I am running Python 2.5.
This is my folder tree:
ptdraft/
nib.py
simulations/
life/
life.py
(I also have __init__.py in each folder, omitted here for readability)
How do I import the nib module from inside the life module? I am hoping it is possible to do without tinkering with sys.path.
Note: The main module being run is in the ptdraft folder.
You could use relative imports (python >= 2.5):
from ... import nib
(What’s New in Python 2.5) PEP 328: Absolute and Relative Imports
EDIT: added another dot '.' to go up two packages
I posted a similar answer also to the question regarding imports from sibling packages. You can see it here.
Solution without sys.path hacks
Summary
Wrap the code into one folder (e.g. packaged_stuff)
Create a setup.py script where you use setuptools.setup().
Pip install the package in editable state with pip install -e <myproject_folder>
Import using from packaged_stuff.modulename import function_name
Setup
I assume the same folder structure as in the question
.
└── ptdraft
├── __init__.py
├── nib.py
└── simulations
├── __init__.py
└── life
├── __init__.py
└── life.py
I call the . the root folder, and in my case it is located in C:\tmp\test_imports.
Steps
Add a setup.py to the root folder
--
The contents of the setup.py can be simply
from setuptools import setup, find_packages
setup(name='myproject', version='1.0', packages=find_packages())
Basically "any" setup.py would work. This is just a minimal working example.
Use a virtual environment
If you are familiar with virtual environments, activate one, and skip to the next step. Usage of virtual environments are not absolutely required, but they will really help you out in the long run (when you have more than 1 project ongoing..). The most basic steps are (run in the root folder)
Create virtual env
python -m venv venv
Activate virtual env
. venv/bin/activate (Linux) or ./venv/Scripts/activate (Win)
Deactivate virtual env
deactivate (Linux)
To learn more about this, just Google out "python virtualenv tutorial" or similar. You probably never need any other commands than creating, activating and deactivating.
Once you have made and activated a virtual environment, your console should give the name of the virtual environment in parenthesis
PS C:\tmp\test_imports> python -m venv venv
PS C:\tmp\test_imports> .\venv\Scripts\activate
(venv) PS C:\tmp\test_imports>
pip install your project in editable state
Install your top level package myproject using pip. The trick is to use the -e flag when doing the install. This way it is installed in an editable state, and all the edits made to the .py files will be automatically included in the installed package.
In the root directory, run
pip install -e . (note the dot, it stands for "current directory")
You can also see that it is installed by using pip freeze
(venv) PS C:\tmp\test_imports> pip install -e .
Obtaining file:///C:/tmp/test_imports
Installing collected packages: myproject
Running setup.py develop for myproject
Successfully installed myproject
(venv) PS C:\tmp\test_imports> pip freeze
myproject==1.0
Import by prepending mainfolder to every import
In this example, the mainfolder would be ptdraft. This has the advantage that you will not run into name collisions with other module names (from python standard library or 3rd party modules).
Example Usage
nib.py
def function_from_nib():
print('I am the return value from function_from_nib!')
life.py
from ptdraft.nib import function_from_nib
if __name__ == '__main__':
function_from_nib()
Running life.py
(venv) PS C:\tmp\test_imports> python .\ptdraft\simulations\life\life.py
I am the return value from function_from_nib!
Relative imports (as in from .. import mymodule) only work in a package.
To import 'mymodule' that is in the parent directory of your current module:
import os
import sys
import inspect
currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
parentdir = os.path.dirname(currentdir)
sys.path.insert(0, parentdir)
import mymodule
edit: the __file__ attribute is not always given. Instead of using os.path.abspath(__file__) I now suggested using the inspect module to retrieve the filename (and path) of the current file
It seems that the problem is not related to the module being in a parent directory or anything like that.
You need to add the directory that contains ptdraft to PYTHONPATH
You said that import nib worked with you, that probably means that you added ptdraft itself (not its parent) to PYTHONPATH.
You can use OS depending path in "module search path" which is listed in sys.path .
So you can easily add parent directory like following
import sys
sys.path.insert(0,'..')
If you want to add parent-parent directory,
sys.path.insert(0,'../..')
This works both in python 2 and 3.
Don't know much about python 2.
In python 3, the parent folder can be added as follows:
import sys
sys.path.append('..')
