On Wed, 21 Jun 2017, J C Nash wrote:
Using a more stable nonlinear modeling tool will also help, but key is to get
the periodicity right.
The model is linear up to omega after transformation as Don and I noted.
Taking a guess that 2*pi/240 = 0.0262 is about right for omega:
rsq <-
Using a more stable nonlinear modeling tool will also help, but key is to get
the periodicity right.
y=c(16.82, 16.72, 16.63, 16.47, 16.84, 16.25, 16.15, 16.83, 17.41, 17.67,
17.62, 17.81, 17.91, 17.85, 17.70, 17.67, 17.45, 17.58, 16.99, 17.10)
t=c(7, 37, 58, 79, 96, 110, 114, 127, 146, 156,
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
On Tue, 20 Jun 2017, lily li wrote:
Hi R users,
I have a question about fitting a cosine curve. I don't know how to set the
approximate starting values.
See
Y.L. Tong (1976) Biometrics 32:85-94
The method is known as `cosinor' analysis. It takes advantage of the
*intrinsic*
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
wrote:
>
> If you
Thanks. I will do a trial first. Also, is it okay to have the datasets that
have only part of the cycle, or better to have equal or more than one
cycle? That is to say, I cannot have the complete datasets sometimes.
On Tue, Jun 20, 2017 at 7:37 PM, Don Cohen
wrote:
What I did was to plot your initial values, then plot the smoothed
values and guess the constants. That is, I got an "eyeball" fit to the
smoothed values. As I have described this as "gross cheating" in the
past, you should either split your data, estimate on one subset and
then test on another,
Thanks, that is cool. But would there be a way that can approximate the
curve by trying more starting values automatically?
On Tue, Jun 20, 2017 at 5:45 PM, Jim Lemon wrote:
> Hi lily,
> You can get fairly good starting values just by eyeballing the curves:
>
> plot(y)
>
Hi lily,
You can get fairly good starting values just by eyeballing the curves:
plot(y)
lines(supsmu(1:20,y))
lines(0.6*cos((1:20)/3+0.6*pi)+17.2)
Jim
On Wed, Jun 21, 2017 at 9:17 AM, lily li wrote:
> Hi R users,
>
> I have a question about fitting a cosine curve. I don't
Hi R users,
I have a question about fitting a cosine curve. I don't know how to set the
approximate starting values. Besides, does the method work for sine curve
as well? Thanks.
Part of the dataset is in the following:
y=c(16.82, 16.72, 16.63, 16.47, 16.84, 16.25, 16.15, 16.83, 17.41, 17.67,
10 matches
Mail list logo