How to check if a Python module exists without importing it

Question:

How can I know if a Python module exists, without importing it?

Importing something that might not exist (not what I want) results in:

try:
    import eggs
except ImportError:
    pass
Asked By: yarbelk

||

Answers:

TL;DR) Use importlib.util.find_spec(module_name) (Python 3.4+).

Python2: imp.find_module

To check if import can find something in Python 2, using imp:

import imp
try:
    imp.find_module('eggs')
    found = True
except ImportError:
    found = False

To find dotted imports, you need to do more:

import imp
try:
    spam_info = imp.find_module('spam')
    spam = imp.load_module('spam', *spam_info)
    imp.find_module('eggs', spam.__path__) # __path__ is already a list
    found = True
except ImportError:
    found = False

You can also use pkgutil.find_loader (more or less the same as the Python 3 part:

import pkgutil
eggs_loader = pkgutil.find_loader('eggs')
found = eggs_loader is not None

Python 3

Python 3 ≤ 3.3: importlib.find_loader

You should use importlib. I went about doing this like:

import importlib
spam_loader = importlib.find_loader('spam')
found = spam_loader is not None

My expectation being, if you can find a loader for it, then it exists. You can also be a bit more smart about it, like filtering out what loaders you will accept. For example:

import importlib
spam_loader = importlib.find_loader('spam')
# only accept it as valid if there is a source file for the module - no bytecode only.
found = issubclass(type(spam_loader), importlib.machinery.SourceFileLoader)

Python 3 ≥ 3.4: importlib.util.find_spec

In Python 3.4 importlib.find_loader Python documentation was deprecated in favour of importlib.util.find_spec. The recommended method is the importlib.util.find_spec. There are others like importlib.machinery.FileFinder, which is useful if you’re after a specific file to load. Figuring out how to use them is beyond the scope of this.

import importlib.util
spam_spec = importlib.util.find_spec("spam")
found = spam_spec is not None

This also works with relative imports, but you must supply the starting package, so you could also do:

import importlib.util
spam_spec = importlib.util.find_spec("..spam", package="eggs.bar")
found = spam_spec is not None
spam_spec.name == "eggs.spam"

While I’m sure there exists a reason for doing this – I’m not sure what it would be.

Warning

When trying to find a submodule, it will import the parent module (for ALL of the above methods)!

food/
  |- __init__.py
  |- eggs.py

## __init__.py
print("module food loaded")

## eggs.py
print("module eggs")

were you then to run
>>> import importlib
>>> spam_spec = importlib.util.find_spec("food.eggs")
module food loaded
ModuleSpec(name='food.eggs', loader=<_frozen_importlib.SourceFileLoader object at 0x10221df28>, origin='/home/user/food/eggs.py')

Comments are welcome on getting around this

Acknowledgements

  • @rvighne for importlib
  • @lucas-guido for Python 3.3+ deprecating find_loader
  • @enpenax for pkgutils.find_loader behaviour in Python 2.7
Answered By: yarbelk

Use one of the functions from pkgutil, for example:

from pkgutil import iter_modules

def module_exists(module_name):
    return module_name in (name for loader, name, ispkg in iter_modules())
Answered By: ArtOfWarfare

After using yarbelk’s response, I’ve made this so I don’t have to import ìmp.

try:
    __import__('imp').find_module('eggs')
    # Make things with a supposed existing module
except ImportError:
    pass

It is useful in Django‘s settings.py file, for example.

Answered By: Zulu

Here is a way to check if a module is loaded from the command line:

Linux/UNIX script file method: make a file module_help.py:

#!/usr/bin/env python

help('modules')

Then make sure it’s executable: chmod u+x module_help.py

And call it with a pipe to grep:

./module_help.py | grep module_name

Invoke the built-in help system. (This function is intended for interactive use.) If no argument is given, the interactive help system starts on the interpreter console. If the argument is a string, then the string is looked up as the name of a module, function, class, method, keyword, or documentation topic, and a help page is printed on the console. If the argument is any other kind of object, a help page on the object is generated.

