В Thu, 31 Aug 2023 14:59:07 +
Christophe Bousquet пишет:
> So I get no output for tools::Rcmd('SHLIB --version').
>
> I tried to run tools::Rcmd('SHLIB --help') or tools::Rcmd('check
> --help'), but no output neither.
>
> I had more success with tools::Rcmd('BATCH --help'):
The job of Rcmd
> This can be considered good news. You have just successfully performed
> the job that is normally done by R CMD SHLIB when installing source
> packages or running inline C++ code. The environment variables, at
> least inside your running R session, are completely fine.
>
> Now we need to find out
В Thu, 31 Aug 2023 14:01:54 +
Christophe Bousquet пишет:
> So when I run the commands, I get this output. I honestly have no
> clue whether this can be considered as something useful or not :-/
>
> ```
> > tools:::.shlib_internal(c('-n', 'hello.c'))
> make cmd is
> make -f "C:/PROGRA~1/R
> So starting a new Rcmd.exe process fails for some reason.
>
> If you take the same R session where the environment variables are
> right and Sys.which() resolves Make and GCC and try to run
> tools:::.shlib_internal(c('-n', 'hello.c')) or
> tools:::.shlib_internal('hello.c'), does it do somethin
On 8/31/23 1:27 AM, Spencer Graves wrote:
Hello, All:
I want to simulate future observations from fits to
heteroscedastic data. A simple example is as follows:
(DF3_2 <- data.frame(y=c(1:3, 10*(1:3)),
gp=factor(rep(1:2, e=3
# I want to fit 4 models
# an
В Thu, 31 Aug 2023 11:57:06 +
Christophe Bousquet пишет:
> > tools::Rcmd('SHLIB -n hello.c')
> > tools::Rcmd('SHLIB hello.c')
> >
> > What do the commands print? Does the second command fail?
>
> I basically get no output from the two commands, apart from a new
> blank R prompt.
So starti
> When installing packages containing code to compile, R eventually calls > R
> CMD SHLIB. Same thing happens with inline C++: it gets stored in a
> temporary file, compiled into a *.dll using R CMD SHLIB and then loaded
> using dyn.load().
>
> Write the following into a file named hello.c:
>
>
Add a break. Something like:
If (is.character(x)) break
If you have nested loops then a similar statement is needed for each level.
"break" only exits the innermost loop.
Tim
-Original Message-
From: R-help On Behalf Of Jeff Reichman
Sent: Wednesday, August 30, 2023 9:46 PM
To: r-help@
Thanks, Greg. I appreciate your helpful assistance.
Kindest Regards,
*Stephen Dawson, DSL*
/Executive Strategy Consultant/
Business & Technology
+1 (865) 804-3454
http://www.shdawson.com
On 8/30/23 16:52, Greg Snow wrote:
Stephen, I see lots of answers with packages and resources, but not
bo
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