How do I import all the submodules of a Python namespace package?

Question

A Python namespace package can be spread over many directories, and zip files or custom importers. What's the correct way to iterate over all the importable submodules of a namespace package?

Accepted Answer

Here is a way that works well for me. Create a new submodule `all.py`, say, in one of the packages in the namespace.

If you write

```
import mynamespace.all
```

you are given the object for the `mynamespace` module. This object contains all of the loaded modules in the namespace, irrespective of where they were loaded, since there is only one instance of `mynamespace` around.

So, just load all the packages in the namespace in `all.py`!

```
# all.py

from pkgutil import iter_modules

# import this module's namespace (= parent) package
pkg = __import__(__package__)

# iterate all modules in pkg's paths,
# prefixing the returned module names with namespace-dot,
# and import the modules by name
for m in iter_modules(pkg.__path__, __package__ + '.'):
    __import__(m.name)
```

Or in a one-liner that keeps the `all` module empty, if you care about that sort of thing:

```
# all.py

(lambda: [__import__(_.name) for _ in __import__('pkgutil').iter_modules(__import__(__package__).__path__, __package__ + '.')])()  # noqa
```

After importing the `all` module from your namespace, you then actually receive a fully populated namespace module:

```
import mynamespace.all

mynamespace.mymodule1  # works
mynamespace.mymodule2  # works
...
```

Of course, you can use the same mechanism to enumerate or otherwise process the modules in the namespace, if you do not want to import them immediately.

How do I import all the submodules of a Python namespace package?

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