Interactive method: in the console, load python

>>> help('module_name')

If found, quit reading by typing q.
To exit the Python interpreter interactive session, press Ctrl + D

Windows script file method, also Linux/UNIX compatible, and better overall:

#!/usr/bin/env python

import sys

help(sys.argv[1])

Calling it from the command like:

python module_help.py site

Would output:

Help on module site:

NAME
site – Append module search paths for third-party packages to sys.path.

FILE
/usr/lib/python2.7/site.py

MODULE DOCS
http://docs.python.org/library/site

DESCRIPTION

:

And you’d have to press q to exit interactive mode.

Using it for an unknown module, e.g.,

python module_help.py lkajshdflkahsodf

Would output:

no Python documentation found for ‘lkajshdflkahsodf’

and exit.

Answered By: go_as

go_as’s answer as a one-liner:

 python -c "help('modules');" | grep module
Answered By: user2268226

Python 2, without relying on ImportError

Until the current answer is updated, here is the way for Python 2

import pkgutil
import importlib

if pkgutil.find_loader(mod) is not None:
    return importlib.import_module(mod)
return None

Why another answer?

A lot of answers make use of catching an ImportError. The problem with that is, that we cannot know what throws the ImportError.

If you import your existent module and there happens to be an ImportError in your module (e.g., typo on line 1), the result will be that your module does not exist.

It will take you quite the amount of backtracking to figure out that your module exists and the ImportError is caught and makes things fail silently.

Answered By: enpenax

There isn’t any way to reliably check if "dotted module" is importable without importing its parent package. Saying this, there are many solutions to problem "how to check if a Python module exists".

The below solution addresses the problem that an imported module can raise an ImportError even if it exists. We want to distinguish that situation from such in which the module does not exist.

Python 2:

import importlib
import pkgutil
import sys

def find_module(full_module_name):
    """
    Returns module object if module `full_module_name` can be imported.

    Returns None if module does not exist.

    Exception is raised if (existing) module raises exception during its import.
    """
    module = sys.modules.get(full_module_name)
    if module is None:
        module_path_tail = full_module_name.split('.')
        module_path_head = []
        loader = True
        while module_path_tail and loader:
            module_path_head.append(module_path_tail.pop(0))
            module_name = ".".join(module_path_head)
            loader = bool(pkgutil.find_loader(module_name))
            if not loader:
                # Double check if module realy does not exist
                # (case: full_module_name == 'paste.deploy')
                try:
                    importlib.import_module(module_name)
                except ImportError:
                    pass
                else:
                    loader = True
        if loader:
            module = importlib.import_module(full_module_name)
    return module

Python 3:

import importlib

def find_module(full_module_name):
    """
    Returns module object if module `full_module_name` can be imported.

    Returns None if module does not exist.

    Exception is raised if (existing) module raises exception during its import.
    """
    try:
        return importlib.import_module(full_module_name)
    except ImportError as exc:
        if not (full_module_name + '.').startswith(exc.name + '.'):
            raise
Answered By: Marcin Raczyński

In django.utils.module_loading.module_has_submodule:


