Problem importing python file from folder above - python-3.x

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

Related

Running a python file within a different directory

How could I run the entirety of test.py from main.py. Both main.py and test.py is allocated within the application folder. The test.py file is within the app folder. How would I be able to achieve this, the code I have below does not work?
Directories:
application folder
├── appFolder
│ └──test.py
└── main.py
Main.py:
from .appFolder import test
from subprocess import call
call(["Python3","test.py"])
You don't need to import anything, just reference the folder name in main.py. To make it more robust, you should probably use a relative file otherwise you might get some odd results depending on where main.py is called from.
import os.path
from subprocess import call
d = os.path.dirname(os.path.realpath(__file__)) # application folder
call(["python3", f"{d}/appFolder/test.py"])
See also: https://stackoverflow.com/a/9271479/1904146

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('..')

ModuleNotFoundError: No module named in python 3 even after hacking sys.path()

I have this file structure:
/home/test
├── dirA
│   └── ClassA.py
└── dirB
└── Main.py
With the following content in the files:
ClassA.py:
class ClassA:
def __str__(self):
return 'Hi'
Main.py:
from dirA.ClassA import ClassA
class Main:
def main():
a = ClassA()
if __name__ == '__main__':
Main.main()
I change the current dir to:
$ cd /home/test/dirB
This works:
$ PYTHONPATH=/home/test python Main.py
This doesn't:
$ python Main.py
Traceback (most recent call last):
File "Main.py", line 1, in <module>
from dirA.ClassA import ClassA
ModuleNotFoundError: No module named 'dirA'
Adding this lines in Main.py has no effect:
import os, sys
# Get the top level dir.
path = os.path.dirname(os.path.dirname(__file__))
sys.path.append(path)
The module still can't be found! There are plenty of similar questions but I couldn't make this work programmatically (skipping the PYTHONPATH env var.) I understand that dirs are not modules, files are but this works in PyCharm (is the IDE fixing PYTHONPATH?)
You need to make sure that you've altered your sys.path before you attempt to load any package that might depend on the altered path - otherwise your script will fail the moment it encounters and import statement. In other words, make sure your Main.py begins as:
import os
import sys
path = os.path.join(os.path.dirname(__file__), os.pardir)
sys.path.append(path)
from dirA.ClassA import ClassA
To ensure that the last import statement operates on the altered path.
Thanks for all the help. Per suggestion, I added the appended path as the first statement. That still didn't work. But then I used:
sys.path.append(sys.path.append(os.path.dirname(os.path.dirname(os.path.abspath(__file__)))))
And this suddenly worked. Still not sure why the abspath makes a difference as the printed path I got was already absolute for __file__.
Building on the answer given by #zwer, we need to check if the script is being run in interactive mode or not. If it is, then we can use the approach mentioned by #zwer. If it is not, we can use the hard coded way (which is not always foolproof.
# https://stackoverflow.com/questions/16771894/python-nameerror-global-name-file-is-not-defined
if '__file__' in vars():
print("We are running the script non interactively")
path = os.path.join(os.path.dirname(__file__), os.pardir)
sys.path.append(path)
else:
print('We are running the script interactively')
sys.path.append("..")

Importing submodules

I am new to python and i m having a really bad time to overcome a problem with the importing system.
Lets say i have the file system presented below:
/src
/src/main.py
/src/submodules/
/src/submodules/submodule.py
/src/submodules/subsubmodules
/src/submodules/subsubmodules/subsubmodule.py
All the folders (src, submodules, subsubmodules) have and empty __init__.py file.
In submodule.py i have:
from subsubmodules import subsubmodule
In main.py i have:
from submodules import submodule
When i run submodule.py python accepts the import. But when i run main.py python raises error for the import of subsubmodule.py because /src/submodules/subsubmodules/ folder is not in the path.
Only solution is to change the import of submodule.py to
from submodules.subsubmodules import subsubmodule
This seems to me as an awful solution because after that i cannot run submodule.py and i m sure that something else is the key to that.
An other solution is to add the following code to the __init__.py file:
import os
import sys
import inspect
cmd_subfolder = os.path.split(inspect.getfile(inspect.currentframe()))[0]
if cmd_subfolder not in sys.path:
sys.path.insert(0, cmd_subfolder)
Is there any way to do this using just the importing system of python and not other methods that do it manually using, for example sys.path or other modules like os, inspect etc..?
How can i import modules without caring about the modules they import?
You can run subsubmodule.py as
python3 -m submodule.subsubmodules.subsubmodule
If you want a shorter way to invoke it, you're free to add a shell or Python script for that on the top level of your package.
This is how imports work in Python 3; there are reasons for that.
You can avoid this issue by using sys.path in your program.
sys.path.insert(0, './lib')
import subsubmodule
For this code, you can put all your imports to a lib folder.
You can read the official documentation on Python packages where this is explained in depth.

Python 3.5 - Smart module imports in the file tree

I was wondering if it was possible for modules in a project to be smart about their imports...
Say I have the following data structure :
/parent-directory
/package
__init__.py
main.py
/modules
__init__.py
one.py
/submodules-one
__init__.py
oneone.py
onetwo.py
two.py
Files higher in the hierarchy are supposed to import Classes from those lower in the hierarchy.
For instance main.py has
import modules.one
import modules.two
Now I'd like to be able to directly run not only main.py, but also one.py (and all the others)
Except that it doesn't work like I hoped :
If I run from main.py, I need to have in one.py
import modules.submodules-one.oneone
import modules.submodules-one.onetwo
But if I run from one.py, I'll get an error, and I need to have instead
import submodules-one.oneone
import submodules-one.onetwo
I've found a hacky way to get around it using in one.py :
if __name__ == '__main__':
import submodules-one.oneone
import submodules-one.onetwo
else:
import modules.submodules-one.oneone
import modules.submodules-one.onetwo
But isn't there a better solution?
P.S.: I also have an additional complication, I'm using pint,
which to work properly only needs to have a single instance of the unit registry, so I have in the top ____init____.py :
from pint import UnitRegistry
ur = UnitRegistry()
And obviously
from .. import ur
will fail if running from one of the files of the subfolders.
Thank you in advance for your answer.

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