R help - May 2010 - 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

array chip <arrayprofile <at> yahoo.com> writes:
> Hi, I would like to use the curve() function to draw the 
> predicted curve from an nls() object. for example:
> 
  Is there a reason you don't want to use plot() and predict() ???
  
> dd<-read.table("dd.txt",sep='\t',header=T,row.names=1)
  note that this is not a reproducible example ...
> 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)
> 
  predframe <- data.frame(x=seq(1,20000,length.out=101))
  predframe <- cbind(predframe,y=predict(obj,newdata=predframe))
  with(predframe,plot(y~x,type="l"))

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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