Question

In a Python package directory of my own creation, I have an __init__.py file that says:

from _foo import *

In the same directory there is a _foomodule.so which is loaded by the above. The shared library is implemented in C++ (using Boost Python). This lets me say:

import foo
print foo.MyCppClass

This works, but with a quirk: the class is known to Python by the full package path, which makes it print this:

foo._foo.MyCppClass

So while MyCppClass exists as an alias in foo, foo.MyCppClass is not its canonical name. In addition to being a bit ugly, this also makes help() a bit lame: help(foo) will say that foo contains a module _foo, and only if you say help(foo._foo) do you get the documentation for MyCppClass.

Is there something I can do differently in __init__.py or otherwise to make it so Python sees foo.MyCppClass as the canonical name?

I’m using Python 2.7; it would be great if the solution worked on 2.6 as well.

Answer 1

I had the same problem. You can change the module name in your Boost.Python definition:

BOOST_PYTHON_MODULE(_foo)
{
  scope().attr("__name__") = "foo";
  ...
}

The help issue is a separate problem. I think you need to add each item to __all__ to get it exported to help.

When I do both of these, the name of foo.MyCppClass is just that — foo.MyCppClass — and help(foo) gives documentation for MyCppClass.