[R] automate curve drawing on nls() object

[Shi, Tao]

In this case, Ben's approach is the way to go.  

I'm curious how you fit nls without knowing the model formula beforehand?

...Tao





----- Original Message ----
> From: array chip <arrayprofile at yahoo.com>
> To: r-help at r-project.org; TaoShi <shidaxia at yahoo.com>
> Sent: Tue, May 18, 2010 5:22:47 PM
> Subject: Re: [R] automate curve drawing on nls() object
> 
> well, this is not going automate enough because you have to know how the model 
> (formula) looks like, and how many parameters there are in the model beforehand 
> to do what you are suggesting.

Thanks


--- On Tue, 5/18/10, 
> Shi, Tao <
> href="mailto:shidaxia at yahoo.com">shidaxia at yahoo.com> wrote:

> 
> From: Shi, Tao <
> href="mailto:shidaxia at yahoo.com">shidaxia at yahoo.com>
> Subject: Re: 
> [R] automate curve drawing on nls() object
> To: "array chip" <
> ymailto="mailto:arrayprofile at yahoo.com" 
> href="mailto:arrayprofile at yahoo.com">arrayprofile at yahoo.com>, 
> ymailto="mailto:r-help at r-project.org" 
> href="mailto:r-help at r-project.org">r-help at r-project.org
> Date: 
> Tuesday, May 18, 2010, 7:42 PM
> I can't directly answer your 
> question
> regarding 'expression', but can you just replace b, c,d, 
> and
> e with coef(obj)[1], coef(obj)[2], ...
> etc.   You still can 
> automate the whole
> process this way, right?
> 
> 
> 
> 
> 
> 
> ----- Original Message ----
> > From: array 
> chip <
> href="mailto:arrayprofile at yahoo.com">arrayprofile at yahoo.com>
> > 
> To: 
> href="mailto:r-help at r-project.org">r-help at r-project.org
> > Sent: 
> Tue, May 18, 2010 4:13:33 PM
> > Subject: [R] automate curve drawing on 
> nls() object
> > 
> > Hi, I would like to use the curve() 
> function to draw
> the predicted curve from an 
> > nls() object. 
> for 
> > example:
> 
> 
> dd<-read.table("dd.txt",sep='\t',header=T,row.names=1)
> 
> obj<-nls(y~c+(d-c)/(1+(x/e)^b),data=dd,start=list(b=-1,
> 
> > 
> c=0, d=100, e=150))
> coef(obj)
>           b  
> >        
>   c   
>        d    
> >        e 
> 
> -1.1416422   0.6987028 102.8613176 
> > 135.9373131
> 
> curve(0.699+(102.86-0.699)/(1+(x/135.94)^(-1.1416)),1,20000)
> 
> 
> Now 
> > I am going to have a lot of datasets to do this, so
> 
> certainly I would like to 
> > automate this. Suppose that I can create 
> a character
> string for the formula, but 
> > I am not sure how 
> to pass that character string into
> the curve() to make it 
> > 
> work. The help page of curve() says the first argument
> is "an expression 
> written 
> > as a function of x, or alternatively the name of a
> 
> function which will be 
> > plotted". I tried the following, for 
> example:
> 
> substitute(expression(c + 
> > (d - c)/(1 + 
> (x/e)^b)),as.list(coef(obj)))
> will 
> > return:
> 
> "expression(0.698704171233635 + (102.861317499063 - 
> > 
> 0.698704171233635)/(1 +
> 
> (x/135.937317917920)^-1.14164217993857))"
> 
> so I 
> > 
> tried:
> curve(substitute(expression(c + (d - c)/(1 + (x/e)^b)), 
> 
> > as.list(coef(obj))), 1,20000)
> 
> but it returns an 
> error:
> "Error in 
> > xy.coords(x, y, xlabel, ylabel, log) : 
> 
>   'x' and 'y' lengths 
> > differ"
> 
> Any 
> suggestions?
> 
> 
> A related question:
> 
> If I 
> do 
> > this:
> substitute(expression(c + (d - c)/(1 + 
> 
> > (x/e)^b)),as.list(coef(obj)))
> 
> I will get: 
> > 
> 
> "expression(0.698704171233635 + (102.861317499063 -
> 
> 0.698704171233635)/(1 + 
> > 
> (x/135.937317917920)^-1.14164217993857))"
> 
> But if I 
> 
> > do:
> 
> substitute(parse(text=as.character(obj$call$formula[3]),srcfile=NULL),as.list(coef(obj)))
> 
> 
> I 
> > only get:
> "parse(text = 
> as.character(obj$call$formula[3]), srcfile =
> NULL)" 
> > as a 
> result, not have b,c,d,e replaced by coefficient
> 
> > 
> values.
> 
> where
>     
> > 
> 
> parse(text=as.character(obj$call$formula[3]),srcfile=NULL)
> returns the 
> 
> > wanted expression:
> "expression(c + (d - c)/(1 + 
> (x/e)^b))"
> 
> Why is 
> > that?
> 
> 
> Thanks
> 
> John
> 
> 
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