...and then one is able to import modules from it
If adding your module folder to the PYTHONPATH didn't work, You can modify the sys.path list in your program where the Python interpreter searches for the modules to import, the python documentation says:
When a module named spam is imported, the interpreter first searches for a built-in module with that name. If not found, it then searches for a file named spam.py in a list of directories given by the variable sys.path. sys.path is initialized from these locations:
the directory containing the input script (or the current directory).
PYTHONPATH (a list of directory names, with the same syntax as the shell variable PATH).
the installation-dependent default.
After initialization, Python programs can modify sys.path. The directory containing the script being run is placed at the beginning of the search path, ahead of the standard library path. This means that scripts in that directory will be loaded instead of modules of the same name in the library directory. This is an error unless the replacement is intended.
Knowing this, you can do the following in your program:
import sys
# Add the ptdraft folder path to the sys.path list
sys.path.append('/path/to/ptdraft/')
# Now you can import your module
from ptdraft import nib
# Or just
import ptdraft
Here is an answer that's simple so you can see how it works, small and cross-platform.
It only uses built-in modules (os, sys and inspect) so should work
on any operating system (OS) because Python is designed for that.
Shorter code for answer - fewer lines and variables
from inspect import getsourcefile
import os.path as path, sys
current_dir = path.dirname(path.abspath(getsourcefile(lambda:0)))
sys.path.insert(0, current_dir[:current_dir.rfind(path.sep)])
import my_module # Replace "my_module" here with the module name.
sys.path.pop(0)
For less lines than this, replace the second line with import os.path as path, sys, inspect,
add inspect. at the start of getsourcefile (line 3) and remove the first line.
- however this imports all of the module so could need more time, memory and resources.
The code for my answer (longer version)
from inspect import getsourcefile
import os.path
import sys
current_path = os.path.abspath(getsourcefile(lambda:0))
current_dir = os.path.dirname(current_path)
parent_dir = current_dir[:current_dir.rfind(os.path.sep)]
sys.path.insert(0, parent_dir)
import my_module # Replace "my_module" here with the module name.
It uses an example from a Stack Overflow answer How do I get the path of the current
executed file in Python? to find the source (filename) of running code with a built-in tool.
from inspect import getsourcefile
from os.path import abspath
Next, wherever you want to find the source file from you just use:
abspath(getsourcefile(lambda:0))
My code adds a file path to sys.path, the python path list
because this allows Python to import modules from that folder.
After importing a module in the code, it's a good idea to run sys.path.pop(0) on a new line
when that added folder has a module with the same name as another module that is imported
later in the program. You need to remove the list item added before the import, not other paths.
If your program doesn't import other modules, it's safe to not delete the file path because
after a program ends (or restarting the Python shell), any edits made to sys.path disappear.
Notes about a filename variable
My answer doesn't use the __file__ variable to get the file path/filename of running
code because users here have often described it as unreliable. You shouldn't use it
for importing modules from parent folder in programs used by other people.
Some examples where it doesn't work (quote from this Stack Overflow question):
• it can't be found on some platforms • it sometimes isn't the full file path
py2exe doesn't have a __file__ attribute, but there is a workaround
When you run from IDLE with execute() there is no __file__ attribute
OS X 10.6 where I get NameError: global name '__file__' is not defined
Here is more generic solution that includes the parent directory into sys.path (works for me):
import os.path, sys
sys.path.append(os.path.join(os.path.dirname(os.path.realpath(__file__)), os.pardir))
The pathlib library (included with >= Python 3.4) makes it very concise and intuitive to append the path of the parent directory to the PYTHONPATH:
import sys
from pathlib import Path
sys.path.append(str(Path('.').absolute().parent))
In a Jupyter Notebook (opened with Jupyter LAB or Jupyter Notebook)
As long as you're working in a Jupyter Notebook, this short solution might be useful:
%cd ..
import nib
It works even without an __init__.py file.
I tested it with Anaconda3 on Linux and Windows 7.
I found the following way works for importing a package from the script's parent directory. In the example, I would like to import functions in env.py from app.db package.
.
