Generally nlsr package has better reliability in getting parameter estimates
because it tries to use automatic derivatives rather than a rather poor
numerical
estimate, and also uses a Levenberg-Marquardt stabilization of the linearized
model. However, nls() can sometimes be a bit more flexible.
Thanks Jeff
Bernard
Sent from my iPhone so please excuse the spelling!"
> On Apr 5, 2020, at 3:14 PM, Jeff Newmiller wrote:
>
> stats::nlm?
>
>> On April 5, 2020 11:53:10 AM PDT, Bernard Comcast
>> wrote:
>> Any recommendations on an R package to fit data to a nonlinear model
>> Y=f(x) with
err... stats::nls...
On April 5, 2020 12:14:15 PM PDT, Jeff Newmiller
wrote:
>stats::nlm?
>
>On April 5, 2020 11:53:10 AM PDT, Bernard Comcast
> wrote:
>>Any recommendations on an R package to fit data to a nonlinear model
>>Y=f(x) with a single x and y variable?
>>
>>I want to be able to gener
stats::nlm?
On April 5, 2020 11:53:10 AM PDT, Bernard Comcast
wrote:
>Any recommendations on an R package to fit data to a nonlinear model
>Y=f(x) with a single x and y variable?
>
>I want to be able to generate parameter uncertainty estimates and
>prediction uncertainties if possible.
>
>Berna
Hi Waltenegus,
you should consider to show us your code and the data you used to fit the
curve. If you can't or if you prefer you can simply show the code and some
sample data on which the code can be run.
You'll find nice and useful tool in nlstools package.
Hope this help
Giuseppe
-
Gius
You could use the summation convention built into mgcv:gam for this. See
?linear.functional.terms
for details, but here is some example code, both for the exact match,
you describe, and a noisy version. best, Simon
library(mgcv)
f2 <- function(x) 0.2*x^11*(10*(1-x))^6+10*(10*x)^3*(1-x)^10 ## te
This sounds like possibly using logsplines may be what you want. See
the 'oldlogspline' function in the 'logspline' package.
On Thu, Apr 12, 2012 at 7:45 AM, Michael Haenlein
wrote:
> Dear all,
>
> This is probably more related to statistics than to [R] but I hope someone
> can give me an idea h
ovided some technical and practical information which I could learn
>> from and be very thankful for.
>>
>> Regards,
>> Joseph
>>
>> On Fri, Apr 30, 2010 at 11:35 PM, Greg Snow
>> wrote:
>> >
>> >> -Original Message-----
>>
nal Message-
> From: Kyeong Soo (Joseph) Kim [mailto:kyeongsoo@gmail.com]
> Sent: Friday, April 30, 2010 5:24 PM
> To: Greg Snow
> Cc: r-help@r-project.org
> Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
>
> I have already learned a lot from the
lto:r-help-boun...@r-
>> project.org] On Behalf Of Kyeong Soo (Joseph) Kim
>> Sent: Friday, April 30, 2010 4:10 AM
>> To: kMan
>> Cc: r-help@r-project.org
>> Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
>
> [snip]
>
>> By the way, I won
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Kyeong Soo (Joseph) Kim
> Sent: Friday, April 30, 2010 4:10 AM
> To: kMan
> Cc: r-help@r-project.org
> Subject: Re: [R] Curve Fitting/Regression with
You can use nls2 to try many starting values. It works just like nls but:
- if you give it a two row data frame as the start value it will
create a grid between the upper and lower values of each parameter and
then run an optimization starting at each such point on the grid
returning the best
- i
Thomas,
I think the issue of having reasonable starting values is inherent in all
nonlinear optimization problems (unless they have some additional
properties like convexity, for example). Using a different algorithm may
or may not help. In fact, a vast majority of existing algorithms
guaran
gsoo@gmail.com]
> Sent: Friday, April 30, 2010 4:10 AM
> To: kMan
> Cc: r-help@r-project.org
> Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
>
> Dear Keith,
>
> Thanks for the suggestion and taking your time to respond to it.
>
> But, you misunde
n wrote:
>> > Dear Joseph,
>> >
>> > If you do not need to make any inferences, that is, you
>> just want it to look pretty, then drawing a curve by hand is
>> as good a solution as any. Plus, there is no reason for
>> expert testimony to say that the cu
eithC.
-Original Message-
From: Kyeong Soo (Joseph) Kim [mailto:kyeongsoo@gmail.com]
Sent: Friday, April 30, 2010 4:10 AM
To: kMan
Cc: r-help@r-project.org
Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
Dear Keith,
Thanks for the suggestion and taking your time to re
rve does not mean anything.
> >
> > Sincerely,
> > KeithC.
> >
> > -Original Message-
> > From: Kyeong Soo (Joseph) Kim [mailto:kyeongsoo@gmail.com]
> > Sent: Tuesday, April 27, 2010 2:33 PM
> > To: Gabor Grothendieck
> > Cc: r
m: Kyeong Soo (Joseph) Kim [mailto:kyeongsoo@gmail.com]
> Sent: Tuesday, April 27, 2010 2:33 PM
> To: Gabor Grothendieck
> Cc: r-help@r-project.org
> Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
>
> Frankly speaking, I am not looking for such a framework.
>
-
From: Kyeong Soo (Joseph) Kim [mailto:kyeongsoo@gmail.com]
Sent: Tuesday, April 27, 2010 2:33 PM
To: Gabor Grothendieck
Cc: r-help@r-project.org
Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
Frankly speaking, I am not looking for such a framework.
