init.py not working as expected in Python 3 [duplicate]

Question

I have the below module structure:

/home/Dir1/Dir2/Mod.py

Inside /home i have the script Test.py which I am executing and has the below code :

import Dir1.Dir2 
dummy()

Again the Directory: Dir1 has an __init__.py file which has the below code:

print('dir1 init')
x = 1

The directory Dir2 has the below __init__.py file having the following code:

from Mod import dummy
print('dir2 init')
y = 2

There is a final file Mod.py which is defined as follows:

print ("in Mod.py")
z = 3
def dummy():
    print ("Hello World")

Since Test.py is the entry point in my application this is the below error I get:

dir1 init
Traceback (most recent call last):
  File "Test.py", line 1, in <module>
    import Dir1.Dir2
  File "C:\Users\bhatsubh\Desktop\Everything\Codes\Python\Dir1\Dir2\__init__.py"
, line 1, in <module>
    from Mod import dummy
ImportError: No module named 'Mod'

Why is the __init__.py file inside Dir2 not able to find the file Mod.py?

Basically what I am trying to do here is make the function dummy inside Mod.py public without the user knowing that there is a module by that name.

Am I doing something silly here?

Mike Müller · Accepted Answer · 2017-01-24 23:06:51Z

0

Change your Test.py to use dummy with the fully qualified name:

import Dir1.Dir2

Dir1.Dir2.dummy()

and your Dir1/Dir2/.__init__.py to use a relative import with a leading . in front of Mod:

from .Mod import dummy
print('dir2 init')
y = 2

to make it work:

python Test.py 
dir1 init
in Mod.py
dir2 init
Hello World

Note: Per convention all package (i.e. directory) and module (i.e. py file) names should be lower case.

edited Jan 24, 2017 at 23:06

answered Jan 24, 2017 at 23:01

Mike Müller

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