:* 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(c
And that 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
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]
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] +
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
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
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
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))
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 500
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
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 a way
)
--
Best, JN
On 07/12/2012 06:00 AM, r-help-requ...@r-project.org wrote:
From: Jonas Stein n...@jonasstein.de
To: r-h...@stat.math.ethz.ch
Subject: [R] nls problem: singular gradient
Message-ID: e8h0d9-ao4@news.jonasstein.de
Content-Type: text/plain
Why fails
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)
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 as
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 Duncan,
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
den hoff j.van_den_h...@hzdr.de
To: r-help@r-project.org
Subject: [R] nls problem
Message-ID: op.wgvcolzs24o...@marco.fz-rossendorf.de
Content-Type: text/plain; charset=iso-8859-15; format=flowed;
delsp=yes
hi list,
used versions: 2.12.1 and 2.14.0 under ubuntu and macosx.
I
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:
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.
__
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
--
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
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
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
4
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 5.45484486
4 1 0.4986 5.193124598
5 1 0.7451
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
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
-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 fact that T2 and V2 are of different lengths seems like a
likely culprit. Other than
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:
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:
-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
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 initial values.
And I will read
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