Python get current package name

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

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__))) 

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:

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 :

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Источник

Get Module File Information in Python

In Python sometimes you need to figure out where a function lives on the hard disk, or what file is calling your function. The os module provides tools for getting file and directory information and the inspect module provides the ability to inspect the stack and see what module is calling your function.

Get file information about a module

The os module in Python provides many useful ways for getting the file name of where a module lives. There are functions to get file names, directory names, and get absolute paths. Check out this example:

# Example assumes this file is named test.py
# and lives in /home/nanodano/
# https://docs.python.org/3/library/os.path.html

import os

# Prints the full path of where the os.py module resides
print(os.__file__)


# The name of file where this print statement is written
print(__file__)
# test.py

# Get the full absolute path using os.path.abspath()
print(os.path.abspath(__file__))
# /home/nanodano/test.py

# Get the directory name of a file with os.path.dirname()
print(os.path.dirname(__file__))
# Relative to current directory
# Prints an empty string if being run from current directory
# Prints .. if being called from a deeper directory

# Combine to get absolute path of the directory containing the python file
print(os.path.abspath(os.path.dirname(__file__)))
# /home/nanodano

## Other functions that might be useful
# os.path.join() # Join directory names using system specific dir separator
# os.sep() # Path separator character (system specific)
# os.getcwd() # Full absolute path
# os.walk() # Generator to walk dirs recursively
# os.listdir() # Current dir or specific dir
 
Get module and file name of caller
 
In some cases, your function is getting called by code from other .py files. If you want to figure out where your function is being called from, you can inspect the stack. The inspect module provides a way to get this data.
 
import inspect

# Import this to other module and call it
def print_caller_info(): 
# Get the full stack 
stack = inspect.stack() 
 
# Get one level up from current 
previous_stack_frame = stack[1] 
print(previous_stack_frame.filename) # Filename where caller lives
 
# Get the module object of the caller 
calling_module = inspect.getmodule(stack_frame[0]) 
print(calling_module) 
print(calling_module.__file__)


if __name__ == '__main__': 
print_caller_info()
 
Conclusion
 
After reading this, you should know how to get the relative and absolute filename of the current Python modules as well as modules that are calling your function.
 
Reference links
 
Источник