Python命名空间更改`import package.sub_module; 从包导入*`

Question

The Python documentation says

Consider this code:

import sound.effects.echo
import sound.effects.surround
from sound.effects import *

In this example, the echo and surround modules are imported in the current namespace because they are defined in the sound.effects package when the from...import statement is executed. (This also works when __all__ is defined.)

I try the following code

# package/
#     __init__.py
#     sub_module.py

import package.sub_module
from package import *
print(sub_module)

When package/__init__.py is empty, the code works fine. However, when package/__init__.py contains __all__ = [] , print(sub_module) will raise NameError . What is (This also works when all is defined.) from the documentation means?

---

The codes:

package/
    __init__.py
    sub_module.py # empty file
main.py

In main.py:

import package.sub_module
from package import *
print(sub_module)

When package/__init__.py is empty, executing python3 main.py gets <module 'package.sub_module' from '/path/to/package/sub_module.py'

When package/__init__.py contains __all__ = [] , executing python3 main.py gets

Traceback (most recent call last):
File "main.py", line 3, in <module>
    print(sub_module)
NameError: name 'sub_module' is not defined

Answer 1

If a module package defines __all__ , it is the list of module names that are imported by from package import *

So if you define __all__ as empty list, from package import * will import nothing.

Try defining it like this:

__all__ = ['sub_module']

Also note that you don't have to do from package import * to use sub_module

You can also just do:

import package.sub_module
print(package.sub_module)

Answer 2

Solution : you have __all__ set to empty list ie from package import * basically imports nothing

set it to __all__ = ['submodule'] in __init__.py

---

What exactly is __all__ ?

In simplest words all help customizing the from package import * ie with all we can set what will be imported and what not.

From the docs :

The public names defined by a module are determined by checking the module's namespace for a variable named all ; if defined, it must be a sequence of strings which are names defined or imported by that module. The names given in all are all considered public and are required to exist. If all is not defined, the set of public names includes all names found in the module's namespace which do not begin with an underscore character ('_'). all should contain the entire public API. It is intended to avoid accidentally exporting items that are not part of the API (such as library modules which were imported and used within the module).

---

One important thing to note here is - Imports without * are not affected by __all__ ie Members that are not mentioned in __all__ are accessible from outside the module using direct import - from <module> import <member> .

An Example : the following code in a module.py explicitly exports the symbols foo and bar :

__all__ = ['foo', 'bar']

waz = 5
foo = 10
def bar(): return 'bar'

These symbols can then be imported like so:

from foo import *

print foo
print bar

# now import `waz` will trigger an exception, 
# as it is not in the `__all__`, hence not a public member.
print waz

Answer 3

If you define __all__ , then only the attributes mentioned there will be imported via * , while the excluded ones have to be imported explicitly . So either use

from package import submodule

or if you really want to use the (discouraged!) from package import * , declare

__all__ = ['submodule']

in package . Note how tedious it will become to keep this up-to-date...