How to import a module given its name as string?

Question

I'm writing a Python application that takes a command as an argument, for example:

$ python myapp.py command1

I want the application to be extensible, that is, to be able to add new modules that implement new commands without having to change the main application source. The tree looks something like:

myapp/
    __init__.py
    commands/
        __init__.py
        command1.py
        command2.py
    foo.py
    bar.py

So I want the application to find the available command modules at runtime and execute the appropriate one.

Python defines an __import__() function, which takes a string for a module name:

__import__(name, globals=None, locals=None, fromlist=(), level=0)

The function imports the module name, potentially using the given globals and locals to determine how to interpret the name in a package context. The fromlist gives the names of objects or submodules that should be imported from the module given by name.

Source: https://docs.python.org/3/library/functions.html#__import__

So currently I have something like:

command = sys.argv[1]
try:
    command_module = __import__("myapp.commands.%s" % command, fromlist=["myapp.commands"])
except ImportError:
    # Display error message

command_module.run()

This works just fine, I'm just wondering if there is possibly a more idiomatic way to accomplish what we are doing with this code.

Note that I specifically don't want to get in to using eggs or extension points. This is not an open-source project and I don't expect there to be "plugins". The point is to simplify the main application code and remove the need to modify it each time a new command module is added.

Answer 1

With Python older than 2.7/3.1, that's pretty much how you do it.

For newer versions, see importlib.import_module for Python 2 and Python 3.

You can use exec if you want to as well.

Or using __import__ you can import a list of modules by doing this:

>>> moduleNames = ['sys', 'os', 're', 'unittest'] 
>>> moduleNames
['sys', 'os', 're', 'unittest']
>>> modules = map(__import__, moduleNames)

Ripped straight from Dive Into Python.

Answer 2

The recommended way for Python 2.7 and 3.1 and later is to use importlib module:

importlib.import_module(name, package=None)

Import a module. The name argument specifies what module to import in absolute or relative terms (e.g. either pkg.mod or ..mod). If the name is specified in relative terms, then the package argument must be set to the name of the package which is to act as the anchor for resolving the package name (e.g. import_module('..mod', 'pkg.subpkg') will import pkg.mod).

e.g.

my_module = importlib.import_module('os.path')

Answer 3

Note: imp is deprecated since Python 3.4 in favor of importlib

As mentioned the imp module provides you loading functions:

imp.load_source(name, path)
imp.load_compiled(name, path)

I've used these before to perform something similar.

In my case I defined a specific class with defined methods that were required. Once I loaded the module I would check if the class was in the module, and then create an instance of that class, something like this:

import imp
import os

def load_from_file(filepath):
    class_inst = None
    expected_class = 'MyClass'

    mod_name,file_ext = os.path.splitext(os.path.split(filepath)[-1])

    if file_ext.lower() == '.py':
        py_mod = imp.load_source(mod_name, filepath)

    elif file_ext.lower() == '.pyc':
        py_mod = imp.load_compiled(mod_name, filepath)

    if hasattr(py_mod, expected_class):
        class_inst = getattr(py_mod, expected_class)()

    return class_inst

Answer 4

Nowadays you should use importlib.

Import a source file

The docs actually provide a recipe for that, and it goes like:

import sys
import importlib.util

file_path = 'pluginX.py'
module_name = 'pluginX'

spec = importlib.util.spec_from_file_location(module_name, file_path)
module = importlib.util.module_from_spec(spec)
spec.loader.exec_module(module)

# check if it's all there..
def bla(mod):
    print(dir(mod))
bla(module)

This way you can access the members (e.g, a function "hello") from your module pluginX.py -- in this snippet being called module -- under its namespace; E.g, module.hello().

If you want to import the members (e.g, "hello") you can include module/pluginX in the in-memory list of modules:

sys.modules[module_name] = module

from pluginX import hello
hello()

Import a package

Importing a package (e.g., pluginX/__init__.py) under your current dir is actually straightforward:

import importlib

pkg = importlib.import_module('pluginX')

# check if it's all there..
def bla(mod):
    print(dir(mod))
bla(pkg)

Answer 5

Use the imp module, or the more direct __import__() function.

Answer 6

If you want it in your locals:

>>> mod = 'sys'
>>> locals()['my_module'] = __import__(mod)
>>> my_module.version
'2.6.6 (r266:84297, Aug 24 2010, 18:46:32) [MSC v.1500 32 bit (Intel)]'

same would work with globals()

Answer 7

You can use exec:

exec("import myapp.commands.%s" % command)

Answer 8

Similar as @monkut 's solution but reusable and error tolerant described here http://stamat.wordpress.com/dynamic-module-import-in-python/:

import os
import imp

def importFromURI(uri, absl):
    mod = None
    if not absl:
        uri = os.path.normpath(os.path.join(os.path.dirname(__file__), uri))
    path, fname = os.path.split(uri)
    mname, ext = os.path.splitext(fname)

    if os.path.exists(os.path.join(path,mname)+'.pyc'):
        try:
            return imp.load_compiled(mname, uri)
        except:
            pass
    if os.path.exists(os.path.join(path,mname)+'.py'):
        try:
            return imp.load_source(mname, uri)
        except:
            pass

    return mod

Answer 9

The below piece worked for me:

>>>import imp; 
>>>fp, pathname, description = imp.find_module("/home/test_module"); 
>>>test_module = imp.load_module("test_module", fp, pathname, description);
>>>print test_module.print_hello();

if you want to import in shell-script:

python -c '<above entire code in one line>'