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[alpine zodiac]

Let's say we have the following file system:

• Root
• A
  • a.py
  • b.py
  • c.py
• main.py

where b.py and c.py use a.py under their own namespaces, and main.py uses everything in A under the A namespace.

How would someone use import in each file for the above to work as mapped?

[charred flame]

@alpine zodiac

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[dire minnow]

# Root/A/__init__.py
from . import a, b, c
# Root/A/b.py  and  Root/A/c.py
import a
# Root/main.py
import A
```jez I'm messing all the stuff up today

[alpine zodiac]

I tried this:

# Root/A/b.py and Root/A/c.py
from a import *
# Root/main.py
import A

It presumably imports everything, but main.py throws an error saying nothing from A exists.

How do you work the __init__.py file?

[dire minnow]

that's cause you can't import a folder
if you did import A.a or from A import b, then they'd work
notice the __init__.py I have

[dire minnow]

basically, think of the __init__.py in A as the script that'll be ran when you do import A

from . import a, b, c means "from the same directory as __init__.py (which is A in this case, which also contains a.py b.py c.py), import a b and c"

[dire minnow]

so once you have that,

# main.py
import A
#      ^ runs __init__.py in A, which imports a, b, c into A
A.b # ok

[alpine zodiac]

Traceback (most recent call last):
  File "X:\Root\main.py", line 1, in <module>
    import A
  File "X:\Root\A\__init__.py", line 1, in <module>
    from . import a, b, c
  File "X:\Root\A\b.py", line 1, in <module>
    from a import *
ModuleNotFoundError: No module named 'a'

[dire minnow]

ah right
replace from a import * with from .a import * (again the . signifies to search in the same dir as that file)

[alpine zodiac]

from a import * is implicit from startup?

So if I started IDLE inside A then it would work?

[dire minnow]

fyi, that error happens because import a and from a import * looks for a in the "current working directory" (cwd), in this case Root is probably your cwd
if you have a terminal that'd be this part

C:\Users\User\Desktop> py main.py
^^^^^^^^^^^^^^^^^^^^^

so you can either leverage "relative imports," i.e. the . thing, or you can do

# Root/A/b.py  Root/A/c.py
import a  # error, there's no file named `a.py` in `Root/`

import A.a  # ok, the file `Root/A/a.py` exists
from A.a import *  # ok, same as above

[dire minnow]

well if you did that, then you won't be able to find main.py

[alpine zodiac]

I know.

Seems if I put test functions in each file under a.py, b.py, and c.py naming them after their file and try running it from main.py, I would need to call them as A.a.a(), A.b.b(), and A.c.c(). How do I get it imported so that I call those test functions, for instance, as A.a()?

[dire minnow]

you can control that in __init__.py, e.g.

# Root/A/__init__.py
from .a import *
from .b import *
from .c import *

[alpine zodiac]

Oh.

This there a way to iterate through each file in the folder and import them in that way?

If .a from from .a import * was a string then I would know how, but I don't know what .a is.

[dire minnow]

personally I've never done it, but look into importlib.import_module to dynamically import some module by their path, and use something like pathlib.Path.glob to find the file paths in a specific folder

[alpine zodiac]

Thanks.

[charred flame]

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