└── my_application
└── alembic
└── env.py
└── app
├── __init__.py
└── db
import os
import sys
currentdir = os.path.dirname(os.path.realpath(__file__))
parentdir = os.path.dirname(currentdir)
sys.path.append(parentdir)
Above mentioned solutions are also fine. Another solution to this problem is
If you want to import anything from top level directory. Then,
from ...module_name import *
Also, if you want to import any module from the parent directory. Then,
from ..module_name import *
Also, if you want to import any module from the parent directory. Then,
from ...module_name.another_module import *
This way you can import any particular method if you want to.
Two line simplest solution
import os, sys
sys.path.insert(0, os.getcwd())
If parent is your working directory and you want to call another child modules from child scripts.
You can import all child modules from parent directory in any scripts and execute it as
python child_module1/child_script.py
For me the shortest and my favorite oneliner for accessing to the parent directory is:
sys.path.append(os.path.dirname(os.getcwd()))
or:
sys.path.insert(1, os.path.dirname(os.getcwd()))
os.getcwd() returns the name of the current working directory, os.path.dirname(directory_name) returns the directory name for the passed one.
Actually, in my opinion Python project architecture should be done the way where no one module from child directory will use any module from the parent directory. If something like this happens it is worth to rethink about the project tree.
Another way is to add parent directory to PYTHONPATH system environment variable.
Though the original author is probably no longer looking for a solution, but for completeness, there one simple solution. It's to run life.py as a module like this:
cd ptdraft
python -m simulations.life.life
This way you can import anything from nib.py as ptdraft directory is in the path.
I think you can try this in that specific example, but in python 3.6.3
import sys
sys.path.append('../')
same sort of style as the past answer - but in fewer lines :P
import os,sys
parentdir = os.path.dirname(__file__)
sys.path.insert(0,parentdir)
file returns the location you are working in
In a Linux system, you can create a soft link from the "life" folder to the nib.py file. Then, you can simply import it like:
import nib
I have a solution specifically for git-repositories.
First I used sys.path.append('..') and similar solutions. This causes especially problems if you are importing files which are themselves importing files with sys.path.append('..').
I then decided to always append the root directory of the git repository. In one line it would look like this:
sys.path.append(git.Repo('.', search_parent_directories=True).working_tree_dir)
Or in more details like this:
import os
import sys
import git
def get_main_git_root(path):
main_repo_root_dir = git.Repo(path, search_parent_directories=True).working_tree_dir
return main_repo_root_dir
main_repo_root_dir = get_main_git_root('.')
sys.path.append(main_repo_root_dir)
For the original question: Based on what the root directory of the repository is, the import would be
import ptdraft.nib
or
import nib
Our folder structure:
/myproject
project_using_ptdraft/
main.py
ptdraft/
__init__.py
nib.py
simulations/
__init__.py
life/
__init__.py
life.py
The way I understand this is to have a package-centric view.
The package root is ptdraft, since it's the top most level that contains __init__.py
All the files within the package can use absolute paths (that are relative to package root) for imports, for example
in life.py, we have simply:
import ptdraft.nib
However, to run life.py for package dev/testing purposes, instead of python life.py, we need to use:
cd /myproject
python -m ptdraft.simulations.life.life
Note that we didn't need to fiddle with any path at all at this point.
Further confusion is when we complete the ptdraft package, and we want to use it in a driver script, which is necessarily outside of the ptdraft package folder, aka project_using_ptdraft/main.py, we would need to fiddle with paths:
import sys
sys.path.append("/myproject") # folder that contains ptdraft
import ptdraft
import ptdraft.simulations
and use python main.py to run the script without problem.
Helpful links:
https://tenthousandmeters.com/blog/python-behind-the-scenes-11-how-the-python-import-system-works/ (see how __init__.py can be used)
https://chrisyeh96.github.io/2017/08/08/definitive-guide-python-imports.html#running-package-initialization-code
https://stackoverflow.com/a/50392363/2202107
https://stackoverflow.com/a/27876800/2202107
Work with libraries.
Make a library called nib, install it using setup.py, let it reside in site-packages and your problems are solved.
You don't have to stuff everything you make in a single package. Break it up to pieces.
I had a problem where I had to import a Flask application, that had an import that also needed to import files in separate folders. This is partially using Remi's answer, but suppose we had a repository that looks like this:
.
