[R] automate curve drawing on nls() object

[array chip]

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 <shidaxia at yahoo.com> wrote:

> From: Shi, Tao <shidaxia at yahoo.com>
> Subject: Re: [R] automate curve drawing on nls() object
> To: "array chip" <arrayprofile at yahoo.com>, 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 <arrayprofile at yahoo.com>
> > To: 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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