# [R] Finding starting values for the parameters using nls() or nls2()

Bert Gunter bgunter.4567 at gmail.com
Mon Oct 10 00:40:17 CEST 2016

```Well... (inline -- and I hope this isn't homework!)

On Sun, Oct 9, 2016 at 3:05 PM, Andrew Robinson
<A.Robinson at ms.unimelb.edu.au> wrote:
> Here are some things to try.  Maybe divide Area by 1000 and retention
> by 100.  Try plotting the data and superimposing the line that
> corresponds to the 'fit' from nls2.  See if you can correct it with
> some careful guesses.
>
> Getting suitable starting parameters for non-linear modeling is one of
> the black arts of statistical fitting. ...
>
> Andrew

True. But it's usually worthwhile thinking about the math a bit before guessing.

Note that the model can be linearized to:

log(log(Retention)) = b0 + b1*Area^th

So a plot of log(log(Retention)) vs Area may be informative and useful
for finding starting values. e.g., for a grid of th's, do linear
regression fits .

However, when I look at that plot, it seems pretty linear with a
negative slope. This suggests that you may have an overparametrization
problem . i.e. fix th =1 and use the b0 and b1 from the above
regression for starting values.

Do note that this strategy isn't foolproof, as it ignores that the
error term is additive in the above transformed metric, rather than
the original. This can sometimes mislead. But this is just a
heuristic.

Cheers,
Bert

>
> On 9 October 2016 at 22:21, Pinglei Gao <gaopinglei at 163.com> wrote:
>> Hi,
>>
>> I have some data that i'm trying to fit a double exponential model: data.
>> Frame (Area=c (521.5, 689.78, 1284.71, 2018.8, 2560.46, 524.91, 989.05,
>> 1646.32, 2239.65, 2972.96, 478.54, 875.52, 1432.5, 2144.74, 2629.2),
>>
>> Retention=c (95.3, 87.18, 44.94, 26.36, 18.12, 84.68, 37.24, 33.04, 23.46,
>> 9.72, 97.92, 71.44, 44.52, 24.44, 15.26) ) and the formula of the double
>> exponential is: exp (b0*exp (b1*x^th)).
>>
>>
>>
>> I failed to guess the initial parameter values and then I learned a measure
>> to find starting values from Nonlinear Regression with R (pp. 25-27):
>>
>>
>>
>>> cl<-data.frame(Area =c(521.5, 689.78, 1284.71, 2018.8, 2560.46, 524.91,
>> 989.05, 1646.32, 2239.65, 2972.96, 478.54, 875.52, 1432.5, 2144.74, 2629.2),
>>
>> + Retention =c(95.3, 87.18, 44.94, 26.36, 18.12, 84.68, 37.24, 33.04, 23.46,
>> 9.72, 97.92, 71.44, 44.52, 24.44, 15.26) )
>>
>>> expFct <- function(Area, b0, b1,th) {exp(b0*exp(b1*Area^th))}
>>
>>> grid.Disperse <- expand.grid(list(b0 = seq(0.01,4, by = 0.01), th =
>> c(0.02),b1 = seq(0.01, 4, by = 0.01)))
>>
>>> Disperse.m2a <- nls2(Retention ~expFct(Area, b0, b1,th), data = cl, start
>> = grid.Disperse, algorithm = "brute-force")
>>
>>> Disperse.m2a
>>
>> Nonlinear regression model
>>
>>   model: Retention ~ expFct(Area, b0, th, b1)
>>
>>    data: cl
>>
>> b0   th   b1
>>
>> 3.82 0.02 0.01
>>
>> residual sum-of-squares: 13596
>>
>> Number of iterations to convergence: 160000
>>
>> Achieved convergence tolerance: NA
>>
>>
>>
>> I got no error then I use the output as starting values to nls2 ():
>>
>>> nls.m2<- nls2(Retention ~ expFct(Area, b0, b1, th), data = cl, start =
>> list(b0 = 3.82, b1 = 0.02, th = 0.01))
>>
>> Error in (function (formula, data = parent.frame(), start, control =
>> nls.control(),  :
>>
>>
>>
>>
>> Why? Did I do something wrong or misunderstand something?
>>
>>
>>
>> Later, I found another measure from Modern Applied Statistics with S (pp.
>> 216-217):
>>
>>
>>
>>> negexp <- selfStart(model = ~ exp(b0*exp(b1*x^th)),initial =
>> negexp.SSival, parameters = c("b0", "b1", "th"),
>>
>> + template = function(x, b0, b1, th) {})
>>
>>> Disperse.ss <- nls(Retention ~ negexp(Area, b0, b1, th),data = cl, trace =
>> T)
>>
>>          b0          b1          th
>>
>>    4.208763  144.205455 1035.324595
>>
>> Error in qr.default(.swts * attr(rhs, "gradient")) :
>>
>>  NA/NaN/Inf (arg1) can not be called when the external function is called.
>>
>>
>>
>> Error happened again. How can I fix it? I am desperate.
>>
>>
>>
>> Best regards,
>>
>>
>>
>> Pinglei Gao
>>
>>
>>
>>
>>         [[alternative HTML version deleted]]
>>
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>
>
>
> --
> Andrew Robinson
> Deputy Director, CEBRA, School of Biosciences
> Reader & Associate Professor in Applied Statistics  Tel: (+61) 0403 138 955
> School of Mathematics and Statistics                        Fax: +61-3-8344 4599
> University of Melbourne, VIC 3010 Australia
> Email: a.robinson at ms.unimelb.edu.au
> Website: http://www.ms.unimelb.edu.au/~andrewpr
>
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