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Why do I get an import error when a child file is trying to access a parent file

My directory is structured like this
>project
>tests
>test_to_json.py
>app.py
>x.db
in my test_to_json.py I want to access a class from app.py
How would I import it? I've tried from ..app import myClass but that gives me ImportError: attempted relative import with no known parent package. Is there something I'm missing?

You cannot use .. relative pathing in python. That specific type of relative python is simply not allowed. . is allowed, though. The resolution for this problem is usually done by converting your project into a python package.
Extensive tutorials for doing so can be found here, but I will give an example of how to convert your project into a package.
Step 1
The new file structure should look like this:
>project
>tests
>__init__.py #Note this file
>test_to_json.py
>__init__.py #Note this file
>setup.py #Note this file
>app.py
>x.db
Step 2
Write your setup.py.
Here is an generic setup.py that should work for your project:
from setuptools import setup, find_packages
setup(
name='project_package', #the name of your desired import
version='0.0.1',
author='An Awesome Coder',
packages=find_packages(),
description='An awesome package that does something',
install_requires=[], # a list of python dependencies for your package
)
find_packages() looks for all the __init__.py files in your package to identify submodules.
Step 3
Install your package. In the folder with your new setup.py, run pip install -e . This will install your package on your computer, basically adding the files to your python system path.
Step 4
From any python terminal on your computer you should now be able to import your package using the package_name you specified.
import project_package
from project_package.app import myClass
myClass()
.
.
.

What does an import statement with "..." in python mean? [duplicate]

I want to inherit from a class in a file that lies in a directory above the current one.
Is it possible to relatively import that file?

from ..subpkg2 import mod
Per the Python docs: When inside a package hierarchy, use two dots, as the import statement doc says:
When specifying what module to import you do not have to specify the absolute name of the module. When a module or package is contained within another package it is possible to make a relative import within the same top package without having to mention the package name. By using leading dots in the specified module or package after from you can specify how high to traverse up the current package hierarchy without specifying exact names. One leading dot means the current package where the module making the import exists. Two dots means up one package level. Three dots is up two levels, etc. So if you execute from . import mod from a module in the pkg package then you will end up importing pkg.mod. If you execute from ..subpkg2 import mod from within pkg.subpkg1 you will import pkg.subpkg2.mod. The specification for relative imports is contained within PEP 328.
PEP 328 deals with absolute/relative imports.

import sys
sys.path.append("..") # Adds higher directory to python modules path.

#gimel's answer is correct if you can guarantee the package hierarchy he mentions. If you can't -- if your real need is as you expressed it, exclusively tied to directories and without any necessary relationship to packaging -- then you need to work on __file__ to find out the parent directory (a couple of os.path.dirname calls will do;-), then (if that directory is not already on sys.path) prepend temporarily insert said dir at the very start of sys.path, __import__, remove said dir again -- messy work indeed, but, "when you must, you must" (and Pyhon strives to never stop the programmer from doing what must be done -- just like the ISO C standard says in the "Spirit of C" section in its preface!-).
Here is an example that may work for you:
import sys
import os.path
sys.path.append(
os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir)))
import module_in_parent_dir

Import module from a directory which is exactly one level above the current directory:
from .. import module

