[R] non-linear plot parameters

[Peter Ehlers]

Just a small fix to my solution; inserted below.

On 2010-08-27 3:51, Peter Ehlers wrote:
> On 2010-08-26 15:52, Marlin Keith Cox wrote:
>> I agree. I typically do not use non-linear functions, so am seeing the
>> "art" in describing functions of non-linear plots. One last thing. I
>> tried
>> to use a self-starting Weibull function with the posted data and received
>> the following error.
>>
>> model<-nls(Level~ SSweibull(Time,Asym,Drop,lrc,pwr))
>> Error in qr.default(.swts * attr(rhs, "gradient")) :
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>> I do not understand the error statement.
>
> The problem is the zeros in your independent variable (see
> the definition of coefficient 'lrc'). You can try using
> SSweibull(Time + 0.000001, ...). But a better way would
> be to use
>
> init <- getInitial(Level~ SSweibull(Time,Asym,Drop,lrc,pwr))

This needs a 'data=' argument for getInitial(). I had put
the variables in a data.frame, then modified the code by
deleting 'data=dat', forgetting that getInitial() does not
default to looking in the global environment for its data.
Easily fixed with

  init <- getInitial(Level~ SSweibull(Time,Asym,Drop,lrc,pwr),
                     data=.GlobalEnv)

or, of course, with an appropriate data.frame.


   -Peter Ehlers

>
> (which will often give the converged values). You can
> follow up with
>
> fm <- nls(Level ~ Asym-Drop*exp(-exp(lrc)*Time^pwr), start = init)
>
> -Peter Ehlers
>
>>
>> kc
>> On Thu, Aug 26, 2010 at 1:44 PM, Bert Gunter<gunter.berton at gene.com>
>> wrote:
>>
>>> My opinions only below; consume at your own risk.
>>>
>>> On Thu, Aug 26, 2010 at 2:20 PM, Marlin Keith Cox<marlinkcox at gmail.com>
>>> wrote:
>>>> The background you requested are energetic level (joules) in a group of
>>>> starved fish over a time period of 45 days. Weekly, fish (n=5) were
>>> removed
>>>> killed and measured for energy. This was done at three temperatures. I
>>> am
>>>> comparing the rates at which the fish consume stored body energy at
>>>> each
>>> of
>>>> the three temperatures. Initial data looks like the colder fish
>>>> have different rates (as would be expected) than do warmer fish. In all
>>>> cases the slope is greatest at the beginning of the curve and flattens
>>> after
>>>> several weeks. This is what is interesting - where in time the line
>>>> starts to flatten out.
>>>>
>>>> By calculating a non-linear equation of a line, I was hoping to use the
>>>> first and second derivatives of the function to compare and explain
>>>> differences between the three temperature.
>>>
>>> Bad idea. Derivatives from fitted curves are generally pretty
>>> imprecisely determined. And you don't need them: If the curves are
>>> being (adequately/appropriately) parameterized as Weibull, then all
>>> the information is in the parameters anyway, which can be directly
>>> modeled, fitted, and compared as functions of temperature -- provided
>>> that the design permits this (i.e. provides sufficient precision for
>>> the characterizations/comparisons).
>>>
>>> If you don't know how to do this, seek further statistical help.
>>>
>>> --
>>> Bert Gunter
>>> Genentech Nonclinical Statistics
>>>
>>>
>>>>
>>>> The data originally posted was an example of one of the curves
>>> experienced.
>>>>
>>>> kc
>>>>
>>>> On Thu, Aug 26, 2010 at 9:48 AM, David Winsemius<dwinsemius at comcast.net
>>>> wrote:
>>>>
>>>>>
>>>>> On Aug 26, 2010, at 1:35 PM, Marlin Keith Cox wrote:
>>>>>
>>>>> I need the parameters estimated for a non-linear equation, an example
>>> of
>>>>>> the
>>>>>> data is below.
>>>>>>
>>>>>>
>>>>>> # rm(list=ls()) I really wish people would add comments to
>>> destructive
>>>>>> pieces of code.
>>>>>>
>>>>>
>>>>> Time<-c( 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4,
>>>>>> 4, 4, 5, 5, 5, 5, 5, 8, 8, 8, 8, 8)
>>>>>> Level<-c( 100, 110, 90, 95, 87, 60, 65, 61, 55, 57, 40, 41,
>>> 50,
>>>>>> 47,
>>>>>> 44, 44, 42, 38, 40, 37, 37, 35, 40, 34, 32, 20, 22, 25,
>>> 27,
>>>>>> 29)
>>>>>> plot(Time,Level,pch=16)
>>>>>>
>>>>>
>>>>> You did not say what sort of "non-linear equation" would best suit,
>>>>> nor
>>> did
>>>>> you offer any background regarding the domain of study. There must be
>>> many
>>>>> ways to do this. After looking at the data, a first pass looks like
>>> this:
>>>>>
>>>>>> lm(log(Level) ~Time )
>>>>>
>>>>> Call:
>>>>> lm(formula = log(Level) ~ Time)
>>>>>
>>>>> Coefficients:
>>>>> (Intercept) Time
>>>>> 4.4294 -0.1673
>>>>>
>>>>>> exp(4.4294)
>>>>> [1] 83.88107
>>>>>> points(unique(Time), exp(4.4294 -unique(Time)*0.1673), col="red",
>>> pch=4)
>>>>>
>>>>> Maybe a Weibull model would be more appropriate.
>>>>>
>>>>>
>>>>> --
>>>>>
>>>>> David Winsemius, MD
>>>>> West Hartford, CT
>>>>>
>>>>>
>>>>
>>>>
>>>
>>>> [[alternative HTML version deleted]]
>>> >
>>>> ______________________________________________
>>>> R-help at r-project.org mailing list
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
>>>
>>>> and provide commented, minimal, self-contained, reproducible code.
>>>>
>>>
>>>
>>>
>>> --
>>> Bert Gunter
>>> Genentech Nonclinical Biostatistics
>>> 467-7374
>>> http://devo.gene.com/groups/devo/depts/ncb/home.shtml
>>>
>>
>>
>