rl <-
2005 Nov 08
1
Can someone Help in nls() package
Hello R-Community,
we are running aprogram to fit Non-linear differential equations to Aphid
population Data and to estimate the birth and death parameters,
here is the code:
dat<-data.frame(Time=c(0:60),Cur=c(5,6.2,59,39,38,44,20.4,19.4,34.2,35.4,38.2,48.2,55.4,113.2,
97,112,115,126,136.6,140.6,147.2,151.6,157.8,170,202,210.4,221.2,224.4,248.2,266,
2008 May 06
2
NLS plinear question
Hi All.
I've run into a problem with the plinear algorithm in nls that is confusing
me.
Assume the following reaction time data over 15 trials for a single unit.
Trials are coded from 0-14 so that the intercept represents reaction time in
the first trial.
trl RT
0 1132.0
1 630.5
2 1371.5
3 704.0
4 488.5
5 575.5
6 613.0
7 824.5
8 509.0
9
2003 May 08
1
nls, restrict parameter values
Hi,
I posted a question (bellow) a few weeks ago and had a reply (thanks
Christian) that partly solves the problem, but I still would like to be able
to restrict some of the independent variables in a nls model to be always
>0, (is there a way to do it)??
Thanks,
Angel
>From: "Christian Ritz" <ritz at dina.kvl.dk>
>To: "Angel -" <angel_lul at
2010 Jan 13
1
Problem fitting a non-linear regression model with nls
Hi,
I'm trying to make a regression of the form :
formula <- y ~ Asym_inf + Asym_sup * ( (1 / (1 + (n1 * (exp( (tmid1-x)
/ scal1) )^(1/n1) ) ) ) - (1 / (1 + (n2 * (exp( (tmid2-x) / scal2)
)^(1/n2) ) ) ) )
which is a sum of the generalized logistic model proposed by richards.
with data such as these:
x <- c(88,113,128,143,157,172,184,198,210,226,240,249,263,284,302,340)
y <-