How to load a module that is a directory up
preface: I did a substantial rewrite of a previous answer with the hopes of helping ease people into python's ecosystem, and hopefully give everyone the best change of success with python's import system.
This will cover relative imports within a package, which I think is the most probable case to OP's question.
Python is a modular system
This is why we write import foo to load a module "foo" from the root namespace, instead of writing:
foo = dict(); # please avoid doing this
with open(os.path.join(os.path.dirname(__file__), '../foo.py') as foo_fh: # please avoid doing this
exec(compile(foo_fh.read(), 'foo.py', 'exec'), foo) # please avoid doing this
Python isn't coupled to a file-system
This is why we can embed python in environment where there isn't a defacto filesystem without providing a virtual one, such as Jython.
Being decoupled from a filesystem lets imports be flexible, this design allows for things like imports from archive/zip files, import singletons, bytecode caching, cffi extensions, even remote code definition loading.
So if imports are not coupled to a filesystem what does "one directory up" mean? We have to pick out some heuristics but we can do that, for example when working within a package, some heuristics have already been defined that makes relative imports like .foo and ..foo work within the same package. Cool!
If you sincerely want to couple your source code loading patterns to a filesystem, you can do that. You'll have to choose your own heuristics, and use some kind of importing machinery, I recommend importlib
Python's importlib example looks something like so:
import importlib.util
import sys
# For illustrative purposes.
file_path = os.path.join(os.path.dirname(__file__), '../foo.py')
module_name = 'foo'
foo_spec = importlib.util.spec_from_file_location(module_name, file_path)
# foo_spec is a ModuleSpec specifying a SourceFileLoader
foo_module = importlib.util.module_from_spec(foo_spec)
sys.modules[module_name] = foo_module
foo_spec.loader.exec_module(foo_module)
foo = sys.modules[module_name]
# foo is the sys.modules['foo'] singleton
Packaging
There is a great example project available officially here: https://github.com/pypa/sampleproject
A python package is a collection of information about your source code, that can inform other tools how to copy your source code to other computers, and how to integrate your source code into that system's path so that import foo works for other computers (regardless of interpreter, host operating system, etc)
Directory Structure
Lets have a package name foo, in some directory (preferably an empty directory).
some_directory/
foo.py # `if __name__ == "__main__":` lives here
My preference is to create setup.py as sibling to foo.py, because it makes writing the setup.py file simpler, however you can write configuration to change/redirect everything setuptools does by default if you like; for example putting foo.py under a "src/" directory is somewhat popular, not covered here.
some_directory/
foo.py
setup.py
.
#!/usr/bin/env python3
# setup.py
import setuptools
setuptools.setup(
name="foo",
...
py_modules=['foo'],
)
.
python3 -m pip install --editable ./ # or path/to/some_directory/
"editable" aka -e will yet-again redirect the importing machinery to load the source files in this directory, instead copying the current exact files to the installing-environment's library. This can also cause behavioral differences on a developer's machine, be sure to test your code!
There are tools other than pip, however I'd recommend pip be the introductory one :)
I also like to make foo a "package" (a directory containing __init__.py) instead of a module (a single ".py" file), both "packages" and "modules" can be loaded into the root namespace, modules allow for nested namespaces, which is helpful if we want to have a "relative one directory up" import.
some_directory/
foo/
__init__.py
setup.py
.
#!/usr/bin/env python3
# setup.py
import setuptools
setuptools.setup(
name="foo",
...
packages=['foo'],
)
I also like to make a foo/__main__.py, this allows python to execute the package as a module, eg python3 -m foo will execute foo/__main__.py as __main__.
some_directory/
foo/
__init__.py
__main__.py # `if __name__ == "__main__":` lives here, `def main():` too!
setup.py
.
#!/usr/bin/env python3
# setup.py
import setuptools
setuptools.setup(
name="foo",
...
packages=['foo'],
...
entry_points={
'console_scripts': [
# "foo" will be added to the installing-environment's text mode shell, eg `bash -c foo`
'foo=foo.__main__:main',
]
},
)
Lets flesh this out with some more modules:
Basically, you can have a directory structure like so:
some_directory/
bar.py # `import bar`
foo/
__init__.py # `import foo`
__main__.py
baz.py # `import foo.baz
spam/
__init__.py # `import foo.spam`
eggs.py # `import foo.spam.eggs`
setup.py
setup.py conventionally holds metadata information about the source code within, such as:
what dependencies are needed to install named "install_requires"
what name should be used for package management (install/uninstall "name"), I suggest this match your primary python package name in our case foo, though substituting underscores for hyphens is popular
licensing information
maturity tags (alpha/beta/etc),
audience tags (for developers, for machine learning, etc),
single-page documentation content (like a README),
shell names (names you type at user shell like bash, or names you find in a graphical user shell like a start menu),
a list of python modules this package will install (and uninstall)
a defacto "run tests" entry point python ./setup.py test
Its very expansive, it can even compile c extensions on the fly if a source module is being installed on a development machine. For a every-day example I recommend the PYPA Sample Repository's setup.py
If you are releasing a build artifact, eg a copy of the code that is meant to run nearly identical computers, a requirements.txt file is a popular way to snapshot exact dependency information, where "install_requires" is a good way to capture minimum and maximum compatible versions. However, given that the target machines are nearly identical anyway, I highly recommend creating a tarball of an entire python prefix. This can be tricky, too detailed to get into here. Check out pip install's --target option, or virtualenv aka venv for leads.
back to the example
how to import a file one directory up:
From foo/spam/eggs.py, if we wanted code from foo/baz we could ask for it by its absolute namespace:
import foo.baz
If we wanted to reserve capability to move eggs.py into some other directory in the future with some other relative baz implementation, we could use a relative import like:
import ..baz

