How do I access the value passed into __init__.py by .run() - python-3.x

My relevant file structure looks like this.
Project
-launcher.py
-lib
--bot
---__init__.py
launcher.py runs __init__.py in lib.bot like this:
#launcher.py
from lib.bot import bot
VERSION = "0.0.4"
bot.run(VERSION)
Here is where the issue is. I am trying to access the value of VERSION inside of __init__.py. I have tried using sys.argv but when I do that like this:
import sys
VERSION = sys.argv
print(VERSION)
it simply prints ['launcher.py'].
I have tried lots of other ways to access this value but, try as I might: I am unable.
Any thoughts?

# __init__.py
VERSION = '0.0.4'
# launcher.py
from lib.bot import VERSION # access VERSION like this
print(VERSION)

Related

Sublime Text 3 + Build (ctrl + b) [duplicate]

I am running Python 2.5.
This is my folder tree:
ptdraft/
nib.py
simulations/
life/
life.py
(I also have __init__.py in each folder, omitted here for readability)
How do I import the nib module from inside the life module? I am hoping it is possible to do without tinkering with sys.path.
Note: The main module being run is in the ptdraft folder.
You could use relative imports (python >= 2.5):
from ... import nib
(What’s New in Python 2.5) PEP 328: Absolute and Relative Imports
EDIT: added another dot '.' to go up two packages
I posted a similar answer also to the question regarding imports from sibling packages. You can see it here.
Solution without sys.path hacks
Summary
Wrap the code into one folder (e.g. packaged_stuff)
Create a setup.py script where you use setuptools.setup().
Pip install the package in editable state with pip install -e <myproject_folder>
Import using from packaged_stuff.modulename import function_name
Setup
I assume the same folder structure as in the question
.
└── ptdraft
├── __init__.py
├── nib.py
└── simulations
├── __init__.py
└── life
├── __init__.py
└── life.py
I call the . the root folder, and in my case it is located in C:\tmp\test_imports.
Steps
Add a setup.py to the root folder
--
The contents of the setup.py can be simply
from setuptools import setup, find_packages
setup(name='myproject', version='1.0', packages=find_packages())
Basically "any" setup.py would work. This is just a minimal working example.
Use a virtual environment
If you are familiar with virtual environments, activate one, and skip to the next step. Usage of virtual environments are not absolutely required, but they will really help you out in the long run (when you have more than 1 project ongoing..). The most basic steps are (run in the root folder)
Create virtual env
python -m venv venv
Activate virtual env
. venv/bin/activate (Linux) or ./venv/Scripts/activate (Win)
Deactivate virtual env
deactivate (Linux)
To learn more about this, just Google out "python virtualenv tutorial" or similar. You probably never need any other commands than creating, activating and deactivating.
Once you have made and activated a virtual environment, your console should give the name of the virtual environment in parenthesis
PS C:\tmp\test_imports> python -m venv venv
PS C:\tmp\test_imports> .\venv\Scripts\activate
(venv) PS C:\tmp\test_imports>
pip install your project in editable state
Install your top level package myproject using pip. The trick is to use the -e flag when doing the install. This way it is installed in an editable state, and all the edits made to the .py files will be automatically included in the installed package.
In the root directory, run
pip install -e . (note the dot, it stands for "current directory")
You can also see that it is installed by using pip freeze
(venv) PS C:\tmp\test_imports> pip install -e .
Obtaining file:///C:/tmp/test_imports
Installing collected packages: myproject
Running setup.py develop for myproject
Successfully installed myproject
(venv) PS C:\tmp\test_imports> pip freeze
myproject==1.0
Import by prepending mainfolder to every import
In this example, the mainfolder would be ptdraft. This has the advantage that you will not run into name collisions with other module names (from python standard library or 3rd party modules).
Example Usage
nib.py
def function_from_nib():
print('I am the return value from function_from_nib!')
life.py
from ptdraft.nib import function_from_nib
if __name__ == '__main__':
function_from_nib()
Running life.py
(venv) PS C:\tmp\test_imports> python .\ptdraft\simulations\life\life.py
I am the return value from function_from_nib!
Relative imports (as in from .. import mymodule) only work in a package.
To import 'mymodule' that is in the parent directory of your current module:
import os
import sys
import inspect
currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
parentdir = os.path.dirname(currentdir)
sys.path.insert(0, parentdir)
import mymodule
edit: the __file__ attribute is not always given. Instead of using os.path.abspath(__file__) I now suggested using the inspect module to retrieve the filename (and path) of the current file
It seems that the problem is not related to the module being in a parent directory or anything like that.
You need to add the directory that contains ptdraft to PYTHONPATH
You said that import nib worked with you, that probably means that you added ptdraft itself (not its parent) to PYTHONPATH.
You can use OS depending path in "module search path" which is listed in sys.path .
So you can easily add parent directory like following
import sys
sys.path.insert(0,'..')
If you want to add parent-parent directory,
sys.path.insert(0,'../..')
This works both in python 2 and 3.
Don't know much about python 2.
In python 3, the parent folder can be added as follows:
import sys
sys.path.append('..')
...and then one is able to import modules from it
If adding your module folder to the PYTHONPATH didn't work, You can modify the sys.path list in your program where the Python interpreter searches for the modules to import, the python documentation says:
When a module named spam is imported, the interpreter first searches for a built-in module with that name. If not found, it then searches for a file named spam.py in a list of directories given by the variable sys.path. sys.path is initialized from these locations:
