How to reference python package when filename contains a period

Question:

I am using django and I have a file named models.admin.py and I want to do the following idea in models.py:

from "models.admin" import *

however, I get a syntax error for having double quotes. But if I just do 

from models.admin import *

then I get “ImportError: No module named admin”

Is there any way to import from a python file that has a period in its name?

Asked By: Alexander Bird

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Source

Answers:

The file is called models/admin.py. (Source)

That is, it should be called admin.py in a directory called models.

Then you can import using from models.admin import *, assuming that it is in your Python path.

Answered By: Mark Byers

No, you can’t import a python file as a module if its name contains a period (or a question mark, or exclamation mark, etc). A python module’s name (not including the .py) must be a valid python name (ie can be used as a variable name).

Answered By: Gabriel Reid

You are not referencing files in the import statement, you are referencing modules and packages.

Please read the docs, they are very clear on that matter.

Anyway, since you are using django, the usual approach won’t work. If you want to keep models in separate files, rather than in models.py, you have to take extra steps, outlined, for example, here.

Edit:
Well, I don’t really know what the questioneer means when he mentions admin and whether or not it is related to the admin interface of django. My points still stand.

Answered By: shylent

If you really want to, you can import a module with an unusual filename (e.g., a filename containing a ‘.’ before the ‘.py’) using the imp module:

>>> import imp
>>> a_b = imp.load_source('a.b', 'a.b.py')
>>> a_b.x
"I was defined in a.b.py!"

However, that’s generally a bad idea. It’s more likely that you’re trying to use packages, in which case you should create a directory named “a”, containing a file named “b.py”; and then “import a.b” will load a/b.py.

Answered By: Edward Loper

Actually, you can import a module with an invalid name. But you’ll need to use imp for that, e.g. assuming file is named models.admin.py, you could do

import imp
with open('models.admin.py', 'rb') as fp:
    models_admin = imp.load_module(
        'models_admin', fp, 'models.admin.py',
        ('.py', 'rb', imp.PY_SOURCE)
    )

But read the docs on imp.find_module and imp.load_module before you start using it.

Answered By: Cat Plus Plus

Like below

Assume dir structure is like this:

C:.
│   script.py
│   
└───Parent
    └───Child
        ├───1.1
        │       main.py
        │       
        └───1.2

**assume you want to import main.py in script.py **

your main.py looks like below

def my_function():
  print("Hello from a function")

your script.py looks like below

from os import path
import importlib
from os.path import dirname
import sys
import importlib.util


def getPath():
    # your logic to get to the path
    return path.join(dirname(__file__),'Parent','Child','1.1','main.py')

file_path = getPath() 
module_name = 'main'

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

#call functions like this
module.my_function()

Check out this gist

Answered By: SciTech Enthusiast

In my case, I am using grafanalib, and the filename has to be xx.dashboard.py based on the doc. However, I do want to import this file to simplify the uploading step.

I got warning when I use import imp:

the imp module is deprecated in favour of importlib and slated for removal in Python 3.12; see the module’s documentation for alternative uses

Here is the simple demo using importlib and pathlib:

foo.bar.py and main.py are in the same foler.

# foo.bar.py

num = 42

# main.py

import importlib.machinery
import pathlib

module = importlib.machinery.SourceFileLoader(
    "foo_bar",
    pathlib.Path(__file__).parent.joinpath("foo.bar.py").resolve().as_posix(),
).load_module()
print(module.num)  # 42
Answered By: Hongbo Miao