Displaying 20 results from an estimated 27 matches for "nlsr".
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2023 Nov 06
2
non-linear regression and root finding
Dear friends - I have a function for the charge in a fluid (water)
buffered with HEPES and otherwise only containing Na and Cl so that [Na]
- [Cl] = SID (strong ion difference) goes from -1 mM to 1 mM. With known
SID and total HEPES concentration I can calculate accurately the pH if I
know 3 pK values for HEPES by finding the single root with uniroot
Now, the problem is that there is some
2017 Jun 18
0
R_using non linear regression with constraints
...are useful,
but it takes some time to sort them all out and make sense of the problem as a whole. Right now I'm getting ready
to go to UseR!, so probably won't have time to spend sorting things out, though I'll have a go if I get time.
David: Did you get a crash with the Mac version of nlsr? (You have the latest version I uploaded to CRAN.)
I don't have Mac, just Linux and a very ancient Win XP. I suspect the latter is too old to take seriously. I use
Win-Builder for package checks. I don't think there's anything seriously wrong with nlsr functions, but there is
more to b...
2017 Jun 18
3
R_using non linear regression with constraints
...# Green point is brute-force solution
# Red point is optimizer solution for myfun2
##----------end
On Sun, 18 Jun 2017, J C Nash wrote:
> I ran the following script. I satisfied the constraint by
> making a*b a single parameter, which isn't always possible.
> I also ran nlxb() from nlsr package, and this gives singular
> values of the Jacobian. In the unconstrained case, the svs are
> pretty awful, and I wouldn't trust the results as a model, though
> the minimum is probably OK. The constrained result has a much
> larger sum of squares.
>
> Notes:
> 1) nls...
2017 Jun 18
0
R_using non linear regression with constraints
I ran the following script. I satisfied the constraint by
making a*b a single parameter, which isn't always possible.
I also ran nlxb() from nlsr package, and this gives singular
values of the Jacobian. In the unconstrained case, the svs are
pretty awful, and I wouldn't trust the results as a model, though
the minimum is probably OK. The constrained result has a much
larger sum of squares.
Notes:
1) nlsr has been flagged with a check er...
2023 Aug 19
1
Determining Starting Values for Model Parameters in Nonlinear Regression
Thank you so much Dr. Nash, I truly appreciate your kind and valuable
contribution.
Cheers,
Paul
El El s?b, 19 de ago. de 2023 a la(s) 3:35 p. m., J C Nash <
profjcnash at gmail.com> escribi?:
> Why bother. nlsr can find a solution from very crude start.
>
> Mixture <- c(17, 14, 5, 1, 11, 2, 16, 7, 19, 23, 20, 6, 13, 21, 3, 18, 15,
> 26, 8, 22)
> x1 <- c(69.98, 72.5, 77.6, 79.98, 74.98, 80.06, 69.98, 77.34, 69.99,
> 67.49, 67.51, 77.63,
> 72.5, 67.5, 80.1, 69.99, 72.49, 64....
2017 Jun 18
3
R_using non linear regression with constraints
https://cran.r-project.org/web/views/Optimization.html
(Cran's optimization task view -- as always, you should search before posting)
In general, nonlinear optimization with nonlinear constraints is hard,
and the strategy used here (multiplying by a*b < 1000) may not work --
it introduces a discontinuity into the objective function, so
gradient based methods may in particular be
2023 Jan 26
1
Potential bug in fitted.nls
Doesn't nls() expect that the lengths of vectors on both sides of the
formula match (if both are supplied)? Perhaps it should check for that.
-Bill
On Thu, Jan 26, 2023 at 12:17 AM Dave Armstrong <darmst46 at uwo.ca> wrote:
> Dear Colleagues,
>
> I recently answered [this question]() on StackOverflow that identified
> what seems to be unusual behaviour with
2023 Aug 20
1
Determining Starting Values for Model Parameters in Nonlinear Regression
...te:
> Thank you so much Dr. Nash, I truly appreciate your kind and valuable contribution.
>
> Cheers,
> Paul
>
> El El s?b, 19 de ago. de 2023 a la(s) 3:35 p.?m., J C Nash <profjcnash at gmail.com <mailto:profjcnash at gmail.com>> escribi?:
>
> Why bother. nlsr can find a solution from very crude start.
>
> Mixture <- c(17, 14, 5, 1, 11, 2, 16, 7, 19, 23, 20, 6, 13, 21, 3, 18, 15, 26, 8, 22)
> x1 <- c(69.98, 72.5, 77.6, 79.98, 74.98, 80.06, 69.98, 77.34, 69.99, 67.49, 67.51, 77.63,
> ? ? ? ? ?72.5, 67.5, 80.1, 69.99, 72.49,...
2017 Feb 17
4
Wish List: Extensions to the derivatives table
The derivative table resides in the function D. In S+ that table is extensible because it is written in the S language. R is faster but less flexible, since that table is programmed in C. It would be useful if R provided a mechanism for extending the derivative table, or barring that, provided a broader table. Currently unsupported mathematical functions of one argument include expm1, log1p,
2023 Aug 19
1
Determining Starting Values for Model Parameters in Nonlinear Regression
Dear friends,
Hope you are all doing well and having a great weekend. I have data that
was collected on specific gravity and spectrophotometer analysis for 26
mixtures of NG (nitroglycerine), TA (triacetin), and 2 NDPA (2 -
nitrodiphenylamine).
In the dataset, x1 = %NG, x2 = %TA, and x3 = %2 NDPA.
The response variable is the specific gravity, and the rest of the
variables are the predictors.
