Proper ways to set the path of my app in Python - python

I have a question in how to properly create a path in Python (Python 3.x).
I developed a small scraping app in Python with the following directory structure.
root
├── Dockerfile
├── README.md
├── tox.ini
├── src
│   └── myapp
│   ├── __init__.py
│   ├── do_something.py
│   └── do_something_else.py
└── tests
├── __init__.py
├── test_do_something.py
└── test_do_something_else.py
When I want to run my code, I can go to the src directory and do with
python do_something.py
But, because do_something.py has an import statement from do_something_else.py, it fails like:
Traceback (most recent call last):
File "src/myapp/do_something.py", line 1, in <module>
from src.myapp.do_something_else import do_it
ModuleNotFoundError: No module named 'src'
So, I eventually decided to use the following command to specify the python path:
PYTHONPATH=../../ python do_something.py
to make sure that the path is seen.
But, what are the better ways to feed the path so that my app can run?
I want to know this because when I run pytest via tox, the directory that I would run the command tox would be at the root so that tox.ini is seen by tox package. If I do that, then I most likely run into a similar problem due to the Python path not properly set.
Questions I want to ask specifically are:
where should I run my main code when creating my own project like this? root as like python src/myapp/do_something.py? Or, go to the src/myapp directory and run like python do_something.py?
once, the directory where I should execute my program is determined, what is the correct way to import modules from other py file? Is it ok to use from src.myapp.do_something_else import do_it (this means I must add path from src directory)? Or, different way to import?
What are ways I can have my Python recognize the path? I am aware there are several ways to make the pass accessible as below:
a. write export PYTHONPATH=<path_of_my_choice>:$PYTHONPATH to make the
path accessible temporarily, or write that line in my .bashrc to make it permanent (but it's hard to reproduce when I want to automate creating Python environment via ansible or other automation tools)
b. write import sys; sys.path.append(<root>) to have the root as an accessible path
c. use pytest-pythonpath package (but this is not really a generic answer)
Thank you so much for your inputs!
my environment
OS: MacOS and Amazon Linux 2
Python Version: 3.7
Dependency in Python: pytest, tox

I would suggest to use setup.py to make this a python package. Then you can install it in development mode python setup.py develop. This way it will be available in your python environment w/o needing to specify the PYTHONPATH.
For testing, you can simply install the package python setup.py install.
Hope that helps.

Two simple steps should make it happen. Python experts can comment if this is a good way to do it (especially going by the concluding caution raised towards the end of this post).
I would have done it like below.
First I would have put a "__init__.py" in root so that hierarchy looks like below. This way python will treat the folder as a package.
root
├── Dockerfile
├── README.md
├── tox.ini
├── __init__.py
├── src
│ └── myapp
│ ├── __init__.py
│ ├── do_something.py
│ └── do_something_else.py
└── tests
├── __init__.py
├── test_do_something.py
└── test_do_something_else.py
Then in "do_something.py", I would have added these lines at the top. In the second line please put the full path to the "root" directory.
import sys
sys.path += ['/home/SomeUserName/SomeFolderPath/root']
from src.myapp.do_something_else import do_it
Please note that the second line will essentially modify the sys.path by adding the root folder path (I guess until the interpreter quits). If this is not what you can afford then I am sorry.

Related

Python How to run scripts from a subdirectory?

I have such structure of project:
lib/
...
scripts/
...
I have many Python scripts in the scripts/ directory. All of them contains relative imports: from lib import ...
So, how can I easy run scripts from the root of project /, without changing scripts (without write chdir in each script)?
Maybe can I use some __init__ file to change work dir? Or maybe can I use special command to run python scripts with root folder? Any other ways?
Your question is not clear to me and its to much for a comment... Your structure is like that?
root/
├── lib/
│ ├── __init__.py
│ ├── lib_foo.py
│ ├── lib_bar.py
├── scripts/
│ ├── script_util.py
│ └── script_yeah.py
└── main.py
And your program starts always at main.py? Or do you have python files with main also in the scripts folder ?
Never use chdir except if you have a very good reason. Add init files as detailed in the other answer and run your script from the parent directory (say root) as
$ python -m scripts.yourscript # note no .py
If you must run the scripts from their own directory and not use the suggestion from Mr_and_Mrs_D, then the simplest way to handle your use case is to manipulate the search path for the default module finder:
import sys
sys.path.append('..')
from lib import foo
sys.path.pop()
print(foo.bar())
It was a decision in Python to prevent explicitly loading a module from a relative folder above the executing process root. You can do it by creating a new 'finder' for importlib that returns a ModuleSpec, but the amount of code needed is somewhat excessive.

