'Import a function from a module in another folder in parent directory
I have been trying unsuccessfully to solve this for hours now. This is my folder structure.
/parent_folder
main.py
module1/
script1.py
module2/
script2.py
script2.py has only this inside:
def subtract_numbers(x, y):
return x - y
I want script1.py to be able to call this function. I have:
from ..module2.script2 import subtract_numbers
result_subtraction = subtract_numbers(5, 5)
print(result_subtraction)
I get ImportError: attempted relative import with no known parent package
I have tried many different permutations in the import line in scrip1.py but i get the same error. I also have to note that i have __init__.py files in the two folders.
How exactly can i call the function in script2.py?
Solution 1:[1]
This is what worked for me.
I exported the directory of the parent directory (/parent_folder) to the PYTHONPATH.
export PYTHONPATH=$PYTHONPATH:/home/username/Desktop/parent_folder
Then, inside file module1/script1.py, i had this line changed:
from module2.script2 import subtract_numbers
Now i can call the script script1.py that calls the function declared in module2/script2.py with python script1.py and it will work.
I also have to note that i have __init__.py files everywhere (in the parent directory and both subfolders), but i am now sure if this is vital.
Solution 2:[2]
Relative imports cannot go back to a higher level than the one from which the python call was originated. So, your problem here is that you are invoking script1.py directly from the module1 directory. I guess something like this:
user:/path/to/parent/module1$ python script1.py
So you will need to make your call to script1.py from a level where you can actually see script2.py.
First, change your relative import to absolute import in script1.py:
from module2.script2 import subtract_numbers
Then, move back to the parent directory of module1 and run the module as a script from there (note the location in the prompt):
user:/path/to/parent$ python -m module1.script1
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | user1584421 |
| Solution 2 |