2018 Apr 07
0
Obtain gradient at multiple values for exponential decay model
I have never found the R symbolic differentiation helpful because my
functions are typically quite complicated, but was prompted by Steve
Ellison's suggestion to try it out in this case:
################# reprex (see reprex package)
graphdta <- read.csv( text =
"t,c
0,100
40,78
80,59
120,38
160,25
200,21
240,16
280,12
320,10
360,9
400,7
", header = TRUE )
nd <- c( 100, 250,
2023 Nov 06
1
non-linear regression and root finding
...the Jacobian of the model function cannot be inverted and fails with
>> the message "Singular gradient". I wish that R could have a more
>> reliable built-in nonlinear least squares solver. (I could also be
>> holding it wrong.) Meanwhile, we have excellent CRAN packages nlsr and
>> minpack.lm:
>>
>> minpack.lm::nlsLM(
>> ? pHobs ~ pHm(SID, pK1, pK2, pK3),
>> ? data.frame(pHobs = pHobs, SID = SID),
>> ? start = c(pK1 = pK1, pK2 = pK2, pK3 = pK3),
>> ? # the following is also needed to avoid MINPACK failing to fit
>> ? lowe...
2023 Nov 06
1
non-linear regression and root finding
...where where crossprod()
of the Jacobian of the model function cannot be inverted and fails with
the message "Singular gradient". I wish that R could have a more
reliable built-in nonlinear least squares solver. (I could also be
holding it wrong.) Meanwhile, we have excellent CRAN packages nlsr and
minpack.lm:
minpack.lm::nlsLM(
pHobs ~ pHm(SID, pK1, pK2, pK3),
data.frame(pHobs = pHobs, SID = SID),
start = c(pK1 = pK1, pK2 = pK2, pK3 = pK3),
# the following is also needed to avoid MINPACK failing to fit
lower = rep(-1, 3), upper = rep(9, 3)
)
# Nonlinear regression model
# model:...
2023 Nov 06
2
non-linear regression and root finding
...d()
> of the Jacobian of the model function cannot be inverted and fails with
> the message "Singular gradient". I wish that R could have a more
> reliable built-in nonlinear least squares solver. (I could also be
> holding it wrong.) Meanwhile, we have excellent CRAN packages nlsr and
> minpack.lm:
>
> minpack.lm::nlsLM(
> pHobs ~ pHm(SID, pK1, pK2, pK3),
> data.frame(pHobs = pHobs, SID = SID),
> start = c(pK1 = pK1, pK2 = pK2, pK3 = pK3),
> # the following is also needed to avoid MINPACK failing to fit
> lower = rep(-1, 3), upper = rep(9, 3...
2017 Jun 21
1
fitting cosine curve
..., 16.99, 17.10)
t=c(7, 37, 58, 79, 96, 110, 114, 127, 146, 156, 161, 169, 176, 182,
190, 197, 209, 218, 232, 240)
lidata <- data.frame(y=y, t=t)
#I use the method to fit a curve, but it is different from the real curve,
#which can be seen in the figure.
linFit <- lm(y ~ cos(t))
library(nlsr)
#fullFit <- nls(y ~ A*cos(omega*t+C) + B,
#start=list(A=coef(linFit)[1],B=coef(linFit)[2],C=0,omega=.4))
#omega cannot be set to 1, don't know why.
fullFit <- nlxb(y ~ A*cos(omega*t+C) + B, data=lidata,
start=list(A=coef(linFit)[1],B=coef(linFit)[2],C=0,omega=.04), trace=TRUE)
co <-...
2018 Apr 06
3
Obtain gradient at multiple values for exponential decay model
> On Apr 6, 2018, at 8:03 AM, David Winsemius <dwinsemius at comcast.net> wrote:
>
>
>> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote:
>>
>>> Sent: Friday, April 06, 2018 at 5:55 AM
>>> From: "David Winsemius" <dwinsemius at comcast.net>
>>>
>>>
>>> Not correct. You already have
2023 Aug 21
2
Interpreting Results from LOF.test() from qpcR package
...n from the qpcR package and got the following
result:
> LOF.test(nlregmod3)
$pF
[1] 0.97686
$pLR
[1] 0.77025
Can I conclude from the LOF.test() results that my nonlinear regression
model is significant/statistically significant?
Where my nonlinear model was fitted as follows:
nlregmod3 <- nlsr(formula=y ~ theta1 - theta2*exp(-theta3*x), data =
mod14data2_random,
start = list(theta1 = 0.37,
theta2 = -exp(-1.8),
theta3 = 0.05538))
And the data used to fit this model is the following:
dput(mod14data2_random)
structure(list(index = c(14L, 27L, 37L,...
2017 Jun 21
1
fitting cosine curve
If you know the period and want to fit phase and amplitude, this is
equivalent to fitting a * sin + b * cos
> >>> > I don't know how to set the approximate starting values.
I'm not sure what you meant by that, but I suspect it's related to
phase and amplitude.
> >>> > Besides, does the method work for sine curve as well?
sin is the same as cos with
2017 Jun 21
0
fitting cosine curve
I'm trying the different parameters, but don't know what the error is:
Error in nlsModel(formula, mf, start, wts) :
singular gradient matrix at initial parameter estimates
Thanks for any suggestions.
On Tue, Jun 20, 2017 at 7:37 PM, Don Cohen <don-r-help at isis.cs3-inc.com>
wrote:
>
> If you know the period and want to fit phase and amplitude, this is
> equivalent to
2017 Oct 20
0
nls() and loop
?tryCatch
--
Sent from my phone. Please excuse my brevity.
On October 20, 2017 7:37:12 AM PDT, Evangelina Viotto <evangelinaviotto at gmail.com> wrote:
>Hello I?m need fitt growth curve with data length-age. I want to
>evaluate
>which is the function that best predicts my data, to do so I compare
>the
>Akaikes of different models. I'm now need to evaluate if changing the