Sendt:* 2. april 2020 18:50
> *Til:* Troels Ring
> *Cc:* r-help mailing list
> *Emne:* Re: [R] nls problem
>
>
>
> Roundoff/cancelation error: compare the following. The first is
> equivalent to your function, the last to fitted().
>
>
>
> > with(aedf
alone saves me.
BW
Troels
Fra: William Dunlap
Sendt: 2. april 2020 18:50
Til: Troels Ring
Cc: r-help mailing list
Emne: Re: [R] nls problem
Roundoff/cancelation error: compare the following. The first is equivalent to
your function, the last to fitted().
> with(aedf, t(cbind(1,
Yes, I was waiting to see how long before it would be noticed that this is
not the sort of problem for which nls() is appropriate.
And I'll beat the drum again that nls() uses a simple (and generally
deprecated) forward difference derivative approximation that gets into
trouble a VERY high proport
Roundoff/cancelation error: compare the following. The first is equivalent
to your function, the last to fitted().
> with(aedf, t(cbind(1, pH, pH^2) %*% round(coef(m), digits=2)))
[,1] [,2] [,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,
Simpler:
FPG2 <- function(x, model){
as.vector(cbind(1, x, x^2) %*% coef(model))
}
Hope this helps,
Rui Barradas
Às 12:56 de 02/04/20, Rui Barradas escreveu:
Hello,
Sorry, disregard my previous e-mail.
Instead of your FPG function try
FPG <- function(pH, model) {
coef(model)[1] + co
Hello,
Sorry, disregard my previous e-mail.
Instead of your FPG function try
FPG <- function(pH, model) {
coef(model)[1] + coef(model)[2]*pH + coef(model)[3]*pH^2
}
FPG(aedf$pH, m)
fitted(m)
Hope this helps,
Rui Barradas
Às 12:30 de 02/04/20, Rui Barradas escreveu:
Hello,
Tip: in a for
Hello,
Tip: in a formula, pH^2 = pH*pH is an interaction.
pH^2 = pH*pH = pH + pH + pH:pH = pH
Try I(pH^2)
Hope this helps,
Rui Barradas
Às 12:07 de 02/04/20, Troels Ring escreveu:
Dear friends - I'm on Win10 with R 6.3.1 and have a very simple problem with
nls which apparently gives a decen
Dear friends - I'm on Win10 with R 6.3.1 and have a very simple problem with
nls which apparently gives a decent fit to the parable below, even without
starting values. But when I then think I know the meaning of the three
parameters a, b, and d it goes very wrong. I guess I am again overlooking
so
Would you show the exact message you get please.
-- Bert
On Thu, Jan 3, 2013 at 5:22 AM, Karine HEERAH <
karine.hee...@locean-ipsl.upmc.fr> wrote:
> Hi,
>
> I am using the nls function and it stops because the number of iterations
> exceeded 50, but i used the nls.control argument to allow for 5
Hi,
I am using the nls function and it stops because the number of
iterations exceeded 50, but i used the nls.control argument to allow for
500 iterations. Do you have any idea why it's not working?
fm1 <- nls(npe ~ SSgompertz(npo, Asym, b2, b3),
data=f,control=nls.control(maxiter=500))
T
ace=TRUE)
--
Best, JN
On 07/12/2012 06:00 AM, r-help-requ...@r-project.org wrote:
> From: Jonas Stein
> To:
> Subject: [R] nls problem: singular gradient
> Message-ID:
> Content-Type: text/plain
>
> Why fails nls with "singular gradient&qu
On Jul 11, 2012, at 20:34 , Jonas Stein wrote:
>> Take a look at the predicted values at your starting fit: there's a
>> discontinuity at 0.4, which sure makes it look as though overflow is
>> occurring. I'd recommend expanding tanh() in terms of exponentials and
>> rewrite the prediction in
On 07/12/2012 01:39 AM, Duncan Murdoch wrote:
On 12-07-11 2:34 PM, Jonas Stein wrote:
Take a look at the predicted values at your starting fit: there's a
discontinuity at 0.4, which sure makes it look as though overflow is
occurring. I'd recommend expanding tanh() in terms of exponentials and
re
On 12-07-11 2:34 PM, Jonas Stein wrote:
Take a look at the predicted values at your starting fit: there's a
discontinuity at 0.4, which sure makes it look as though overflow is
occurring. I'd recommend expanding tanh() in terms of exponentials and
rewrite the prediction in a way that won't over
> Take a look at the predicted values at your starting fit: there's a
> discontinuity at 0.4, which sure makes it look as though overflow is
> occurring. I'd recommend expanding tanh() in terms of exponentials and
> rewrite the prediction in a way that won't overflow.
>
> Duncan Murdoch
Hi Du
On 11/07/2012 11:04 AM, Jonas Stein wrote:
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.
