R help - Oct 2009 - How to get slope estimates from a four parameter logistic with SSfpl?

Hi,

I was hoping to get some advice on how to derive estimates of slopes from four
parameter logistic models fit with SSfpl.

I fit the model using:

model<-nls(temp~SSfpl(time,a,b,c,d))
summary(model)

I am interested in the values of the lower and upper asymptotes (parameters a
and b), but also in the gradient of the line at the inflection point (c) which I
assume tells me my rate of increase when it is steepest (?).

However, I cannot work out how to derive a slope estimate. I'm guessing it
has something to do with the scaling parameter d but having searched the
internet for hours I have not made any progress, and it is probably quite
simple. Any help would be hugely appreciated!

All the best

Sam

	[[alternative HTML version deleted]]

Weber, Sam wrote:
> Hi,
> 
> I was hoping to get some advice on how to derive estimates of slopes from
four parameter logistic models fit with SSfpl.
> 
> I fit the model using:
> 
> model<-nls(temp~SSfpl(time,a,b,c,d))
> summary(model)
> 
> I am interested in the values of the lower and upper asymptotes (parameters
a and b), but also in the gradient of the line at the inflection point (c) which
I assume tells me my rate of increase when it is steepest (?).
> 
> However, I cannot work out how to derive a slope estimate. I'm guessing
it has something to do with the scaling parameter d but having searched the
internet for hours I have not made any progress, and it is probably quite
simple. Any help would be hugely appreciated!
> 
Sam,

If I understand you correctly, all you want is the derivative
of the fpl wrt the input. You can differentiate the fpl function
or you can use the result provided by predict.nls() in the
"gradient" attribute. The gradient wrt 'c' is the negative
of the slope you want. The maximum slope will occur at time = c.

So this should give you the slope:

  ypred <- predict(model,newdata=data.frame(Time=coef(model)[3]))
  myslope <- -attr(ypred, "gradient")[,3]

  -Peter Ehlers
> All the best
> 
> Sam
> 
> 	[[alternative HTML version deleted]]
> 
> ______________________________________________
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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 
>

Is the following helpful?

  pdd<-deriv(~a+(b-a)/(1+exp((c-t)/d)),"d")
 > pdd
expression({
     .expr1 <- b - a
     .expr2 <- c - t
     .expr4 <- exp(.expr2/d)
     .expr5 <- 1 + .expr4
     .value <- a + .expr1/.expr5
     .grad <- array(0, c(length(.value), 1L), list(NULL, c("d")))
     .grad[, "d"] <- .expr1 * (.expr4 * (.expr2/d^2))/.expr5^2
     attr(.value, "gradient") <- .grad
     .value
})

Or perhaps you want it with respect to "t"?

JN

> Message: 46
> Date: Mon, 19 Oct 2009 14:50:15 +0100
> From: "Weber, Sam" <Sam.Weber at exeter.ac.uk>
> Subject: [R] How to get slope estimates from a four parameter logistic
> 	with SSfpl?
> To: "r-help at R-project.org" <r-help at r-project.org>
> Message-ID:
> 	<5BB78285B60F9B4DB2C6A4DA8F3E788C12C64B3803 at
EXCHMBS04.isad.isadroot.ex.ac.uk>
> 	
> Content-Type: text/plain
> 
> Hi,
> 
> I was hoping to get some advice on how to derive estimates of slopes from
four parameter logistic models fit with SSfpl.
> 
> I fit the model using:
> 
> model<-nls(temp~SSfpl(time,a,b,c,d))
> summary(model)
> 
> I am interested in the values of the lower and upper asymptotes (parameters
a and b), but also in the gradient of the line at the inflection point (c) which
I assume tells me my rate of increase when it is steepest (?).
> 
> However, I cannot work out how to derive a slope estimate. I'm guessing
it has something to do with the scaling parameter d but having searched the
internet for hours I have not made any progress, and it is probably quite
simple. Any help would be hugely appreciated!
> 
> All the best
> 
> Sam