└── service
└── misc
└── categories.csv
└── test
└── app_test.py
app.py
pipeline.py
Then before importing the app object from the app.py file, we change the directory one level up, so when we import the app (which imports the pipeline.py), we can also read in miscellaneous files like a csv file.
import os,sys,inspect
currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
parentdir = os.path.dirname(currentdir)
sys.path.insert(0,parentdir)
os.chdir('../')
from app import app
After having imported the Flask app, you can use os.chdir('./test') so that your working directory is not changed.
It's seems to me that you don't really need to import the parent module. Let's imagine that in nib.py you have func1() and data1, you need to use in life.py
nib.py
import simulations.life.life as life
def func1():
pass
data1 = {}
life.share(func1, data1)
life.py
func1 = data1 = None
def share(*args):
global func1, data1
func1, data1 = args
And now you have the access to func1 and data in life.py. Of course you have to be careful to populate them in life.py before you try to use them,
I made this library to do this.
https://github.com/fx-kirin/add_parent_path
# Just add parent path
add_parent_path(1)
# Append to syspath and delete when the exist of with statement.
with add_parent_path(1):
# Import modules in the parent path
pass
This is the simplest solution that works for me:
from ptdraft import nib
After removing some sys path hacks, I thought it might be valuable to add
My preferred solution.
Note: this is a frame challenge - it's not necessary to do in-code.
Assuming a tree,
project
└── pkg
└── test.py
Where test.py contains
import sys, json; print(json.dumps(sys.path, indent=2))
Executing using the path only includes the package directory
python pkg/test.py
[
"/project/pkg",
...
]
But using the module argument includes the project directory
python -m pkg.test
[
"/project",
...
]
Now, all imports can be absolute, from the project directory. No further skullduggery required.
Although it is against all rules, I still want to mention this possibility:
You can first copy the file from the parent directory to the child directory. Next import it and subsequently remove the copied file:
for example in life.py:
import os
import shutil
shutil.copy('../nib.py', '.')
import nib
os.remove('nib.py')
# now you can use it just fine:
nib.foo()
Of course there might arise several problems when nibs tries to import/read other files with relative imports/paths.
This works for me to import things from a higher folder.
import os
os.chdir('..')
let's say my project's structure looks like:
project
├── important.py
└── files
└── file1.py
and the file important.py contains the class Important.
How can I import the class(Important) from file1, while file1 is the python file which being executed?.
The only solution I found was to add this code, but I wonder if there is a cleaner way:
import sys; sys.path.append("..")
from important import Important
Things I have tried without success:
from project.important import Important
# ModuleNotFoundError: No module named 'project'
# But it does work inside PyCharm (Why is that?)
from ..important import Important
# ValueError: attempted relative import beyond top-level package
And this errors were keep showing even if I added a __init__.py file inside the project's directory.
import to say is that I am looking for a solution that will fit any machine, as I want to share this project on github to the public.
You will need to make a reference to parent folder within sys.path. This can be done explicitly inside the code, like you have done, which is not really unpythonic. It can also be done from outside the code, e.g. by modifying the system variable PYTHONPATH or by installing your module within python.
I strongly discourage to use the absolute path, as suggested by other responses, because then, the code will only work on your machine. It is tolerable for student projects, but it is bad practice in real life development since multiple people will work on it, it will execute on test/production/sandbox servers, etc...
So the approach is correct. However, I still suggest modifying slightly your syntax because there are cases in which it will not work as expected:
all_projects/
└── current_project/
├── important.py
└── files
└── file1.py
$ cd /path/to/all_projects/current_project/files/
$ python file1.py
#### > Ok, thanks to the line sys.path.append("..")
$ cd /path/to/all_projects/current_project/
$ python files/file1.py
#### > Ok, because python implicitly add the execution path to sys.path
$ cd /path/to/all_projects/
$ python current_project/files/file1.py
#### > ModuleNotFoundError: No module named 'important'
Instead, use the following:
import sys, os
sys.path.append(os.path.dirname(sys.path[0]))
Or if file1.py could even be imported from another file, the following is even safer:
import sys, os
sys.path.append(os.path.dirname(os.path.dirname(os.path.realpath(__file__))))
Finally, for good development practice, I also suggest to put these lines into a different file, called e.g. _set_path. The reason is that you can reuse this in future files, and if you want to modify the code architecture, you only need to change one file:
├── file1.py
├── file2.py
├── ...
