Python中“__package__”属性的目的是什么？

Question

All I want to know is what exactly does __package__ mean ? Didn't find any explanation in the official doc, even on SO.

If you could provide some examples I would be very happy.

Answer 1

See the PEP 366 and import system reference documentation :

The major proposed change is the introduction of a new module level attribute, __package__ . When it is present, relative imports will be based on this attribute rather than the module __name__ attribute.

and

• The module's __package__ attribute should be set. Its value must be a string, but it can be the same value as its __name__ . If the attribute is set to None or is missing, the import system will fill it in with a more appropriate value. When the module is a package, its __package__ value should be set to its __name__ . When the module is not a package, __package__ should be set to the empty string for top-level modules, or for submodules, to the parent package's name. See PEP 366 for further details.

So, for a module located in foo/bar/baz.py , __name__ is set to foo.bar.baz , and __package__ is set to foo.bar , while foo/bar/__init__.py will have foo.bar for both the __name__ and __package__ attributes.

Answer 2

All I want to know is what exactly does __package__ mean

It is the mechanism that enables explicit relative imports.

There are three possible categories of values for __package__

• A package name (a string)
• An empty string
• None

Package Name

That is, if a module is in a package, __package__ is set to the package name to enable explicit relative imports. Specifically:

When the module is a package, its __package__ value should be set to its __name__ . When the module is not a package, __package__ should be set [...] for submodules, to the parent package's name.

Empty String

If a module is at root, or top-level, that is, the current module is imported with

import current_module

or when a top-level module is run as the entry point as with:

$ python -m current_module

then __package__ is an empty string. Or as the documentation says :

When the module is not a package, __package__ should be set to the empty string for top-level modules...

None

If a module/script is run by filename, __package__ is None :

When the main module is specified by its filename, then the __package__ attribute will be set to None.

Evidence

First, let's create a file structure with noisy debugging - using Python 3.6:

text = "print(f'{__name__}, __file__: {__file__}, __package__: {repr(__package__)}')"

from pathlib import Path
Path('foo.py').write_text(text)
Path('package').mkdir()
Path('package/__init__.py').write_text(text)
Path('package/__main__.py').write_text(text)
Path('package/bar.py').write_text(text)

# and include a submodule with a relative import:
Path('package/baz.py').write_text(text + '\nfrom . import bar')

Now we see that foo.py executed as a module has an empty string for __package__ , while the script executed by file name as the entry point has None :

$ python -m foo
__main__, __file__: ~\foo.py, __package__: ''
$ python foo.py
__main__, __file__: foo.py, __package__: None

When we execute a package as a module for the entry point, its __init__.py module runs, then its __main__.py runs:

$ python -m package
package, __file__: ~\package\__init__.py, __package__: 'package'
__main__, __file__: ~\package\__main__.py, __package__: 'package'

Similarly, when we execute a submodule as a module for the entry point, the __init__.py module runs, then it runs:

$ python -m package.bar
package, __file__: ~\package\__init__.py, __package__: 'package'
__main__, __file__: ~\package\bar.py, __package__: 'package'

Finally, we see that the explicit relative import, the entire reason for having __package__ , (which happens last here) is enabled:

$ python -m package.baz
package, __file__: ~\package\__init__.py, __package__: 'package'
__main__, __file__: ~\package\baz.py, __package__: 'package'
package.bar, __file__: ~\package\bar.py, __package__: 'package'

_{Note, in the output, I have substituted ~ for the parent directories.}