Here's a three-step, somewhat minimalist version of ThorSummoner's answer for the sake of clarity. It doesn't quite do what I want (I'll explain at the bottom), but it works okay.
Step 1: Make directory and setup.py
filepath_to/project_name/
setup.py
In setup.py, write:
import setuptools
setuptools.setup(name='project_name')
Step 2: Install this directory as a package
Run this code in console:
python -m pip install --editable filepath_to/project_name
Instead of python, you may need to use python3 or something, depending on how your python is installed. Also, you can use -e instead of --editable.
Now, your directory will look more or less like this. I don't know what the egg stuff is.
filepath_to/project_name/
setup.py
test_3.egg-info/
dependency_links.txt
PKG-INFO
SOURCES.txt
top_level.txt
This folder is considered a python package and you can import from files in this parent directory even if you're writing a script anywhere else on your computer.
Step 3. Import from above
Let's say you make two files, one in your project's main directory and another in a sub directory. It'll look like this:
filepath_to/project_name/
top_level_file.py
subdirectory/
subfile.py
setup.py |
test_3.egg-info/ |----- Ignore these guys
... |
Now, if top_level_file.py looks like this:
x = 1
Then I can import it from subfile.py, or really any other file anywhere else on your computer.
# subfile.py OR some_other_python_file_somewhere_else.py
import random # This is a standard package that can be imported anywhere.
import top_level_file # Now, top_level_file.py works similarly.
print(top_level_file.x)
This is different than what I was looking for: I hoped python had a one-line way to import from a file above. Instead, I have to treat the script like a module, do a bunch of boilerplate, and install it globally for the entire python installation to have access to it. It's overkill. If anyone has a simpler method than doesn't involve the above process or importlib shenanigans, please let me know.

Polished answer of #alex-martelli with pathlib:
import pathlib
import sys
_parentdir = pathlib.Path(__file__).parent.parent.resolve()
sys.path.insert(0, str(_parentdir))
import module_in_parent_dir
sys.path.remove(str(_parentdir))

To run python /myprogram/submodule/mymodule.py which imports /myprogram/mainmodule.py, e.g., via
from mainmodule import *
on Linux (e.g., in the python Docker image), I had to add the program root directory to PYTHONPATH:
export PYTHONPATH=/myprogram

It is 2022 and none of the answers really worked for me. Here is what worked in the end
import sys
sys.path.append('../my_class')
import my_class
My directory structure:
src
--my_class.py
notebooks
-- mynotebook.ipynb
I imported my_class from mynotebook.ipynb.

You can use the sys.path.append() method to add the directory containing the package to the list of paths searched for modules. For example, if the package is located two directories above the current directory, you can use the following code:
import sys
sys.path.append("../../")
if the package is location one directory above the current directory, you can use below code:
import sys
sys.path.append("..")

