How do I link the util.py file to the script.py file?
App
Module
__init__.py
util.py
Main
__init__.py
Dir1
script.py
__init__.py
Dir2
script.py
__init__.py
Take this directory layout of my app. It's a bit more complex for requirement reasons to keep the scripts easily separated while sharing the same module across each script. However, it's difficult to find the correct way to import that module to each script due to directory hierarchy.
Environment
Python 2.7, Windows 10, 64Bit
Just add the root directory to the PYTHONPATH. Then you can use absolute imports as well as the relative ones.
# script.py
form Module.util import _
You can put:
import sys
sys.path.insert(1, '.')
in your Module.__init__.py and then just import it in script.py with:
from Modules import util
Additionally you can define a __all__ = [] variable in your __init__ and then you can star import.
Related
My apps are organized like this:
apps/
code/
libglobal/
funglobal.py
tests/
project/
liblocal/
funlocal.py
main.py
In main.py I have:
import liblocal.funlocal
In funlocal.py I try to import funglobal.py with:
from ....code.libglobal import funglobal
When I run
python3 -B tests/project/main.py
I get an error:
from ....code.libglobal import funglobal
ValueError: attempted relative import beyond top-level package
I have read a lot of information about relative imports with python3 and still don't find how to solve this error without changing the apps organization radically. Any solution?
As the script being executed has its __name__ set as __main__ and defines itself to be on the top level of the package, it refuses to recognize scripts in sibling directories.
You can fix this with a sys.path hack:
import sys, os
sys.path.insert(0, os.path.abspath('../..'))
or an interseting alternative with setuptools is presented in this answer.
Have you a __init__.py script in each folder ?
If no, you should create an empty script named __init__.py in each folder.
I have the following directory structure:
app/
bin/
script1.py
script2.py
lib/
module1/
__init__.py
module1a.py
module1b.py
__init__.py
module2.py
Dockerfile
My problem is that I want to execute script1.py and script2.py, but inside those scripts, I want to import the modules in lib/.
I run my scripts from the root app/ directory (i.e. adjacent to Dockerfile) by simply executing python bin/script1.py. When I import modules into my scripts using from lib.module1 import module1a, I get ImportError: No module named lib.module1. When I try to import using relative imports, such as from ..lib.module1 import module1a, I get ValueError: Attempted relative import in non-package.
When I simply fire up the interpreter and run import lib.module1 or something, I have no issues.
How can I get this to work?
In general, you need __init__.py under app and bin, then you can do a relative import, but that expects a package
If you would structure your python code as python package (egg/wheel) then you could also define an entry point, that would become your /bin/ file post install.
here is an example of a package - https://python-packaging.readthedocs.io/en/latest/minimal.html
and this blog explains entry points quite well - https://chriswarrick.com/blog/2014/09/15/python-apps-the-right-way-entry_points-and-scripts/
if so, that way you could just do python setup.py install on your package and then have those entry points available within your PATH, as part of that you would start to structure your code in a way that would not create import issues.
You can add to the Python path at runtime in script1.py:
import sys
sys.path.insert(0, '/path/to/your/app/')
import lib.module1.module1a
you have to add current dir to python path.
use export in terminal
or sys.path.insert in your python script both are ok
Simple question, but could not find the answer.
I've following structure:
./lib1:
main.py
./lib2:
__init__.py utils.py
From the root diretory, I'm running:
python lib1/main.py
and in main.py I want to import lib2/utils.py.
adding import lib2/utils.py fails.
One solution I found is to add:
~/tmp/root$ cat lib1/main.py
import sys,os
sys.path.append(os.getcwd())
import lib2.utils
lib2.utils.foo()
which is good, but I wander if there is other solution. Thanks.
Are lib1 and lib2 separate modules? If yes, the comment by #BrenBarn applies: You need to add the top directory (containing lib1 and lib2 to the Python path (e.g using PYTHONPATH environment variable or appending to sys.path).
If both lib1 and lib2 are part of one module (i.e. there is a __init__.py file in the top directory) you can use relative imports (https://docs.python.org/2.5/whatsnew/pep-328.html).
Your problem is caused by using the wrong directory structure. The main.py script should be in the same top-level directory as the package that it needs to import. So the structure should look like this:
project /
lib2 /
__init__.py
utils.py
other.py
main.py
The directory of the main script will always be added to the start of sys.path, so this will guarantee that any packages in that directory can be always be directly imported, no matter where the script is executed from.
To import the utils module into main.py (or other.py), you should do:
from lib2 import utils
I use
sys.path.append('D:/my_library_folder/')
import mymodule
in order to import some module.
How to add permanently this folder D:/my_library_folder/ to the Python library path, so that I will be able to use only
import mymodule
in the future?
(Even after a reboot, etc.)
just put the folder in site-packages directory. ie:
C:\PythonXY\Lib\site-packages
Note: you need to add an empty file __init__.py to the folder
Files named __init__.py are used to mark directories on disk as a Python package directories.
If you have the files:
C:\PythonXY\Lib\site-packages\<my_library_folder>\__init__.py
C:\PythonXY\Lib\site-packages\<my_library_folder>\module.py
you can import the code in module.py as:
from <my_library_folder> import module
If you remove the __init__.py file, Python will no longer look for submodules inside that directory, so attempts to import the module will fail.
If you have lots of folders, then create the empty __init__.py file in each folder. for eg:
C:\PythonXY\Lib\site-packages\<my_library_folder>\
__init__.py
module.py
subpackage\
__init__.py
submodule1.py
submodule2.py
Set PYTHONPATH environment variable to D:/my_library_folder/
If D:/my_library_folder is a project you're working on and has a setup script, you could also do python setup.py develop. Not entirely related to the question, but I also recommend using virtualenv.
How do I import a module(python file) that resides in the parent directory?
Both directories have a __init__.py file in them but I still cannot import a file from the parent directory?
In this folder layout, Script B is attempting to import Script A:
Folder A:
__init__.py
Script A:
Folder B:
__init__.py
Script B(attempting to import Script A)
The following code in Script B doesn't work:
import ../scriptA.py # I get a compile error saying the "." is invalid
You don't import scripts in Python you import modules. Some python modules are also scripts that you can run directly (they do some useful work at a module-level).
In general it is preferable to use absolute imports rather than relative imports.
toplevel_package/
├── __init__.py
├── moduleA.py
└── subpackage
├── __init__.py
└── moduleB.py
In moduleB:
from toplevel_package import moduleA
If you'd like to run moduleB.py as a script then make sure that parent directory for toplevel_package is in your sys.path.
From the docs:
from .. import scriptA
You can do this in packages, but not in scripts you run directly. From the link above:
Note that both explicit and implicit relative imports are based on the
name of the current module. Since the name of the main module is
always "__main__", modules intended for use as the main module of a
Python application should always use absolute imports.
If you create a script that imports A.B.B, you won't receive the ValueError.
If you want to run the script directly, you can:
Add the FolderA's path to the environment variable (PYTHONPATH).
Add the path to sys.path in the your script.
Then:
import module_you_wanted