Trying to run python code from repo without instructions

I am trying to run a code from this repo without success. There are no instructions on how to run it. I suspect I should run FactcheckingRANLP/Factchecking_clean/classification/lstm_train.py and then run .../lstm_test.py.
The problem is that this code uses import statements as a module, referencing to folders and files that are in different directories, for example, in lstm_train.py:
File "lstm_train.py", line 3, in <module>
from classification.lstm_utils import *
ModuleNotFoundError: No module named 'classification'
This is the tree structure of the classification folder:
.
├── classification
│   ├── __init__.py
│   ├── __init__.pyc
│   ├── lstm_repres.py
│   ├── lstm_test.py
│   ├── lstm_train.py
│   ├── lstm_train.pyc
│   ├── lstm_utils.py
│   ├── lstm_utils.pyc
│   ├── __pycache__
│   │   ├── __init__.cpython-36.pyc
│   │   ├── lstm_train.cpython-36.pyc
│   │   └── lstm_utils.cpython-36.pyc
│   └── svm_run.py
I would like to know how can I make python run lsmt_train/test.py files in such a manner that the import statements contained within them are compiled correctly. I prefer not to modify the code as this could possibly generate a lot of errors..
You could add the path pointing to the classification folder to your python path variable.
I suggest using the sys package:
import sys
sys.path.append('<<<PathToRepo>>>/FactcheckingRANLP/Factchecking_clean')
With the repo classification directory added to your python path, the import statements should work.
EDIT:
Correction; in the initial post I suggested adding the path to .../classification to your path variable, instead the parent folder .../Factchecking_clean is required as the file imports the module 'classification'.
Also, in Lucas Azevedo's answer, the parent directory path is added in the repository lstm_train file itself. While this definitely works, I still think it should be possible without editing the original repository.
I took a look at the repo in question and files like lstm_train.py are scripts which should be executed with the python working directory set as '<<<PathToRepo>>>/FactcheckingRANLP/Factchecking_clean'.
There are a few ways to do so:
You could open the project in a python IDE and configure your execution to use the directory .../Factchecking_clean as the working directory. In pycharm for example this could be done by importing the repo directory .../Factchecking_clean as a project. The following image shows how to set a working directory for execution in pycharm:
I think the repository was developed with this execution configuration set up.
Another possibility is to execute the python script from within another python file. This seems to be rather inconvenient to me, regardless you could do so by creating a separate python file with:
import sys
import os
sys.path.append('<<<PathToRepo>>>/FactcheckingRANLP/Factchecking_clean')
os.chdir('<<<PathToRepo>>>/FactcheckingRANLP/Factchecking_clean')
exec(open('./classification/lstm_train.py').read())
This adds the Factchecking_clean directory to the python path (using sys.path.append()) to be able to import stuff like classification.utils. The working directory is set by os.chdir() and finally exec(open('<<<filepath>>>')).read() executes the lstm_train file with the correct working directory and path variable set up.
Executing the new python file with the code above works for me (without editing the original repository).
However, as scripts like lstm_train.py actually are used to execute specific parts of the code provided in the rest of the repository modules, I think editing these files for experimental purposes is fine. In general, when working with repositories like this I recommend using an IDE (like pycharm) with a correctly set up configuration (method 1).
Ended up having to modify the repository's code, following the suggestion of Jeff.
import sys,os
parent_dir_path = os.path.abspath(__file__+"/../..")
sys.path.append(parent_dir_path)
adding the path until classification doesnt work, as the import is done mentioning the classification folder. The parent directory is the one that has to be added.
__file__ gives the current file name and os.path.abspath resolves the path navigation done with /../..

Python3 'ModuleNotFoundError' when running the code through the terminal, but everything works as intended in a python virtual environment

