Is there a standard way to list names of Python mo

问题:

Is there a straightforward way to list the names of all modules in a package, without using __all__?

For example, given this package:

/testpkg
/testpkg/__init__.py
/testpkg/modulea.py
/testpkg/moduleb.py

I'm wondering if there is a standard or built-in way to do something like this:

>>> package_contents("testpkg")
['modulea', 'moduleb']

The manual approach would be to iterate through the module search paths in order to find the package's directory. One could then list all the files in that directory, filter out the uniquely-named py/pyc/pyo files, strip the extensions, and return that list. But this seems like a fair amount of work for something the module import mechanism is already doing internally. Is that functionality exposed anywhere?

回答1:

Maybe this will do what you're looking for?

import imp
import os
MODULE_EXTENSIONS = ('.py', '.pyc', '.pyo')

def package_contents(package_name):
    file, pathname, description = imp.find_module(package_name)
    if file:
        raise ImportError('Not a package: %r', package_name)
    # Use a set because some may be both source and compiled.
    return set([os.path.splitext(module)[0]
        for module in os.listdir(pathname)
        if module.endswith(MODULE_EXTENSIONS)])

回答2:

Using python2.3 and above, you could also use the pkgutil module:

>>> import pkgutil
>>> [name for _, name, _ in pkgutil.iter_modules(['testpkg'])]
['modulea', 'moduleb']

EDIT: Note that the parameter is not a list of modules, but a list of paths, so you might want to do something like this:

>>> import os.path, pkgutil
>>> import testpkg
>>> pkgpath = os.path.dirname(testpkg.__file__)
>>> print [name for _, name, _ in pkgutil.iter_modules([pkgpath])]

回答3:

import module
help(module)

回答4:

Don't know if I'm overlooking something, or if the answers are just out-dated but;

As stated by user815423426 this only works for live objects and the listed modules are only modules that were imported before.

Listing modules in a package seems really easy using inspect:

>>> import inspect, testpkg
>>> inspect.getmembers(testpkg, inspect.ismodule)
['modulea', 'moduleb']

回答5:

This is a recursive version that works with python 3.6 and above:

import importlib.util
from pathlib import Path
import os
MODULE_EXTENSIONS = '.py'

def package_contents(package_name):
    spec = importlib.util.find_spec(package_name)
    if spec is None:
        return set()

    pathname = Path(spec.origin).parent
    ret = set()
    with os.scandir(pathname) as entries:
        for entry in entries:
            if entry.name.startswith('__'):
                continue
            current = '.'.join((package_name, entry.name.partition('.')[0]))
            if entry.is_file():
                if entry.name.endswith(MODULE_EXTENSIONS):
                    ret.add(current)
            elif entry.is_dir():
                ret.add(current)
                ret |= package_contents(current)


    return ret

回答6:

Based on cdleary's example, here's a recursive version listing path for all submodules:

import imp, os

def iter_submodules(package):
    file, pathname, description = imp.find_module('isc_datasources')
    for dirpath, _, filenames in os.walk(pathname):
        for  filename in filenames:
            if os.path.splitext(filename)[1] == ".py":
                yield os.path.join(dirpath, filename)

回答7:

This should list the modules:

help("modules")

回答8:

print dir(module)

回答9:

def package_contents(package_name):
  package = __import__(package_name)
  return [module_name for module_name in dir(package) if not module_name.startswith("__")]