the directory containing the input script (or the current directory).
PYTHONPATH (a list of directory names, with the same syntax as the shell variable PATH).
the installation-dependent default.
After initialization, Python programs can modify sys.path. The directory containing the script being run is placed at the beginning of the search path, ahead of the standard library path. This means that scripts in that directory will be loaded instead of modules of the same name in the library directory. This is an error unless the replacement is intended.
Knowing this, you can do the following in your program:
import sys
# Add the ptdraft folder path to the sys.path list
sys.path.append('/path/to/ptdraft/')
# Now you can import your module
from ptdraft import nib
# Or just
import ptdraft
Here is an answer that's simple so you can see how it works, small and cross-platform.
It only uses built-in modules (os, sys and inspect) so should work
on any operating system (OS) because Python is designed for that.
Shorter code for answer - fewer lines and variables
from inspect import getsourcefile
import os.path as path, sys
current_dir = path.dirname(path.abspath(getsourcefile(lambda:0)))
sys.path.insert(0, current_dir[:current_dir.rfind(path.sep)])
import my_module # Replace "my_module" here with the module name.
sys.path.pop(0)
For less lines than this, replace the second line with import os.path as path, sys, inspect,
add inspect. at the start of getsourcefile (line 3) and remove the first line.
- however this imports all of the module so could need more time, memory and resources.
The code for my answer (longer version)
from inspect import getsourcefile
import os.path
import sys
current_path = os.path.abspath(getsourcefile(lambda:0))
current_dir = os.path.dirname(current_path)
parent_dir = current_dir[:current_dir.rfind(os.path.sep)]
sys.path.insert(0, parent_dir)
import my_module # Replace "my_module" here with the module name.
It uses an example from a Stack Overflow answer How do I get the path of the current
executed file in Python? to find the source (filename) of running code with a built-in tool.
from inspect import getsourcefile
from os.path import abspath
Next, wherever you want to find the source file from you just use:
abspath(getsourcefile(lambda:0))
My code adds a file path to sys.path, the python path list
because this allows Python to import modules from that folder.
After importing a module in the code, it's a good idea to run sys.path.pop(0) on a new line
when that added folder has a module with the same name as another module that is imported
later in the program. You need to remove the list item added before the import, not other paths.
If your program doesn't import other modules, it's safe to not delete the file path because
after a program ends (or restarting the Python shell), any edits made to sys.path disappear.
Notes about a filename variable
My answer doesn't use the __file__ variable to get the file path/filename of running
code because users here have often described it as unreliable. You shouldn't use it
for importing modules from parent folder in programs used by other people.
Some examples where it doesn't work (quote from this Stack Overflow question):
• it can't be found on some platforms • it sometimes isn't the full file path
py2exe doesn't have a __file__ attribute, but there is a workaround
When you run from IDLE with execute() there is no __file__ attribute
OS X 10.6 where I get NameError: global name '__file__' is not defined
Here is more generic solution that includes the parent directory into sys.path (works for me):
import os.path, sys
sys.path.append(os.path.join(os.path.dirname(os.path.realpath(__file__)), os.pardir))
The pathlib library (included with >= Python 3.4) makes it very concise and intuitive to append the path of the parent directory to the PYTHONPATH:
import sys
from pathlib import Path
sys.path.append(str(Path('.').absolute().parent))
In a Jupyter Notebook (opened with Jupyter LAB or Jupyter Notebook)
As long as you're working in a Jupyter Notebook, this short solution might be useful:
%cd ..
import nib
It works even without an __init__.py file.
I tested it with Anaconda3 on Linux and Windows 7.
I found the following way works for importing a package from the script's parent directory. In the example, I would like to import functions in env.py from app.db package.
.
└── my_application
└── alembic
└── env.py
└── app
├── __init__.py
└── db
import os
import sys
currentdir = os.path.dirname(os.path.realpath(__file__))
parentdir = os.path.dirname(currentdir)
sys.path.append(parentdir)
Above mentioned solutions are also fine. Another solution to this problem is
If you want to import anything from top level directory. Then,
from ...module_name import *
Also, if you want to import any module from the parent directory. Then,
from ..module_name import *
Also, if you want to import any module from the parent directory. Then,
from ...module_name.another_module import *
This way you can import any particular method if you want to.
Two line simplest solution
import os, sys
sys.path.insert(0, os.getcwd())
If parent is your working directory and you want to call another child modules from child scripts.
You can import all child modules from parent directory in any scripts and execute it as
python child_module1/child_script.py
For me the shortest and my favorite oneliner for accessing to the parent directory is:
sys.path.append(os.path.dirname(os.getcwd()))
or:
sys.path.insert(1, os.path.dirname(os.getcwd()))
os.getcwd() returns the name of the current working directory, os.path.dirname(directory_name) returns the directory name for the passed one.
Actually, in my opinion Python project architecture should be done the way where no one module from child directory will use any module from the parent directory. If something like this happens it is worth to rethink about the project tree.
