导入第三方 Python 项目而不将其添加到 sys.path

Question

I am trying to import a third party python 'project' in my own code. I am not referring it as a package because it seems to be prepared for running on its own but not for others to import it and run it somewhere else. For example, that third party project may look like this:

ProjectA/
|- utils/
  |- tools.py
|- lib/
  |- functions.py
|- main.py

and the dependency of files/functions goes like:

In main.py:
- from lib.functions import xxx

In lib/functions.py:
- from utils.tools import yyy

such that

1. running main.py requires the function xxx from lib/functions.py , and
2. running the function xxx in lib/functions.py requires the function yyy in utils.tools.py .

If I run main.py at the top level of Project A, both lib and utils are visible at this level so main.py runs successfully. Or, if I want to run the functions inside main.py somewhere else, I can add path_to_ProjectA to sys.path and everything runs well too.

However, there are occasions in which I prefer not to add this project to sys.path . For example, I might want to use another project ProjectB which also has a utils submodule, or I already have a utils submodule in my own project. In these cases, if I add both ProjectA and ProjectB to sys.path , then only one of utils submodules will get recognized (ie, can be imported).

Therefore, I am wondering if there would be methods to make another's project available for importing. What I can think of is to make the project a 'legit' package, eg, adding __init__.py at each level, changing the way to import packages/functions, etc. (Actually, I do not have much experience in doing this, and I do not have a clear idea about how too.) However, I do not believe this is always practical, especially when that project is sizeable and it may require fair understanding to the design of the project structure.

To summarize, my question is, what should I do so that I can import a third party python project outside that project without adding it to sys.path , even though that project is probably designed to be run only at its top directory level?

---

Edit #1 (20220819)

Following @Lenntror 's comments, I am adding a minimal example here. The goal is to call ProjectA.lib.functions.xxx from the outside of ProjectA without adding ProjectA to sys.path .

(a) Calling ProjectA.lib.functions.xxx inside ProjectA

Structure of the third-party ProjectA :

ProjectA/
|-- lib
|   |-- __init__.py # Empty
|   `-- functions.py
|-- utils
    |-- __init__.py # Empty
    `-- tools.py
`-- main.py

Content of ProjectA/lib/functions.py :

from utils.tools import yyy

def xxx() :
    print("This is xxx().")
    print("Now xxx() calls yyy().")
    yyy()

Content of ProjectA/utils/tools.py :

def yyy() :
    print("This is yyy().")

Content of ProjectA/main.py :

from lib.functions import xxx

def main() :
    print("This is main().")
    print("Now calling xxx().")
    xxx()

if __name__=='__main__' :
    main()

If I run python main.py under the directory ProjectA . It works:

This is main().
Now calling xxx(). 
This is xxx().      
Now xxx() calls yyy().
This is yyy().

(b) Calling ProjectA.lib.functions.xxx outside ProjectA by adding ProjectA to sys.path

Let's say, I have another folder MyProject which is next to (and outside) ProjectA :

./
|-- MyProject                                                               
|   `-- my_script.py    
|                                                    
`-- ProjectA
    |-- ...

Content of MyProject/my_script.py :

import sys
sys.path.append("../ProjectA")
from lib.functions import xxx

def main() :
    print("This is outside of ProjectA.")
    print("Now calling xxx().")
    xxx()

if __name__=='__main__' :
    main()

This also runs successfully:

This is outside of ProjectA.
Now calling xxx().
This is xxx().
Now xxx() calls yyy().
This is yyy().         

(c) Calling ProjectA.lib.functions.xxx by including ProjectA in a sub-directory/module

Now, I have a script in another folder also outside of ProjectA . This time, I include ProjectA inside this folder like this:

MyProject2/
|-- my_script.py
`-- thirdparty
    `-- ProjectA
        |-- ...

Content of MyProject2/my_script.py :

from thirdparty.ProjectA.lib.functions import xxx

def main() :
    print("This is outside of ProjectA.")
    print("Now calling xxx().")
    xxx()

if __name__=='__main__' :
    main()

This gives an error:

Traceback (most recent call last):
  File "my_script.py", line 1, in <module>
    from thirdparty.ProjectA.lib.functions import xxx
  File "/workspace/MyProject2/thirdparty/ProjectA/lib/functions.py", line 1, in <module>
    from utils.tools import yyy
ModuleNotFoundError: No module named 'utils'

Answer 1

So you could have the project file as this:

    ProjectA/
    |- utils/
      |- tools.py
    |- lib/
      |- functions.py
    |- main.py
    
    ProjectB/
    |- utils/
      |- tools.py
    |- lib/
      |- functions.py
    |- main.py

    your_project_file.py

your_project_file.py , ProjectA and ProjectB are sitting inside the same directory.

inside your_project_file.py you could use import statements like this:

from ProjectA.lib.functions import xxx

But I think you figured this out already that this will cause issue from utils.tool . What you need to do, is use relative path. Edit lib/functions.py file and place a dot in front of utils.tool which should become from.utils.tool import yyy

Now it should resolve your problem and import in your_project_file.py should work fine. If not, then you need to define __init__.py file for all folders.