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

ProfJCNash profjcnash at gmail.com
Mon Oct 10 02:14:36 CEST 2016

```I didn't try very hard, but got a solution from .1, 1, .1 with nlxb() from nlmrt. It took a lot
of iterations and looks to be pretty ill-conditioned. Note nlmrt uses analytic derivatives if it
can, and a Marquardt method. It is designed to be a pit bull -- tenacious, not fast.

I'm working on a replacement for this and nls(), but it will be a while. However, I welcome
short scripts like this as tests. My script below

Best, JN

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))}
expf2 <- "Retention ~ 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

library(nlmrt)
test <- nlxb(expf2, start= c(b0=.1, b1=1, th=.1), trace=TRUE, data=cl)

On 16-10-09 07:40 PM, peter dalgaard wrote:
>
>> On 10 Oct 2016, at 00:40 , Bert Gunter <bgunter.4567 at gmail.com> wrote:
>>
>> Well... (inline -- and I hope this isn't homework!)
>>
>
> Pretty much same as I thought.
>
> Fixing th=0.02 in the grid search looks wrong. Bert's plot is pretty linear, so th=1 is a good guesstimate. There's a slight curvature but to reduce it, you would increase th, not decrease it. Running the regression, as Bert suggests, indicates that b0=5.16 and b1= -0.00024 could work as reasonable starting values. Notice that the grid search had "b1 = seq(0.01, 4, by = 0.01)" which is wrong in both sign and scale.
>
> Andrew's suggestion of dividing Retention by 100 is tempting, since it looks like a percentage, but that would make all Y values less than 1 and the double exponential function as written has values that are always bigger than 1. (It is conceivable that the model itself is wrong, though. E.g. it could be that Retention on a scale from 0 to 1 could be modeled as exp(-something), but we really have no idea of the context here.)
>
> (If this was in fact homework, you should now go and write a proper SelfStart initializer routine for this model. Even if it isn't homework, you do need to study the text again, because you have clearly not understood how self-starting models work.)
>
> -pd
>
>>
>>
>>
>> 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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>>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>>>
>>> --
>>> 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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>>>
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