└── _set_path.py
Then, from file1.py, you can use:
import _set_path
from important import Important
(Answer inspired from Python: Best way to add to sys.path relative to the current running script)
People have pointed out the sys.path.append("..") route. While this works there is also an alternative method with os.chdir('..')
You can view a list of your path with the following command python3 -m site. When importing packages Python checks for modules in all of those paths.
The first element in your sys.path is the current working directory.
There may be a scenario where you don't want to have your current working directory be part of the path and want to have one folder structure up added to path.
An "issue" is there are multiple ways of importing the same thing. For instance you have:
project/
├── important.py
└── files
├── file1.py
└── file2.py
By doing sys.path.append("..") and running the program via python3 file1.py you can import file2 via import file2 or from files import file2. This doesn't look nice and you might start write inconsistent code while not understanding how import properly works.
You can stick with sys.path.append("..")if it works. You won't do much wrong with it. It is a common approach many people do. There may just be a special scenario where you might run into problems which is why many people prefer the os.chdir() approach.
For example in both folder, top folder and subfolder, you have python modules which share the same name. You want to import Python modules from one folder up but not the python modules in the current folder.
Example of os.chdir() in action:
Tin#ubuntu:~/Desktop/tmp/test$ python3
Python 3.6.8 (default, Oct 7 2019, 12:59:55)
[GCC 8.3.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> os.getcwd()
'/home/Tin/Desktop/tmp/test'
>>> import helloworld
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ModuleNotFoundError: No module named 'helloworld'
>>> os.chdir('..')
>>> import helloworld
hello world!
>>> os.getcwd()
'/home/Tin/Desktop/tmp'
Now you can import from one directory up and the import is no longer ambiguous.
Note that while I wrote os.chdir('..') you can do what #Tawy did.
import os
os.chdir(os.path.dirname(os.path.dirname(os.path.realpath(__file__))))
# OR:
os.chdir(os.path.join(os.path.dirname(os.path.realpath(__file__)), '..'))
It looks confusing but with this you can internally imagine what your current working directory is and base all your import statements on that. It also gives you a consistent current working directory when you are executing your script from all sorts of subdirectories but expect a certain working directory.
You may also do the mistake of running os.chdir('..') twice which will make you go two folders structure up.
In short:
Least complicated solution is sys.path.append(".."). A cleaner solution would be os.chdir(os.path.join(os.path.dirname(os.path.realpath(__file__)), '..')) with the .. being the relative location to your wanted working directory.
Instead of:
import sys
sys.path.append("..")
You could do:
import sys
sys.path.append("/path/to/project")
Which is the same as the first one, but less confusing since you're adding the absolute path and make it obvious for the user.
Note that you don't need any __init__.py file for the above to work. Also, note that this might not be the only way but it's the one that I find to be the most clean.
Also, if you really hate this solution, you might think of restructuring the project so that you can avoid this scenario.
Although I think this should be pretty simple I still can't make it run.
I have following folder structure:
├── apartment
│ ├── src
│ ├── train_model
│ ├── __init__.py
│ ├── train_model.py
│ ├── utils.py
│ ├── interference.py
│ └── __init__.py
In utils.py I tried:
from src.interference import create_sample
Error: ModuleNotFoundError: No module named 'src'
from .interference import create_sample
Error: ImportError: attempted relative import with no known parent package
from interference import create_features_sample
ModuleNotFoundError: No module named 'interference'
What is the way to make it work? I am not a huge fan of non-pythonic ways as it looks dirty.
The structure starting with a src/ is aimed explicitly at not enabling import by from src.intereference import ..., and you should not put an __init__.py file in src/ folder.
Instead, following nice explanation and examples here: https://blog.ionelmc.ro/2014/05/25/python-packaging/, here is what I recommend:
install the package:
add a setup.py file at root of your folder (this is clearly not as hard as it seems)
maybe create a virtual environment
using pip install -e . (with trailing dot!) command
then simply import your package by from interference import ...
To answer to your primary request, you could update src/__init__.py with from intereference import create_sample, to expose this function at a higher level, then the chained import would work. However, I do not recommend this, as it renders everything very rigid.
You need to add the directory that contains interference to PYTHONPATH.
You can use OS depending path in "module search path" which is listed in sys.path. So you can easily add parent directory like following:
import sys
sys.path.insert(0, '..')
from interference import create_features_sample
Note that the previous code uses a relative path, so you must launch the file inside the same location or it will likely not work.
