95

I have a Python source code file called foobar. It has no .py extension because I also use it as a standalone script, and I don't want to type the .py extension to run it. In the same directory I have another python file that needs to import that code. But import foobar doesn't work; I would have to rename the file to foobar.py.

Is there some other way to import a Python module that doesn't have the .py extension?

5
  • 1
    I'm intrigued. Why do you have a python file without the py extension? Commented Apr 8, 2010 at 16:52
  • 3
    Sometimes it's nice to use python for configuration files (extension as .conf) or to denote a special type of file. In my case, it'd be more of a convenience for an Administrator. Commented Aug 2, 2013 at 3:01
  • 1
    I have a file with configuration that is used both as a python file and as a bash script. I gave it a pysh extension... Commented Jul 11, 2014 at 21:43
  • If that is configuration related things, I recommend using ConfigParser. wiki.python.org/moin/ConfigParserExamples Commented Nov 23, 2015 at 7:53
  • 3
    @voyager One reason is python scripts with .cgi extensions instead of .py extension Commented Jun 15, 2016 at 1:49

8 Answers 8

70

You can use the imp.load_source function (from the imp module), to load a module dynamically from a given file-system path.

import imp
foobar = imp.load_source('foobar', '/path/to/foobar')

This SO discussion also shows some interesting options.

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7 Comments

Fixed. [it is more constructive to suggest an Edit, though]
What is the use of the first argument 'foobar' if it is assigned in the return value?
@AnmolSinghJaggi It sets the module __name__ property; normally that's determined from the filename, but since you're using a non-standard filename (which might not even contain any valid python identifier at all), you have to specify the module name. The variable name in which you store a reference to the created module object is irrelevant, much as if you import foo.bar as baz the module referenced by the variable baz will still have its original __name__.
This has been deprecated since 3.4. Any idea how to import from a file without the .py extension in 3.4+?
For Python 3.4+: this answer is more exact: stackoverflow.com/questions/2601047/…
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44

Here is a solution for Python 3.4+:

from importlib.util import spec_from_loader, module_from_spec
from importlib.machinery import SourceFileLoader 

spec = spec_from_loader("foobar", SourceFileLoader("foobar", "/path/to/foobar"))
foobar = module_from_spec(spec)
spec.loader.exec_module(foobar)

Using spec_from_loader and explicitly specifying a SourceFileLoader will force the machinery to load the file as source, without trying to figure out the type of the file from the extension. This means that you can load the file even though it is not listed in importlib.machinery.SOURCE_SUFFIXES.

If you want to keep importing the file by name after the first load, add the module to sys.modules:

sys.modules['foobar'] = foobar

You can find an implementation of this function in a utility library I maintain called haggis. haggis.load.load_module has options for adding the module to sys.modules, setting a custom name, and injecting variables into the namespace for the code to use.

6 Comments

That module needs a foobar = importlib.nice_import('foobar') helper desperately.
@Ciro. It already has that. I don't think that /some/arbitrary/file.weird_extension qualifies as "nice". That being said, I've started using python code for all my configuration files once I discovered this. It's just so convenient.
What if I want to import * ?
@AlexHarvey. This gives you a module object. You can do something like globals().update(foobar.__dict__) or so, but I would recommend against it.
I'm trying to do this as a single executable and import it just for testing. A bit disappointed at the answers, too much speculation, too few concrete answers... is there a 'nice_import' ? Name it!
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18

Like others have mentioned, you could use imp.load_source, but it will make your code more difficult to read. I would really only recommend it if you need to import modules whose names or paths aren't known until run-time.

What is your reason for not wanting to use the .py extension? The most common case for not wanting to use the .py extension, is because the python script is also run as an executable, but you still want other modules to be able to import it. If this is the case, it might be beneficial to move functionality into a .py file with a similar name, and then use foobar as a wrapper.