Take a look at the predicted values at your starting fit: there's a
discontinuity at 0.4, which sure makes it look a
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)/sm
gt; Date: Tue, 03 Jul 2012 13:54:11 +0200
> From: "joerg van den hoff"
> To: r-help@r-project.org
> Subject: [R] nls problem
> Message-ID:
> Content-Type: text/plain; charset=iso-8859-15; format=flowed;
> delsp=yes
>
> hi list,
>
> used versions: 2.12.1
hi list,
used versions: 2.12.1 and 2.14.0 under ubuntu and macosx.
I recently stumbled over a problem with `nls', which occurs if the model
is not specified explicitly but via an evaluation of a 'call' object.
simple example:
8<--
I am sorry,Andrew,I don't get you.
Please forgive my poor English.
--
View this message in context:
http://r.789695.n4.nabble.com/nls-problem-with-R-tp3494454p3508131.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org
Thanks Mike.
Your suggestion is really helpful.I did with the your instruction , it
really works out.
What's more,can you use this package
http://cran.r-project.org/web/packages/minpack.lm/index.html
it use Levenberg-Marquardt algorithm.
Can this package do with four parameters?
Thanks again
--
Vi
> Date: Thu, 5 May 2011 01:20:33 -0700
> From: sterles...@hotmail.com
> To: r-help@r-project.org
> Subject: Re: [R] nls problem with R
>
> ID1 ID2 t V(t)
> 1 1 0 6.053078443
> 2 1 0.3403 5.56937391
> 3 1 0.4181 5.454844
Apologies, but I don't see a question here ... am I missing something
obvious?
Andrew
On Thu, May 05, 2011 at 01:20:33AM -0700, sterlesser wrote:
> ID1 ID2 t V(t)
> 1 1 0 6.053078443
> 2 1 0.3403 5.56937391
> 3 1 0.4181 5.45484486
>
ID1 ID2 t V(t)
1 1 0 6.053078443
2 1 0.3403 5.56937391
3 1 0.4181 5.45484486
4 1 0.4986 5.193124598
5 1 0.7451 4.31386722
6 1 1.0069 3.645422269
7 1 1.5535 3.587710965
8
the dataset's form is changed after my post
so I repost it here
t
0
0.3403
0.4181
0.4986
0.7451
1.0069
1.5535
1.8049
2.4979
6.4903
13.5049
27.5049
41.5049
V(t)
6.053078443
5.56937391
5.45484486
5.193124598
4.31386722
3.645422269
3.587710965
3.740362689
3.699837726
2.908485019
1.888179494
1.17609
> Date: Wed, 4 May 2011 07:07:44 -0700
> From: sterles...@hotmail.com
> To: r-help@r-project.org
> Subject: Re: [R] nls problem with R
>
> Thanks Andrew.
> I am sorry for some typos that I omit some numbers of T2.
> Based on your suggestion,I think the problem is in the
-project.org] On
Behalf Of sterlesser
Sent: Wednesday, May 04, 2011 10:08 AM
To: r-help@r-project.org
Subject: Re: [R] nls problem with R
Thanks Andrew.
I am sorry for some typos that I omit some numbers of T2.
Based on your suggestion,I think the problem is in the initial values.
And I will read more
Thanks Andrew.
I am sorry for some typos that I omit some numbers of T2.
Based on your suggestion,I think the problem is in the initial values.
And I will read more theory about the non-linear regression.
--
View this message in context:
http://r.789695.n4.nabble.com/nls-problem-with-R-tp3494454p
Thanks Ruben.
Your suggestion about more deeper analysis about the model itself is really
helpful.
I am trying out some new initial values based on the analysis of the special
T2 in the model.
--
View this message in context:
http://r.789695.n4.nabble.com/nls-problem-with-R-tp3494454p3495663.html
-Mensaje original-
> De: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] En nombre de Andrew Robinson
> Enviado el: miércoles, 04 de mayo de 2011 9:15
> Para: sterlesser
> CC: r-help@r-project.org
> Asunto: Re: [R] nls problem with R
>
> The fac
The fact that T2 and V2 are of different lengths seems like a likely
culprit. Other than that, you need to find start points that do not
lead to a singular gradient. There are several books that provide
advice on obtaining initial parameter estimates for non-linear
models. Google Books might hel
the original data are
V2 =c(371000,285000 ,156000, 20600, 4420, 3870, 5500 )
T2=c( 0.3403 ,0.4181 ,0.4986 ,0.7451 ,1.0069 ,1.553)
nls2=nls(V2~v0*(1-epi+epi*exp(-cl*(T2-t0))),start=list(v0=10^7,epi=0.9,cl=6.2,t0=8.7))
after execution error occurs as below
Would appreciate some help on this problem please:
Using nls to solve for 5 parameters to a set of data, and for particular
mixes of the parameters the function can yield NaN and infinity returns.
Is is making nls fall over I have tried starting values, lower and upper
bounds to no avail. New t
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