Finding starting values for the parameters using nls() or nls2()
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. Good luck! And don't forget to check for sensitivity. Andrew
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(), : Singular gradient 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
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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 MSME: http://www.crcpress.com/product/isbn/9781439858028 FAwR: http://www.ms.unimelb.edu.au/~andrewpr/FAwR/ SPuR: http://www.ms.unimelb.edu.au/spuRs/