[R] Odp: nls, how to determine function?

Tue Aug 9 10:38:35 CEST 2011

Hi

> Hi R help,
> 
> I am trying to determine how nls() generates a function based on the
> self-starting SSlogis and what the formula for the function would be.
> I've scoured the help site, and other literature to try and figure
> this out but I still am unsure if I am correct in what I am coming up
> with.

Thanks for providing data and your code

> 
> 
> 
**************************************************************************
> dat <- c(75.44855206,NA,NA,NA,82.70745342,82.5335019,88.56617647,80.
> 00128866,94.15418227,86.63987539,93.91052952,74.10612245,86.62289562,90.
> 47961047,NA,NA,82.45320197,72.14371257,NA,71.44104803,72.59742896,68.
> 36363636,NA,NA,61,NA,NA,71.26502909,NA,85.93333333,84.34248284,79.
> 00522193,79.64223058,97.2074017,88.43700548,96.40413877,95.13511869,92.
> 57379057,93.97498475,NA,97.55995131,89.53321146,97.21728545,93.21980198,
> 77.54054054,95.85392575,86.25684723,97.55325624,80.03950617,NA,91.
> 34023128,92.42906574,88.59433962,65.77272727,89.63772455,NA,NA,NA,NA,74.
> 86344239,83.57594937,70.22516556,65.30543319,NA,NA,67.84852294,60.
> 90909091,54.79303797,NA,52.18735363,33.47003155,NA,41.34693878,24.
> 5047043,NA,NA,NA,NA,9.944444444,13.6875,NA,11.90267176,84.14285714,3.
> 781456954,NA,1.432926829,4.26557377,1.823529412,0.444620253,4.
> 
711155378,NA,6.320284698,0.581632653,0.144578313,3.666666667,0,0,0,0,0,NA,
> 0.032947462,0,0,10.54545455,0,NA,0.561007958,0.75,NA,0.048780488,0.
> 74137931,NA,2.023339318,0,0,0,NA,NA,0.156950673,NA,0.283769634,32.
> 
81818182,NA,NA,0,NA,0,0,0,NA,0.212454212,3.120181406,NA,0.011811024,NA,0,
> 
0.120430108,5.928571429,1.75,0.679292929,0.97,NA,0,NA,NA,1,0.38547486,NA,
> 1.460732984,0.007795889,0.05465288,0.004341534)

> dat.df.1 <- data.frame(dat)
unnecessary

> dat.df.2 <- data.frame(x=x.seq, dat.df=dat.df.1)

some correction
dat.df.2 <- data.frame(x=seq_along(dat), dat=dat)

> fit.dat <-nls(dat ~ SSlogis(x, Asym, xmid,scal), data = dat.df.2,
> start =list(Asym=90, xmid = 75, scal = -6))
> plot(dat.df.2, axes=FALSE, ann=FALSE, ylim=c(0,100))
> lines(dat.df.2$x[complete.cases(dat.df.2)], predict(fit.dat), 
ylim=c(0,100))
> 
> summary(fit.dat)
> 
> 
**************************************************************************
> Formula: dat ~ SSlogis(x, Asym, xmid, scal)
> 
> Parameters:
>      Estimate Std. Error t value Pr(>|t|)
> Asym   85.651      1.716  49.900  < 2e-16 ***
> xmid   72.214      1.036  69.697  < 2e-16 ***
> scal   -6.150      0.850  -7.236  7.9e-11 ***
> ---
> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> 
> Residual standard error: 10.33 on 105 degrees of freedom
> 
> Number of iterations to convergence: 10
> Achieved convergence tolerance: 4.405e-06
>   (45 observations deleted due to missingness)
> 
**************************************************************************
> 
> >From r-help, SSlogis parameters asym, xmid and scal are defined as:
> 
> Asym: a numeric parameter representing the asymptote.
> 
> xmid: a numeric parameter representing the x value at the inflection
> point of the curve. The value of SSlogis will be Asym/2 at xmid.
> 
> scal: a numeric scale parameter on the input axis.
> 
> and it states that the value of SSlogis "is a numeric vector of the
> same length as input. It is the value of the expression
> sym/(1+exp((xmid-input)/scal)). If all of the arguments Asym, xmid,
> and scal are names of objects the gradient matrix with respect to
> these names is attached as an attribute named gradient."
> 
> However, how do I get the actual function for the curve that is
> generated? I don't think it can just be: y=
> asym/((1+e^((xmid-x)/scal)))?

Yes. I think that the best source of information about nonlinear 
regression is book by Bates, Pinheiro - Mixed effect models with S and S+. 
There you can find how to determine starting parameters, how to construct 
and use your own function together with selfstart feature.

> 
> Also, how do you determine the starting parameters to input in for
> asym, xmin, and scal?
> 
> Perhaps I need to start at the beginning and define my own function,
> and not rely on SSlogis to provide it?
> 
> What I want to be able to do is determine a local maximum for my curve
> (the x value at which this curve inflects (the upper inflection)), and
> the x value for the local minimum (the lower inflection curve), and
> the x value counts in between these values. I think in order to do
> this I need to differentiate the function.

Maybe I do not understand well but looking at the picture it seems to me 
that logistic model is fitting your data quite well. You can use also four 
parameter logistic model. 

> fit.dat.2 <-nls(dat ~ SSfpl(x, A, B, xmid,scal), data = dat.df.2, start 
=list(A=85.65, B=0, xmid = 72, scal = -6))
> summary(fit.dat.2)

Formula: dat ~ SSfpl(x, A, B, xmid, scal)

Parameters:
     Estimate Std. Error t value Pr(>|t|) 
A      1.6729     1.5927   1.050    0.296 
B     85.5555     1.7065  50.134  < 2e-16 ***
xmid  71.7628     1.0762  66.679  < 2e-16 ***
scal  -5.8051     0.9162  -6.336 6.13e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 10.32 on 104 degrees of freedom

Number of iterations to convergence: 9 
Achieved convergence tolerance: 7.629e-06 
  (45 observations deleted due to missingness)

As you can see parameter A is insignificant so simple logistic can be used 
too. In that case upper asymptote is 85.6, lower asymptote is zero, 
inflection point is 72 (x where y is in the middle between both 
asymptotes) and scal is rate at which the curve is falling (growing).

There is however some wave in the beginning of your data

fit <-loess(dat ~ x, data = dat.df.2, span=0.3)
lines(dat.df.2$x[complete.cases(dat.df.2)], predict(fit), col=3)

So it is up to you to decide if you are satisfied with getting asymptotic 
values from logistic model or you want to set something more elaborated.

Regards
Petr

> 
> Any insight on this would be greatly appreciated.
> 
> Sincerely,
> 
> Katrina
> 
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