5 Comments

Or instead of wrapping, just symlink foobar.py to foobar (assuming you aren't on Windows)
@whaley, yeah, that would be much cleaner. You could use a .bat for windows to accomplish the same thing.
I've got a neat use case - a readme file, with examples in it, which I'd like doctest to validate. I'm hoping to make a doctest markdown doc that works...
And the answer is (for that use case) - use doctest.loadfile!
The issue with wrappers is that if someone naively copies just the wrapper to a bin/ directory, the program won't work when run from the path.
14

imp.load_source(module_name, path) should do or you can do the more verbose imp.load_module(module_name, file_handle, ...) route if you have a file handle instead

Comments

8

importlib helper function

Here is a convenient, ready-to-use helper to replace imp, with an example, based on what was mentioned at: https://stackoverflow.com/a/43602645/895245

main.py

#!/usr/bin/env python3

import os
import importlib
import sys

def import_path(path):
    module_name = os.path.basename(path).replace('-', '_')
    spec = importlib.util.spec_from_loader(
        module_name,
        importlib.machinery.SourceFileLoader(module_name, path)
    )
    module = importlib.util.module_from_spec(spec)
    spec.loader.exec_module(module)
    sys.modules[module_name] = module
    return module

notmain = import_path('not-main')
print(notmain)
print(notmain.x)

not-main

x = 1

Run:

python3 main.py

Output:

<module 'not_main' from 'not-main'>
1

I replace - with _ because my importable Python executables without extension have hyphens. This is not mandatory, but produces better module names.

This pattern is also mentioned in the docs at: https://docs.python.org/3.7/library/importlib.html#importing-a-source-file-directly

I ended up moving to it because after updating to Python 3.7, import imp prints:

DeprecationWarning: the imp module is deprecated in favour of importlib; see the module's documentation for alternative uses

and I don't know how to turn that off, this was asked at:

Tested in Python 3.7.3.

Comments

2

If you install the script with package manager (deb or alike) another option would be to use setuptools:

"...there’s no easy way to have a script’s filename match local conventions on both Windows and POSIX platforms. For another, you often have to create a separate file just for the “main” script, when your actual “main” is a function in a module somewhere... setuptools fixes all of these problems by automatically generating scripts for you with the correct extension, and on Windows it will even create an .exe file..."

https://pythonhosted.org/setuptools/setuptools.html#automatic-script-creation

Comments

1

While the importlib answer is the correct way to do it, I was looking for something more concise.

You can use the runpy module, with the following caveat (emphasis mine, links removed):

Furthermore, any functions and classes defined by the executed code are not guaranteed to work correctly after a runpy function has returned. If that limitation is not acceptable for a given use case, importlib is likely to be a more suitable choice than this module.

Having said that, as of Python 3.11, the following works perfectly:

import runpy
import types

my_module = types.SimpleNamespace(**runpy.run_path("/path/to/file"))

Comments

0

import imp has been deprecated.

The following is clean and minimal for me:

import sys
import types
import  pathlib

def importFileAs(
        modAsName: str,
        importedFilePath: typing.Union[str,  pathlib.Path],
) -> types.ModuleType:
    """ Import importedFilePath as modAsName, return imported module
by loading importedFilePath and registering modAsName in sys.modules.
importedFilePath can be any file and does not have to be a .py file. modAsName should be python valid.
Raises ImportError: If the file cannot be imported or any Exception: occuring during loading.

Refs:
Similar to: https://stackoverflow.com/questions/19009932/import-arbitrary-python-source-file-python-3-3
    but allows for other than .py files as well through importlib.machinery.SourceFileLoader.
    """
    import importlib.util
    import importlib.machinery

    # from_loader does not enforce .py but  importlib.util.spec_from_file_location() does.
    spec = importlib.util.spec_from_loader(
        modAsName,
        importlib.machinery.SourceFileLoader(modAsName, importedFilePath),
    )
    if spec is None:
        raise ImportError(f"Could not load spec for module '{modAsName}' at: {importedFilePath}")
    module = importlib.util.module_from_spec(spec)

    try:
        spec.loader.exec_module(module)
    except FileNotFoundError as e:
        raise ImportError(f"{e.strerror}: {importedFilePath}") from e

    sys.modules[modAsName] = module
    return module

And then I would use it as so:

aasMarmeeManage = importFileAs('aasMarmeeManage', '/bisos/bpip/bin/aasMarmeeManage.cs')
def g_extraParams(): aasMarmeeManage.g_extraParams()

Comments

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