Another way is to add parent directory to PYTHONPATH system environment variable.
Though the original author is probably no longer looking for a solution, but for completeness, there one simple solution. It's to run life.py as a module like this:
cd ptdraft
python -m simulations.life.life
This way you can import anything from nib.py as ptdraft directory is in the path.
I think you can try this in that specific example, but in python 3.6.3
import sys
sys.path.append('../')
same sort of style as the past answer - but in fewer lines :P
import os,sys
parentdir = os.path.dirname(__file__)
sys.path.insert(0,parentdir)
file returns the location you are working in
In a Linux system, you can create a soft link from the "life" folder to the nib.py file. Then, you can simply import it like:
import nib
I have a solution specifically for git-repositories.
First I used sys.path.append('..') and similar solutions. This causes especially problems if you are importing files which are themselves importing files with sys.path.append('..').
I then decided to always append the root directory of the git repository. In one line it would look like this:
sys.path.append(git.Repo('.', search_parent_directories=True).working_tree_dir)
Or in more details like this:
import os
import sys
import git
def get_main_git_root(path):
main_repo_root_dir = git.Repo(path, search_parent_directories=True).working_tree_dir
return main_repo_root_dir
main_repo_root_dir = get_main_git_root('.')
sys.path.append(main_repo_root_dir)
For the original question: Based on what the root directory of the repository is, the import would be
import ptdraft.nib
or
import nib
Our folder structure:
/myproject
project_using_ptdraft/
main.py
ptdraft/
__init__.py
nib.py
simulations/
__init__.py
life/
__init__.py
life.py
The way I understand this is to have a package-centric view.
The package root is ptdraft, since it's the top most level that contains __init__.py
All the files within the package can use absolute paths (that are relative to package root) for imports, for example
in life.py, we have simply:
import ptdraft.nib
However, to run life.py for package dev/testing purposes, instead of python life.py, we need to use:
cd /myproject
python -m ptdraft.simulations.life.life
Note that we didn't need to fiddle with any path at all at this point.
Further confusion is when we complete the ptdraft package, and we want to use it in a driver script, which is necessarily outside of the ptdraft package folder, aka project_using_ptdraft/main.py, we would need to fiddle with paths:
import sys
sys.path.append("/myproject") # folder that contains ptdraft
import ptdraft
import ptdraft.simulations
and use python main.py to run the script without problem.
Helpful links:
https://tenthousandmeters.com/blog/python-behind-the-scenes-11-how-the-python-import-system-works/ (see how __init__.py can be used)
https://chrisyeh96.github.io/2017/08/08/definitive-guide-python-imports.html#running-package-initialization-code
https://stackoverflow.com/a/50392363/2202107
https://stackoverflow.com/a/27876800/2202107
Work with libraries.
Make a library called nib, install it using setup.py, let it reside in site-packages and your problems are solved.
You don't have to stuff everything you make in a single package. Break it up to pieces.
I had a problem where I had to import a Flask application, that had an import that also needed to import files in separate folders. This is partially using Remi's answer, but suppose we had a repository that looks like this:
.
└── service
└── misc
└── categories.csv
└── test
└── app_test.py
app.py
pipeline.py
Then before importing the app object from the app.py file, we change the directory one level up, so when we import the app (which imports the pipeline.py), we can also read in miscellaneous files like a csv file.
import os,sys,inspect
currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
parentdir = os.path.dirname(currentdir)
sys.path.insert(0,parentdir)
os.chdir('../')
from app import app
After having imported the Flask app, you can use os.chdir('./test') so that your working directory is not changed.
It's seems to me that you don't really need to import the parent module. Let's imagine that in nib.py you have func1() and data1, you need to use in life.py
nib.py
import simulations.life.life as life
def func1():
pass
data1 = {}
life.share(func1, data1)
life.py
func1 = data1 = None
def share(*args):
global func1, data1
func1, data1 = args
And now you have the access to func1 and data in life.py. Of course you have to be careful to populate them in life.py before you try to use them,
I made this library to do this.
https://github.com/fx-kirin/add_parent_path
# Just add parent path
add_parent_path(1)
# Append to syspath and delete when the exist of with statement.
with add_parent_path(1):
# Import modules in the parent path
pass
This is the simplest solution that works for me:
from ptdraft import nib
After removing some sys path hacks, I thought it might be valuable to add
My preferred solution.
Note: this is a frame challenge - it's not necessary to do in-code.
Assuming a tree,
project
└── pkg
└── test.py
Where test.py contains
import sys, json; print(json.dumps(sys.path, indent=2))
Executing using the path only includes the package directory
python pkg/test.py
[
"/project/pkg",
...
]
But using the module argument includes the project directory
python -m pkg.test
[
"/project",
...
]
Now, all imports can be absolute, from the project directory. No further skullduggery required.
Although it is against all rules, I still want to mention this possibility:
You can first copy the file from the parent directory to the child directory. Next import it and subsequently remove the copied file:
for example in life.py:
import os
import shutil
shutil.copy('../nib.py', '.')
import nib
os.remove('nib.py')
# now you can use it just fine:
nib.foo()
Of course there might arise several problems when nibs tries to import/read other files with relative imports/paths.
This works for me to import things from a higher folder.
import os
os.chdir('..')

