On 10/8/2009 6:21 PM, apjawor...@mmm.com wrote:
This is my third attempt to send this message. Hopefully this one will go
through. If anybody got my previous attempts (one with 3 attached PNG
files and one with graph embedded in the message) I apologize for multiple
postings.
I tried to ill
This is expected behavior from the way nls() is written. The nls()
function has a "..." argument, which means that additional arguments are
allowed.
Under "Arguments" the docs say:
|...| Additional optional arguments. None are used at present.
As far as I can see in the code, nothing at al
Hmm, I should have read the .C documentation more carefully still. I can't
use DUP=FALSE when passing character vectors. In that case, since specifying
the style doesn't suppress copying I guess my only option is to use the more
complicated .Call API, which I naturally was hoping to avoid.
Dunca
Steve Jaffe wrote:
>
> Thanks. I did misread the documentation, which says on p 68:
>
> Routines for use with the .C and .Fortran interfaces are described with
> similar data structures [referring to R_CallMethodDef],
> but which have two additional fields for describing the type and ???style???
Potential bug:
I mistyped weights in the call ('weigths') and it did not produce any error=
message. The coefs were exactly the same like without weights, so I was su=
spicious and when weights(nls1) gave NULL, I saw my typo.
Usually the function will say "Unused arguments", which shows you the
Thanks. I did misread the documentation, which says on p 68:
Routines for use with the .C and .Fortran interfaces are described with
similar data structures [referring to R_CallMethodDef],
but which have two additional fields for describing the type and “style” of
each argument
So it is R_CMetho
Hi,
I am using "line" function to plot the line. And I would like to
understand "Tukeyline" algorithm. Since, the line function is calling
the Tukeyline algorithm(which is compiled code) using foreign function
interface, I am not able to look into the source code of this
algorithm. Can somebody h
