R help - Nov 2011 - Obtaining a derivative of nls() SSlogis function

Hello, I am wondering if someone can help me. I have the following function
that I derived using nls() SSlogis. I would like to find its derivative. I
thought I had done this using deriv(), but for some reason this isn't
working out for me.

Here is the function:
asym <- 84.951
xmid <- 66.90742
scal <- -6.3

x.seq <- seq(1, 153,, 153)
nls.fn <- asym/((1+exp((xmid-x.seq)/scal)))

try #1
deriv(nls.fn)
#get an Error in .Internal(deriv.default(expr, namevec, function.arg, tag,
hessian)) : 'namevec' is missing

try #2
deriv(nls.fn, namevec=c("asym", "xmid", "scal"))
#this doesn't seem to give me the expression, and the gradients are zero.

I've tried to do this with Ryacas as well, but I'm lost.

Can anyone help?

Thank you,

Katrina

	[[alternative HTML version deleted]]

On Nov 17, 2011, at 4:40 PM, Katrina Bennett wrote:
> Hello, I am wondering if someone can help me. I have the following  
> function
> that I derived using nls() SSlogis. I would like to find its  
> derivative. I
> thought I had done this using deriv(), but for some reason this isn't
> working out for me.
>
> Here is the function:
> asym <- 84.951
> xmid <- 66.90742
> scal <- -6.3
>
> x.seq <- seq(1, 153,, 153)
> nls.fn <- asym/((1+exp((xmid-x.seq)/scal)))
>
> try #1
> deriv(nls.fn)
> #get an Error in .Internal(deriv.default(expr, namevec,  
> function.arg, tag,
> hessian)) : 'namevec' is missing
>
> try #2
> deriv(nls.fn, namevec=c("asym", "xmid",
"scal"))
> #this doesn't seem to give me the expression, and the gradients are  
> zero.
nls.fn is not a function or an expression. It has been evaluated and  
now it's a vectpr, and not what `deriv` is "expecting". If you
want to
use `deriv` or `D` you must first read the help page:

?deriv

And then construct a proper expresssion, call, or function...

 > nls.expr <- expression( 84.951/((1 + exp(( 66.90742- x)/ -6.3))) )
 > D(nls.expr, "x")
-(84.951 * (exp((66.90742 - x)/-6.3) * (1/6.3))/((1 + exp((66.90742 -
     x)/-6.3)))^2)
 > deriv(nls.expr, "x")
expression({
     .expr4 <- exp((66.90742 - x)/-6.3)
     .expr5 <- 1 + .expr4
     .value <- 84.951/.expr5
     .grad <- array(0, c(length(.value), 1L), list(NULL, c("x")))
     .grad[, "x"] <- -(84.951 * (.expr4 * (1/6.3))/.expr5^2)
     attr(.value, "gradient") <- .grad
     .value
})

-- 
David.
>
> I've tried to do this with Ryacas as well, but I'm lost.
>
> Can anyone help?
>
> Thank you,
>
> Katrina
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT

I think you need to make an expression. I tried
> nls.fn <- asym/((1+exp((xmid-x.seq)/scal)))
Error: object 'asym' not found
> nls.fn <- expression(asym/((1+exp((xmid-x.seq)/scal))))
> D(nls.fn,"asym")
1/((1 + exp((xmid - x.seq)/scal)))
>
Does that help? Maybe there are other approaches too.

JN


On 11/18/2011 06:00 AM, r-help-request at r-project.org
wrote:
> Message: 88
> Date: Thu, 17 Nov 2011 12:40:59 -0900
> From: Katrina Bennett <kebennett at alaska.edu>
> To: r-help at r-project.org
> Subject: [R] Obtaining a derivative of nls() SSlogis function
> Message-ID:
> 	<CA+SNM83UtZMd1HDFGf2BioLPWGwhS-OWRMaHoV4yKQtvSwy4tQ at
mail.gmail.com>
> Content-Type: text/plain
> 
> Hello, I am wondering if someone can help me. I have the following function
> that I derived using nls() SSlogis. I would like to find its derivative. I
> thought I had done this using deriv(), but for some reason this isn't
> working out for me.
> 
> Here is the function:
> asym <- 84.951
> xmid <- 66.90742
> scal <- -6.3
> 
> x.seq <- seq(1, 153,, 153)
> nls.fn <- asym/((1+exp((xmid-x.seq)/scal)))
> 
> try #1
> deriv(nls.fn)
> #get an Error in .Internal(deriv.default(expr, namevec, function.arg, tag,
> hessian)) : 'namevec' is missing
> 
> try #2
> deriv(nls.fn, namevec=c("asym", "xmid",
"scal"))
> #this doesn't seem to give me the expression, and the gradients are
zero.
> 
> I've tried to do this with Ryacas as well, but I'm lost.
> 
> Can anyone help?
> 
> Thank you,
> 
> Katrina

On Thu, Nov 17, 2011 at 4:40 PM, Katrina Bennett <kebennett at alaska.edu>
wrote:
> Hello, I am wondering if someone can help me. I have the following function
> that I derived using nls() SSlogis. I would like to find its derivative. I
> thought I had done this using deriv(), but for some reason this isn't
> working out for me.
>
> Here is the function:
> asym <- 84.951
> xmid <- 66.90742
> scal <- -6.3
>
> x.seq <- seq(1, 153,, 153)
> nls.fn <- asym/((1+exp((xmid-x.seq)/scal)))
>
> try #1
> deriv(nls.fn)
> #get an Error in .Internal(deriv.default(expr, namevec, function.arg, tag,
> hessian)) : 'namevec' is missing
>
> try #2
> deriv(nls.fn, namevec=c("asym", "xmid",
"scal"))
> #this doesn't seem to give me the expression, and the gradients are
zero.
>
> I've tried to do this with Ryacas as well, but I'm lost.
>
> Can anyone help?
>
You can do that with plain R:
> e <- quote(asym/((1+exp((xmid-x.seq)/scal))))
> for(v in all.vars(e)) cat("deriv wrt", v, "is",
format(D(e, v)), "\n")
deriv wrt asym is 1/((1 + exp((xmid - x.seq)/scal)))
deriv wrt xmid is -(asym * (exp((xmid - x.seq)/scal) * (1/scal))/((1 +
exp((xmid -      x.seq)/scal)))^2)
deriv wrt x.seq is asym * (exp((xmid - x.seq)/scal) * (1/scal))/((1 +
exp((xmid -      x.seq)/scal)))^2
deriv wrt scal is asym * (exp((xmid - x.seq)/scal) * ((xmid -
x.seq)/scal^2))/((1 +      exp((xmid - x.seq)/scal)))^2


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