Problem importing python file from folder above

I know theres heaps of questions and answers for this, I tried multitude stackoverflow links but none of these seem to help.
My project structure is:
volume_price_analysis/
README.md
TODO.md
build/
docs/
requirements.txt
setup.py
vpa/
__init__.py
database_worker.py
utils.py
test/
__init__.py
test_utils.py
input/
input_file.txt
I want to load utils.py inside test_utils.py
my test_utils.py is:
import unittest
import logging
import os
from .vpa import utils
class TestUtils(unittest.TestCase):
def test_read_file(self):
input_dir = os.path.join(os.path.join(os.getcwd()+"/test/input"))
file_name = "input_file.txt"
with open(os.path.join(input_dir+"/"+file_name)) as f:
file_contents = f.read()
f.close()
self.assertEqual(file_contents, "Hello World!\n")
if __name__ == '__main__':
unittest.main()
I want to run (say inside test folder):
python3 -m test_utils.py
I can not do that, I get a bunch of errors regarding imports of utils (tried many iterations of . , no ., from this import that etc.. etc..
Why is this so bloody complicated?
I am using Python 3.7 if that helps.
As per this answer, you can do it using importlib,
in spec = importlib.util.spec_from_file_location("module.name", "/path/to/file.py") ,instead of path/to/file, you can use ../utils.py. Also, since you are already importing a package named utils (from importlib), you should call one of them by other name, ie. dont keep module.name as utils or import importlib.utils as something else.
I figured it out, turns out python prefers you to run your code from top level folder, in my case volume_price_analysis folder, all I had to do was make a shell script that calls
python3 -m unittest vpa.test.test_utils
And inside test_utils I can import whatever I want as long as I remember that I am executing the code from main folder so loading utils.py would be
from vpa import utils inside test_utils

No module named xxxx. How to import relative path?

