Thank you. Could you lastly help me with this error. I was trying to use a
self starting function (Weibull).
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)
On Thu, Aug 26, 2010 at 1:42 PM, David Winsemius <dwinsem...@comcast.net>wrote:
>
> On Aug 26, 2010, at 5:20 PM, Marlin Keith Cox 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.
>
> The data originally posted was an example of one of the curves experienced.
>
>
> You might be interest in what happens when you expand the plot a bit using
> the fm1 model:
>
> plot(Level~Time, xlim=c(0,40), ylim=c(0,max(Level)) )
> lines(0:40, predict(fm1, data.frame(Time=0:40 ) ) , col="red", pch=4)
> abline(h=0)
>
> And not forgetting that extrapolation beyond the limits of data is a
> dangerous maneuver. In this instance any Weibull model would predict some
> sort of decay to 0, which may or may not be scientifically plausible.
>
> --
> David.
>
>
> kc
>
> On Thu, Aug 26, 2010 at 9:48 AM, David Winsemius
> <dwinsem...@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
>>
>>
>
>
> --
> M. Keith Cox, Ph.D.
> Alaska NOAA Fisheries, National Marine Fisheries Service
> Auke Bay Laboratories
> 17109 Pt. Lena Loop Rd.
> Juneau, AK 99801
> keith....@noaa.gov
> marlink...@gmail.com
> U.S. (907) 789-6603
>
>
> David Winsemius, MD
> West Hartford, CT
>
>
--
M. Keith Cox, Ph.D.
Alaska NOAA Fisheries, National Marine Fisheries Service
Auke Bay Laboratories
17109 Pt. Lena Loop Rd.
Juneau, AK 99801
keith....@noaa.gov
marlink...@gmail.com
U.S. (907) 789-6603
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