To launch from anywhere, you can use Path from the pathlib module.
from pathlib import Path
import sys
path = str(Path(Path(__file__).parent.absolute()).parent.absolute())
sys.path.insert(0, path)
from interference import create_features_sample
If you want add a custom path to your PYTHONPATH permanently, go to the 'site-packages' folder of your current Python environment and add the file 'custompaths.pth' in which each line should consist of a directory which will then be checked if you try to import the module.
Assuming 'src' being the module you want to import you should add the following line to the .pth file:
your_preceding_path/apartment
Error because of python can't able to find the specific file or package. Python usually find only for the child packages or files, otherwise you need to specify the absolute path.
import sys, os
sys.path.append(os.path.abspath(os.path.join('../..', 'src')))
Please refer the "How to access a module from outside your file folder in Python?"
Have you tried from ..interference import create_sample ?
Or for the whole module from .. import interference.
I've checked here, that I'm using another command, something that is missing from the question at the time of writing this. The command that I'm using is python -m src.train_model.utils from apartment folder.
Thanks to RMPR for helping me.
This problem is similar to this one.
This question already has answers here:
How to do relative imports in Python?
(18 answers)
Closed 4 years ago.
I have a simple project structure like this:
➜ (venv:evernote) evernote_bear_project git:(master) ✗ tree | grep -v pyc
.
├── README.md
...
(snip)
...
├── manage.py
├── sample
│ ├── EDAMTest.py <==== here is an import that won't work
│ └── enlogo.png
└── util
├── __init__.py
├── files.py <====== This is being imported
└── test_files.py
Now I have a relative import in sample/EDAMTest.py:
from ..util.files import *
When I try to run python sample/EDAMTest.py from project root folder in command line, I get an error saying:
ValueError: attempted relative import beyond top-level package
I know this has been asked many times, but I still don't get it.
Since I'm running the script from the project root, in my understanding Python should be able to "know" that when I try to import from ..util.files import *, that it should go up one directory, no?
EDIT
Thanks for all the answers.
So what I understand from the link above is this:
I was running the module sample/EDAMTest.py directly via python sample/EDAMTest.py, and that meant
that the __name__ of the module was __main__
and now the path (sample/) was my "root" so to speak.
So now Python searches only this path and any path that's below it for modules / packages. Hence the error message attempted relative import _beyond top-level package_, so it cannot go one more level up, because it is at the root already.
Also Python cannot look one level "up", since this syntax from ..util.files import * does not go up a level in directory, but in a list of modules / packages it keeps on the "import search path" (sys.path)?
Is that correct?
sys.path.append() is a tweak, if your directory structure is fixed and there is nothing you can do about it.
Otherwise, you can try rearranging the folders. The easiest is moving util under sample, another option is making both of the folders psrt of a larger package.
Also import * is not ensorsed.
The relative import syntax is for importing other modules from the same package, not from the file system.
Depending on what you want, you could...
Fix the package so that the relative import works
Put __init__.py files in the project root and sample directory and run the script from one level up. This doesn't seem like what you want.
Tell python where to find the package
Set the PYTHONPATH environment variable so that python can find the package.
PYTHONPATH='.':$PYTHONPATH python samples/EDAMTest.py
Install util so that python can find it
Add a setup script and use it to install the util package and avoid setting PYTHONPATH.
"The relative import syntax is for importing other modules from the same package, not from the file system.", This is right as stated by George G.
Put __init__.py in your subfolders, which will make them package.
__init__.py can be an empty file but it is often used to perform
setup needed for the package(import things, load things into path, etc).
However you can import File into your __init__.py to make it
available at the package level:
# in your __init__.py
from util import files
# now import File from util package
from util import files
of if you want to import some specific method or class, you can do
from util.files import some_function
Another thing to do is at the package level make util/modules
available with the __all__ variable. When the interpeter sees
an __all__ variable defined in an __init__.py it imports the
modules listed in the __all__ variable when you do:
from package import *
__all__ is a list containing the names of modules that you want to be
imported with import * so looking at our above example again if we
wanted to import the submodules in util the all variable
in util/init.py would be:
__all__ = ['files', 'test_files']
With the __all__ variable populated like that, when you perform
from util import *
it would import files and test_files.