The system I&#
Frankly speaking, I am not looking for such a framework.
The system I'm studying is a communication network (like M/M/1 queue,
but way too complicated to mathematically analyze it using classical
queueing theory) and the conclusion I want to make is qualitative
rather than quantatitive -- a high-l
If you are looking for a framework for statistical inference you could
look at additive models as in the mgcv package which has a book
associated with it if you need more info. e.g.
library(mgcv)
fm <- gam(dist ~ s(speed), data = cars)
summary(fm)
plot(dist ~ speed, cars, pch = 20)
fm.ci <- with(
Hello Gabor,
Many thanks for providing actual examples for the problem!
In fact I know how to apply and generate plots using various R
functions including loess, lowess, and smooth.spline procedures.
My question, however, is whether applying those procedures directly on
the data with multiple ob
This will compute a loess curve and plot it:
example(loess)
plot(dist ~ speed, cars, pch = 20)
lines(cars$speed, fitted(cars.lo))
Also this directly plots it but does not give you the values of the
curve separately:
library(lattice)
xyplot(dist ~ speed, cars, type = c("p", "smooth"))
On Tue,
Joseph:
I believe you need to stop inventing your own statistical methods and
consult a professional statistician. I do not think this list is the proper
place to look for a statistics tutorial when your statistical background
appears to be so inadequate for the task.
Sorry to be so direct -- per
Pascale,
If you do want an nls fit with the associated error structure
assumptions, check ?SSlogis.
fm <- nls(y ~ SSlogis(x, Asy, xmid, scal))
summary(fm)
xx <- seq(123, 248, length = 101)
yy <- predict(fm, list(x = xx))
plot(x, y)
lines(xx, yy)
-Peter Ehlers
Gabor Grothendieck wrote:
A simple y vs log(x) fit seems to work pretty well here:
fit <- lm(y ~ log(x))
summary(fit)
plot(y ~ log(x))
abline(fit)
On Fri, Dec 4, 2009 at 9:06 AM, Pascale Weber wrote:
> Hi to all
>
> This is the first time I am quoting a question and I hope, my question is
> not too basic...
>
> For the
or use nls.lm as in
install.packages("minpack.lm")
library(minpack.lm)
x <- c(2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74)
y <- c(100, 99, 99, 98, 97, 94, 82, 66, 48, 38, 22, 10, 1)
res <- function(p, x, y) y - ff(p,x)
ff <- function(p, x) 100*exp(p[1]*(1-exp(p[2]*x))/p[2])
aa <- nls.lm(par
Dear Dmitry,
Take a look at ?nls and its examples.
HTH,
Jorge
On Tue, May 12, 2009 at 5:44 PM, Dmitry Gospodaryov
wrote:
> I have the data:
> for x: 2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74,
> for y: 100, 99, 99, 98, 97, 94, 82, 66, 48, 38, 22, 10, 1.
> y depends on x by equation: y =
Hi Thanks a lot,
I think you have covered the things I want to do for now so I will try to
implement them as soon I can.
<< A finite Fourier series could be the best tool IF the the multiple
periodicities are all integer fractions of a common scale.>>
This is certainly true for my repetitive
Dear Dr Gkikopoulos:
1. Have you looked at "bioconductor.org"? They have substantive
extensions to R specifically for "genomic data", which I assume would
include chromosome.
2. To "identify periodicities at different timescales", I agree
with Stephen that "spectrum" would l
There are a couple of different goals for this projects
*identify periodicities at different timescales (ie different dT)
*fit data into discrete number of curves, ie 6 different basic functions
should be enough to describe the basic repeating elements in this data (ie 6
different categories of
What is your end goal? If it is to try and account for the
variability of the "timeseries" you may want to look at ?spectrum
If it is to model the periodicity...
Stephen Sefick
On Fri, Apr 3, 2009 at 11:30 AM, trias wrote:
>
> Here is the gif that didn't come through earlier
> http://www.nabble
Here is the gif that didn't come through earlier
http://www.nabble.com/file/p22870832/signal.gif signal.gif
--
View this message in context:
http://www.nabble.com/Curve-fitting%2CFDA-for-biological-data-tp22868069p22870832.html
Sent from the R help mailing list archive at Nabble.com.
_
gregor rolshausen biologie.uni-freiburg.de> writes:
>
> ok. sorry for being blurry.
>
> I have x,y data, that probably fits a asymptotic curve (asymptote at N).
> now I want to fit a curve onto the data, that gives me the N. therefore
> I thought to fit an e-function, namely N(1-e^(y/x)) onto
ok. sorry for being blurry.
I have x,y data, that probably fits a asymptotic curve (asymptote at N).
now I want to fit a curve onto the data, that gives me the N. therefore
I thought to fit an e-function, namely N(1-e^(y/x)) onto the data and
get the N from the fitted curves' equation.
in the
gregor rolshausen wrote:
hello,
I want to fit a curve to a simple x,y dataset - my problem is, that I
want to fit it for the following term:
n(1-e^x/y) - so I get the n constant for my data...
Not an R problem in the first place, but the question arises what
"n(1-e^x/y)" means, its is jus
Your model is singular. Varying m and log(l) have the same
effect: log(ir) = log(k) + m * log(l) * ox
Also with plinear you don't specify the linear coefficients but
rather an X matrix whose coefficients represent them:
If we use this model instead:
ir = k * exp(m * ox)
Then:
> mod0 <- lm(log(
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