How do I create a namespace package in Python?
TL;DR:
On Python 3.3 you don't have to do anything, just don't put any __init__.py
in your namespace package directories and it will just work. On pre-3.3, choose the pkgutil.extend_path()
solution over the pkg_resources.declare_namespace()
one, because it's future-proof and already compatible with implicit namespace packages.
Python 3.3 introduces implicit namespace packages, see PEP 420.
This means there are now three types of object that can be created by an import foo
:
- A module represented by a
foo.py
file - A regular package, represented by a directory
foo
containing an__init__.py
file - A namespace package, represented by one or more directories
foo
without any__init__.py
files
Packages are modules too, but here I mean "non-package module" when I say "module".
First it scans sys.path
for a module or regular package. If it succeeds, it stops searching and creates and initalizes the module or package. If it found no module or regular package, but it found at least one directory, it creates and initializes a namespace package.
Modules and regular packages have __file__
set to the .py
file they were created from. Regular and namespace packages have __path__
set to the directory or directories they were created from.
When you do import foo.bar
, the above search happens first for foo
, then if a package was found, the search for bar
is done with foo.__path__
as the search path instead of sys.path
. If foo.bar
is found, foo
and foo.bar
are created and initialized.
So how do regular packages and namespace packages mix? Normally they don't, but the old pkgutil
explicit namespace package method has been extended to include implicit namespace packages.
If you have an existing regular package that has an __init__.py
like this:
from pkgutil import extend_path
__path__ = extend_path(__path__, __name__)
... the legacy behavior is to add any other regular packages on the searched path to its __path__
. But in Python 3.3, it also adds namespace packages.
So you can have the following directory structure:
├── path1
│ └── package
│ ├── __init__.py
│ └── foo.py
├── path2
│ └── package
│ └── bar.py
└── path3
└── package
├── __init__.py
└── baz.py
... and as long as the two __init__.py
have the extend_path
lines (and path1
, path2
and path3
are in your sys.path
) import package.foo
, import package.bar
and import package.baz
will all work.
pkg_resources.declare_namespace(__name__)
has not been updated to include implicit namespace packages.
There's a standard module, called pkgutil, with which you can 'append' modules to a given namespace.
With the directory structure you've provided:
Package-1/namespace/__init__.py
Package-1/namespace/module1/__init__.py
Package-2/namespace/__init__.py
Package-2/namespace/module2/__init__.py
You should put those two lines in both Package-1/namespace/__init__.py
and Package-2/namespace/__init__.py
(*):
from pkgutil import extend_path
__path__ = extend_path(__path__, __name__)
(* since -unless you state a dependency between them- you don't know which of them will be recognized first - see PEP 420 for more information)
As the documentation says:
This will add to the package's
__path__
all subdirectories of directories onsys.path
named after the package.
From now on, you should be able to distribute those two packages independently.