1

Why python load my module from directory, even if the directory shouldn't be treated as a package (doesn't contains __init__.py)? 

app
 main.py
 fake_module
 module.py

this works, even if fake_module doesn't contains __init__.py

import fake_module.module

If i would create regular fake_module with __init__.py and module.py file somewhere in sys.path, will be module.py loaded from this regular fake_module package?

I am using pycharm and python3.3

asked Jun 15, 2013 at 16:32
7
  • 1
    Are you 100% certain that you are looking at the right fake_module? What does print fake_module.module.__file__ give you? Commented Jun 15, 2013 at 16:33
  • @MartijnPieters: it gives that app/fake_module/module.py Commented Jun 15, 2013 at 16:38
  • @JBernardo: no i didn't Commented Jun 15, 2013 at 16:38
  • 1
    @Krab: and print fake_module.__file__ prints what? Commented Jun 15, 2013 at 16:40
  • 1
    @Krab: What does print type(fake_module) give? Commented Jun 15, 2013 at 16:42

1 Answer 1

7

It looks like the __init__.py is optional as of Python 3.3. See also PEP420.

answered Jun 15, 2013 at 16:45
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1 Comment

Bingo, I forgot about the new implicit namespaces support in 3.3.

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