import sys
import os
import imp

def module_has_submodule(package, module_name):
    """
    check module in package
    django.utils.module_loading.module_has_submodule
    """
    name = ".".join([package.__name__, module_name])
    try:
        # None indicates a cached miss; see mark_miss() in Python/import.c.
        return sys.modules[name] is not None
    except KeyError:
        pass
    try:
        package_path = package.__path__   # No __path__, then not a package.
    except AttributeError:
        # Since the remainder of this function assumes that we're dealing with
        # a package (module with a __path__), so if it's not, then bail here.
        return False
    for finder in sys.meta_path:
        if finder.find_module(name, package_path):
            return True
    for entry in package_path:
        try:
            # Try the cached finder.
            finder = sys.path_importer_cache[entry]
            if finder is None:
                # Implicit import machinery should be used.
                try:
                    file_, _, _ = imp.find_module(module_name, [entry])
                    if file_:
                        file_.close()
                    return True
                except ImportError:
                    continue
            # Else see if the finder knows of a loader.
            elif finder.find_module(name):
                return True
            else:
                continue
        except KeyError:
            # No cached finder, so try and make one.
            for hook in sys.path_hooks:
                try:
                    finder = hook(entry)
                    # XXX Could cache in sys.path_importer_cache
                    if finder.find_module(name):
                        return True
                    else:
                        # Once a finder is found, stop the search.
                        break
                except ImportError:
                    # Continue the search for a finder.
                    continue
            else:
                # No finder found.
                # Try the implicit import machinery if searching a directory.
                if os.path.isdir(entry):
                    try:
                        file_, _, _ = imp.find_module(module_name, [entry])
                        if file_:
                            file_.close()
                        return True
                    except ImportError:
                        pass
                # XXX Could insert None or NullImporter
    else:
        # Exhausted the search, so the module cannot be found.
        return False
Answered By: dibrovsd

You could just write a little script that would try to import all the modules and tell you which ones are failing and which ones are working:

import pip


if __name__ == '__main__':
    for package in pip.get_installed_distributions():
        pack_string = str(package).split(" ")[0]
        try:
            if __import__(pack_string.lower()):
                print(pack_string + " loaded successfully")
        except Exception as e:
            print(pack_string + " failed with error code: {}".format(e))

Output:

zope.interface loaded successfully
zope.deprecation loaded successfully
yarg loaded successfully
xlrd loaded successfully
WMI loaded successfully
Werkzeug loaded successfully
WebOb loaded successfully
virtualenv loaded successfully
...

A word of warning: this will try to import everything, so you’ll see things like PyYAML failed with error code: No module named pyyaml, because the actual import name is just yaml. So as long as you know your imports, this should do the trick for you.

Answered By: User9123

You can also use importlib.util directly

import importlib.util    

def module_exists_without_import(module_name):    
    spec = importlib.util.find_spec(module_name)
    return spec is not None
Answered By: jackotonye

Python 3 >= 3.6: ModuleNotFoundError

The ModuleNotFoundError has been introduced in Python 3.6 and can be used for this purpose:

try:
    import eggs
except ModuleNotFoundError:
    # Error handling
    pass

The error is raised when a module or one of its parents cannot be found. So

try:
    import eggs.sub
except ModuleNotFoundError as err:
    # Error handling
    print(err)

would print a message that looks like No module named 'eggs' if the eggs module cannot be found; but it would print something like No module named 'eggs.sub' if only the sub module couldn’t be found, but the eggs package could be found.

See the documentation of the import system for more information on the ModuleNotFoundError.

Answered By: J.Baoby

A simpler if statement from Ask Ubuntu, How do I check whether a module is installed in Python?:

import sys
print('eggs' in sys.modules)
Answered By: Sylvain

I wrote this helper function:

def is_module_available(module_name):
    if sys.version_info < (3, 0):
        # python 2
        import importlib
        torch_loader = importlib.find_loader(module_name)
    elif sys.version_info <= (3, 3):
        # python 3.0 to 3.3
        import pkgutil
        torch_loader = pkgutil.find_loader(module_name)
    elif sys.version_info >= (3, 4):
        # python 3.4 and above
        import importlib
        torch_loader = importlib.util.find_spec(module_name)

    return torch_loader is not None
Answered By: Fardin Abdi

In case you know the location of file and want to check that the respective Python code file has that module or not, you can simply check via the astor package in Python. Here is a quick example:

"""
Check if a module function exists or not without importing a Python package file
"""
import ast
import astor

tree = astor.parse_file('handler.py')
method_to_check = 'handle'
for item in tree.body:
    if isinstance(item, ast.FunctionDef):
        if item.name == method_to_check:
            print('method exists')
            break
Answered By: Roshan Chauhan
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