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 <arrayprof...@yahoo.com>
> To: r-help@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
______________________________________________
> ymailto="mailto:R-help@r-project.org";
> href="mailto:R-help@r-project.org";>R-help@r-project.org mailing list
> href="https://stat.ethz.ch/mailman/listinfo/r-help"; target=_blank
> >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.
______________________________________________
R-help@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.