I have created a simplified version as to focus solely on getting the relative path to work. This is my file structure:
|
-project
|-package1
| |--page
| |-__init__
|
|-package2
|-test
|-__init__
I am trying to import page into test. However, I get the error that package1 is not a module. Below I have typed all that are in my code. Very simple. I am just trying to import page into test. Is there anything I am missing (file or page set up) that is preventing me from importing?
page.py
one='half'
two='ling'
tests.py
import os
import sys
three = (one+two)
print(three)
Have you tried "from package1 import page" into your test.py? Or "from package1.page import page"?
UPDATE
When import something, Python Interpreter search in the following places:
Built-ins
Current Directory
$PYTHONPATH, environment variable
some other directory related to the installation
The last three make up to be sys.path.
In your case, to import package1 into some script in package2, there's 2 ways:
Add project path into PYTHONPATH.
Dynamically append project path into sys.path
I guess you would appreciate the latter solution, just add
import sys
sys.path.append('..')
in front of everything and it will work.
Plus: It's kind of inconvenient to use the module not inside the current directory though. I've seen only a few actual python project, and What I've seen is some of them use a single main.py in the project root to run the whole project, including test-cases. Maybe this structure is more recommended.
Hope it helps~
Original Answer:
This dir structure works fine on my computer with:
import package.page as page
page.foo() # a function in page
Could I have a guess: your current working directory may be not under your project directory.
To check, test about this:
import os
print(os.getcwd())
If the output is not your current directory, that's my case. I used to mess with this before.
To avoid this, you can:
cd to your directory before running Python
run os.chdir(...) in your code, which is to change your working directory.
If not this case, please provide more information.

ModuleNotFoundError: cannot import local file

I have a module with multiple files structured like this:
/bettermod/
├── __init__.py
├── api.py
├── bettermod.py
├── errors.py
└── loggers.py
From bettermod.py, I'm trying to import two things:
a class called API from api.py
the whole errors.py file
For the first thing, it is quite easy, I just have to do this:
from .api import API
However, for importing the whole errors.py file, I'm encountering a problem; I'm trying to do like this:
from . import errors
which should work, according to this python documentation, but it's raising the following error:
File "/path/to/bettermod/bettermod.py", line 10, in <module>
from . import errors
ModuleNotFoundError: No module named 'bettermod'
Edit: when debugging, I found that __name__ was equal to bettermod.bettermod
From docs:
Note that relative imports are based on the name of the current module.
I cannot tell you what is wrong with certainty, but there is a smell: bettermod package has a bettermod module. You want to do from bettermod.bettermod import MyBetterClass? I doubt it. In Python files ARE namespaces, so choosing your file and directory names is also an API design. Just keep that in mind.
I suspect the problem is masked by the name collision. Try this. If you run python in bettermod directory and say import bettermod you are importing bettermod\bettermod.py. And . is relative to the bettermod.py module. Now try to run python in directory above bettermod package directory. It will work because now . resolves to bettermod package.
Try:
import mymodule
mymodule.__file__
This will tell what mymodule is. For packages, it will show path to __init__.py. This will help you to orient yourself. Also look up PYHTONPATH and how you can use it to make sure you are importing from the right path.

AttributeError when packaging python library

I'm in the process of learning how to package a python library using the official guide. I've started cloning the minimal sample package suggested in the guide here. I've then added the file my_module.py inside the folder sampleproject storing a simple power function. Another function is also stored in /sampleproject/sampleproject/__init__.py. The resulting structure of the library is the following
Finally, I've used pip to successfully install the package in the interpreter. The only thing left is to make sure that I'm able to run the functions stored in the subfolder sampleproject.
import sampleproject
sampleproject.main()
# Output
"Call your main application code here"
This is great. The package is able to run the function in __init__.py. However, the package is not able to find module.py:
import sampleproject
sampleproject.module
# Output
AttributeError: module 'sampleproject' has no attribute 'module'
I've tried to add __init__.py in the main folder and to change the settings in entry_points in setup.py without success. What should I let sampleproject to be able to find the function in module.py?
Your sampleproject.module is a function you would like to execute?
In this case, do as for the sampleproject, add () to execute it:
sampleproject.module()
Otherwise, you can import your package like this:
import sampleproject.module
or:
from sampleproject import module
To be clearer, you would have to import module in your sampleproject __init__.py. Then, when you want to use the package, import it (is some py file at root):
import sampleproject # is enough as it's going to import everything you stated in __init__.py
After that, you can start to use what's in the package you imported with maybe module() if you have a function called module in your package.
init.py discussions
it seems,
you are in sampleproject->module.py
so you need to try,
from sampleproject import module

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