How do I get the current 'package' name? (setup.py)

Question

How do I get the current topmost package, i.e., the name defined in setup.py?

Here is my tree:

.
|-- README.md
|-- the_project_name_for_this_pkg
|   |-- __init__.py
|   |-- __main__.py
|   |-- _config
|   |   `-- foo.conf
|   |-- _data
|   |   `-- logging.yml
|   `-- tests
|       |-- __init__.py
|       `-- test_foo.py   <--- # executing from here
|-- requirements.txt
`-- setup.py

4 directories, 9 files

The only solution I've gotten to work so far is this:

import os
import sys


os.path.basename(sys.path[1])

But this is obviously a bad solution. Other solutions like having a __name__ in my uppermost __init__.py file and using ast.parse to read in the relevant section of setup.py also seems cumbersome.

Other solutions I've tried—by calling them within a unittest.TestCase inheriting class in my tests python [sub]package—include checking sys.modules[__name__], inspect.getmodule & inspect.stack, as well as the answers to these questions:

• Python - Get path of root project structure
• Get full package module name
• Get fully qualified class name of an object in Python
• How can I access the current executing module or class name in Python?
• Get full caller name (package.module.function) (Python recipe)
• https://docs.python.org/2/library/modulefinder.html

BTW: In case you were wondering why I want the package name… it's so I can run things like:

import pkg_resources


version   = pkg_resources.require('the_project_name_for_this_pkg')[0].version
data_file = path.join(resource_filename('the_project_name_for_this_pkg', '__init__.py'),
                      '_config', 'data_file.txt')

Answer 1

Not entirely sure what the larger goal is, but maybe you could be interested in reading about importlib.resources as well as importlib.metadata.

Something like the following:

import importlib.metadata
import importlib.resources

version = importlib.metadata.version('SomeProject')
data = importlib.resources.files('top_level_package.sub_package').joinpath('file.txt').read_text()

And more generally, it is near impossible (or not worth the amount of work) to 100% reliably detect the name of the project (SomeProject) from within the code. It is easier to just hard-code it.

Nevertheless here are some techniques, and ideas to retrieve the name of the project from one of its modules:

• https://bitbucket.org/pypa/distlib/issues/102/getting-the-distribution-that-a-module
• https://stackoverflow.com/a/22845276/11138259
• https://stackoverflow.com/a/56032725/11138259

---

Update:

I believe some function like the following should return the name of the installed distribution containing the current file:

import pathlib
import importlib_metadata

def get_project_name():
    for dist in importlib_metadata.distributions():
        try:
            relative = pathlib.Path(__file__).relative_to(dist.locate_file(''))
        except ValueError:
            pass
        else:
            if relative in dist.files:
                return dist.metadata['Name']
    return None

---

Update (February 2021):

Looks like this could become easier thanks to the newly added packages_distributions() function in importlib_metadata:

• https://importlib-metadata.readthedocs.io/en/stable/using.html#package-distributions
• https://github.com/python/importlib_metadata/pull/287/files

Answer 2

A solution I've been working on:

from os import listdir, path
from contextlib import suppress
import ast



def get_first_setup_py(cur_dir):
    if 'setup.py' in listdir(cur_dir):
        return path.join(cur_dir, 'setup.py')
    prev_dir = cur_dir
    cur_dir = path.realpath(path.dirname(cur_dir))
    if prev_dir == cur_dir:
        raise StopIteration()
    return get_first_setup_py(cur_dir)


setup_py_file_name = get_first_setup_py(path.dirname(__file__))

First pass:

def get_from_setup_py(setup_file): # mostly https://stackoverflow.com/a/47463422
    import importlib.util


    spec = importlib.util.spec_from_file_location('setup', setup_file)
    setup = importlib.util.module_from_spec(spec)
    spec.loader.exec_module(setup)
    # And now access it
    print(setup)

That option did work. So I returned to the ast solution I referenced in the question, and got this second pass to work:

def parse_package_name_from_setup_py(setup_py_file_name):
    with open(setup_py_file_name, 'rt') as f:
        parsed_setup_py = ast.parse(f.read(), 'setup.py')

    # Assumes you have an `if __name__ == '__main__'` block:
    main_body = next(sym for sym in parsed_setup_py.body[::-1]
                     if isinstance(sym, ast.If)).body

    setup_call = next(sym.value
                      for sym in main_body[::-1]
                      if isinstance(sym, ast.Expr)
                      and isinstance(sym.value, ast.Call)
                      and sym.value.func.id in frozenset(('setup',
                                                          'distutils.core.setup',
                                                          'setuptools.setup')))

    package_name = next(keyword
                        for keyword in setup_call.keywords
                        if keyword.arg == 'name'
                        and isinstance(keyword.value, ast.Name))

    # Return the raw string if it is one
    if isinstance(package_name.value, ast.Str):
        return package_name.value.s

    # Otherwise it's a variable defined in the `if __name__ == '__main__'` block:
    elif isinstance(package_name.value, ast.Name):
        return next(sym.value.s
                    for sym in main_body
                    if isinstance(sym, ast.Assign)
                    and isinstance(sym.value, ast.Str)
                    and any(target.id == package_name.value.id
                            for target in sym.targets)
                    )

    else:
        raise NotImplemented('Package name extraction only built for raw strings & '
                             'assigment in the same scope that setup() is called')

Third pass (works for both installed and development versions):

# Originally from https://stackoverflow.com/a/56032725;
# but made more concise and added support whence source
class App(object):
    def get_app_name(self) -> str:
        # Iterate through all installed packages and try to find one
        # that has the app's file in it
        app_def_path = inspect.getfile(self.__class__)
        with suppress(FileNotFoundError):
            return next(
                (dist.project_name
                 for dist in pkg_resources.working_set
                 if any(app_def_path == path.normpath(path.join(dist.location, r[0]))
                        for r in csv.reader(dist.get_metadata_lines('RECORD')))),
                None) or parse_package_name_from_setup_py(
                get_first_setup_py(path.dirname(__file__)))