I am having problems with importing modules/packages with Python. I noticed this problem when I ran it in my terminal (CMD),
rather than my IDE (I use PyCharm). In PyCharm, I use a virtual enviroment setting with Python 3.7 and everything works as a charm
and as intended.
For reference this is how the imports were done in test_suite.py:
...
from tests.scenarios.test_scenario_01 import TestScenario # They work perfectly fine
from tests.scenarios.test_scenario_02 import TestScenario2 # written like this in PyCharm venv Python 3.7, but why?
...
This is a simplified version of my directory (without the unneccesary files):
QA System/
├── locators/
│ ├── locators.py
│ ├── __init__.py
├── pages/
│ ├── pages.py
│ └── __init__.py
└── tests/
├── reports
├── test_scenarios
├── test_scenario_01.py
├── test_scenario_02.py
├── __init__.py
|── test_suite.py
|── __init__.py
However when running the file test_suite.py manually through my CMD (because I want to integrate it with Jenkins
eventually), I get this error (py -3 test_suite.py):
ModuleNotFoundError: No module named 'tests'
Note: I am using the newest Python 3.7
From what I know about Python imports, for a directory to be treated like a python module, there needs to be a '__init__.py' file
included in the same directory.
After a bit of research I found out that it is possible to do a different type of imports in Python 3 and tried it out (putting a .
before the name of the imports). Like this:
from .scenarios.test_scenario_01 import TestScenario
from .scenarios.test_scenario_02 import TestScenario2
But still, it didn't run successfully and this was the error I've gotten:
ModuleNotFoundError: No module named '__main__.scenarios'; '__main__' is not a package
Could you please help me out on this one?
TLDR: Imports work in a Python3.7 venv, but not outside it
I fixed the issue by setting up a PYTHONPATH in system environment variables with the path to the project. As a value, I put the dir to the project. Thanks for the tips.
In case of directory It is mandate to use init.py. In modules it does not require to include init.py. Please, check what is your main directory and which is your modules..Hope this helps....

python -m cannot find my modules

I wrote a package, which lives in /home/michael/python/mcdb-mail-parser/
The file structure in there is:
.
├── __init__.py
├── mcdb_mail_parser
│ ├── __init__.py
│ ├── MCDBAttachment.py
│ ├── MCDBEmail.py
│ ├── Options.py
├── mcdb-mail-parser.conf.sample
├── mcdb-mail-parser.py
├── README.md
mcdb-mail-parser.py imports from the mcdb_mail_parser subdirectory.
if I run the scripts from the source directory (/home/michael/python/src/mcdb_mail_parser) it works fine because the mcdb_mail_parser directory is immediately available in the current directory. However, I need to run it from the home directory of another user (via a cronjob, or from another script via subprocess), python complains it cannot find the module:
I tried to execute it with python3 -m /home/michael/python/src/mcdb_mail_parser, but it complains:
michael#d8:~$ python3 -m /home/michael/python/mcdb-mail-parser/
/usr/bin/python3: No module named /home/michael/python/mcdb-mail-parser/
I am not sure where to go from here. I think that it's a path issue. I could add /home/michael/python/src/mcdb_mail_parser to the system path, or perhaps python path, but that seems like the wrong solution. I certainly don't want to hard code paths into any scripts either.
How do I tell python: "Run the mcdb-mail-parser.py script from /home/michael/python/src/mcdb_mail_parser directory?
Closing notes
The accepted answer was useful, and so was the link they provided. Here's what I eventually did:
1. I moved the contents of mcdb_mail_parser from the subdirectory into the same directory as README.md, thus removing one level of complexity.
2. I added the import statements to __init__.py as suggested.
3. Python complained it couldn't find __main__.py, so I renamed mcdb-mail-parser.py to __main__.py
List the modules in the __init__.py that's in the sub directory and then have the Import in mcdb-mail-parser.py reference that directory
Very similar to this previous StackOverflow Post ->
Loading all modules in a folder in Python

Import issues in Python

I am having an import error problem in a test script. It looks like its something to do with the directory structure.
I have the following folder structure:
A
├── F1
│   ├── __init__.py
│   └── Src
│   └── F2
│   └── __init__.py
└── tests1
└── tests1
└── test_script.py
A/F1/Src/F2
F1 has "__init__py" in its level
F2 has "__init__.py" in its level
In the same level as F1, there is another folder "tests1"
tests1/tests1/test_script.py
in test_script.py, I have a line which says
from F1.src.F2 import C
With the above, I get an error saying, no module named "F1.src.F2"
Does someone know what is going on here?
from F1.src.F2 import C is an absolute import. In order for it to work, "F1" has to be located somewhere on your Python path (sys.path). Generally this also includes the current directory if you ran Python on the command line.
So if the A directory is not one of the directories on your Python path and is not your current working directory, there is no reason the import would work.
Module names are case sensitive. You have Src in one place and src in another, but I'm not sure that reflects your actual directory structure or just what you typed here.
Using a relative import will not work if you are running test_script.py as a script (Which is what it sounds like.) So, what you really want to do is make sure that either you run the script from the A directory, or go whole hog, make your project into a legit package with a setup.py and use a test runner such as tox.
I just had to create a shared library with the "egg" file.
As simple as that but it occurred to me late!

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