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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