R help - Dec 2009 - How to export a function from a package and access it only by specifying the namespace?

I have the following test package.

$ ls
DESCRIPTION  man  NAMESPACE  R
$ cat DESCRIPTION
Package: try.package
Type: Package
Title: What the package does (short line)
Version: 1.0
Date: 2009-10-26
Author: Who wrote it
Maintainer: Who to complain to <yourfault at somewhere.net>
Description: More about what it does (maybe more than one line)
License: What license is it under?
LazyLoad: yes
$ cat NAMESPACE
export(
    my_test_g
    )
$ ls R
my_test_f.R  randomxx.R
$ for f in R/*.R; do echo $f; cat $f;done
R/my_test_f.R
my_test_f<-function() {
  print("Hello")
}
R/randomxx.R
my_test_g<-function() {
  print("Helloggg")
}


----------------------------------------

I install the package by

R CMD INSTALL -l /path/to/R_user --no-docs my_package

`my_package' is the directory where the package is in.

Then I try the package 'try.package' in an R session. I'm wondering
why neither 'my_test_f' and 'try.package::my_test_f' work. Why
'my_test_g' can be accessed with 'try.package::' and without
'try.package::'?

Is there a way to make  'my_test_g' accessible only by specifying the
namespace 'try.package::'?
> library(try.package)
> try.package::my_test_g
function ()
{
    print("Helloggg")
}
<environment: namespace:try.package>
> my_test_g
function ()
{
    print("Helloggg")
}
<environment: namespace:try.package>
> my_test_f
Error: object "my_test_f" not found
> try.package::my_test_f
Error: 'my_test_f' is not an exported object from
'namespace:try.package'

Peng Yu wrote:
> 
> Then I try the package 'try.package' in an R session. I'm
wondering
> why neither 'my_test_f' and 'try.package::my_test_f' work. 
> 
The error message you got below clearly explains this-- you did not export
my_test_f in your NAMESPACE file.  To access unexported functions, you must
use the ':::' operator:

  try.package:::my_test_f()



Peng Yu wrote:
> 
> Why 'my_test_g' can be accessed with 'try.package::' and
without
> 'try.package::'?
> 
Because you exported it in the NAMESPACE file.



Peng Yu wrote:
> 
> Is there a way to make  'my_test_g' accessible only by specifying
the
> namespace 'try.package::'?
> 
No.

The purpose of the '::' operator is for those cases where multiple
packages
are loaded that each export a function with the same name.  This is known as
"masking" and the last loaded package will contribute the dominant
function-- i.e. the function the gets called when the user types
"functionName()" and not "packageName::functionName()".  The
"::" operator
allows the selection of functions that are masked by the dominant function.

If you really want to conceal a function from user-level code, don't export
it and it will only be accessible via the ":::" operator.



Peng Yu wrote:
> 
>> library(try.package)
>> try.package::my_test_g
> function ()
> {
>     print("Helloggg")
> }
> <environment: namespace:try.package>
>> my_test_g
> function ()
> {
>     print("Helloggg")
> }
> <environment: namespace:try.package>
>> my_test_f
> Error: object "my_test_f" not found
>> try.package::my_test_f
> Error: 'my_test_f' is not an exported object from
'namespace:try.package'
> 
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