Python is a modular system
Python doesn't rely on a file system
To load python code reliably, have that code in a module, and that module installed in python's library.
Installed modules can always be loaded from the top level namespace with import <name>
There is a great sample project available officially here: https://github.com/pypa/sampleproject
Basically, you can have a directory structure like so:
the_foo_project/
setup.py
bar.py # `import bar`
foo/
__init__.py # `import foo`
baz.py # `import foo.baz`
faz/ # `import foo.faz`
__init__.py
daz.py # `import foo.faz.daz` ... etc.
.
Be sure to declare your setuptools.setup() in setup.py,
official example: https://github.com/pypa/sampleproject/blob/master/setup.py
In our case we probably want to export bar.py and foo/__init__.py, my brief example:
setup.py
#!/usr/bin/env python3
import setuptools
setuptools.setup(
...
py_modules=['bar'],
packages=['foo'],
...
entry_points={},
# Note, any changes to your setup.py, like adding to `packages`, or
# changing `entry_points` will require the module to be reinstalled;
# `python3 -m pip install --upgrade --editable ./the_foo_project
)
.
Now we can install our module into the python library;
with pip, you can install the_foo_project into your python library in edit mode,
so we can work on it in real time
python3 -m pip install --editable=./the_foo_project
# if you get a permission error, you can always use
# `pip ... --user` to install in your user python library
.
Now from any python context, we can load our shared py_modules and packages
foo_script.py
#!/usr/bin/env python3
import bar
import foo
print(dir(bar))
print(dir(foo))

Python3 Imports Issue

So... I am attempting to teach myself Python.
In such, I am attempting to build something that I appear to have no clue about...
I have a "workingdir" structure such as:
/
-- classes/
-- -- install
-- myfile
In myfile I am simply attempting to "import" the file install by using:
import classes.install
Which fails with: ImportError: No module named 'classes.install'
I have attempted the following as well, and all end the same way, with the same error:
import .classes.install
sys.path.append(os.path.join(os.path.dirname(__file__), '..'))
import classes.install
As well as putting an empty __init__.py file inside the classes directory
the file install simply contains:
class gyo_install():
inst = False
# check if we have everything we need installed.
def __init__():
print("Hello World")
What am I doing wrong? I've searched and searched and searched, everything I see points to the same solutions I've attempted, and none of them work.

Python looks for files with a .py extension when importing modules. So a file named myfile will not be recognized simply by the command import myfile. The pythonic way to ensure that the interpreter will find the module is to ensure it has a .py extension. Renaming myfile to myfile.py and install to install.py and then changing the import command to
import classes.install
should solve the problem.

Create __init__.py inside install directory.
Explanation: You can import from a file that is in your current directory or from a package. A package is a directory with __init__.py inside. In fact, a package can contain only this single file.
You can read the documentation for further information.

Can't import my own modules in Python

I'm having a hard time understanding how module importing works in Python (I've never done it in any other language before either).
Let's say I have:
myapp/__init__.py
myapp/myapp/myapp.py
myapp/myapp/SomeObject.py
myapp/tests/TestCase.py
Now I'm trying to get something like this:
myapp.py
===================
from myapp import SomeObject
# stuff ...
TestCase.py
===================
from myapp import SomeObject
# some tests on SomeObject
However, I'm definitely doing something wrong as Python can't see that myapp is a module:
ImportError: No module named myapp

In your particular case it looks like you're trying to import SomeObject from the myapp.py and TestCase.py scripts. From myapp.py, do
import SomeObject
since it is in the same folder. For TestCase.py, do
from ..myapp import SomeObject
However, this will work only if you are importing TestCase from the package. If you want to directly run python TestCase.py, you would have to mess with your path. This can be done within Python:
import sys
sys.path.append("..")
from myapp import SomeObject
though that is generally not recommended.
In general, if you want other people to use your Python package, you should use distutils to create a setup script. That way, anyone can install your package easily using a command like python setup.py install and it will be available everywhere on their machine. If you're serious about the package, you could even add it to the Python Package Index, PyPI.

The function import looks for files into your PYTHONPATH env. variable and your local directory. So you can either put all your files in the same directory, or export the path typing into a terminal::
export PYTHONPATH="$PYTHONPATH:/path_to_myapp/myapp/myapp/"

exporting path is a good way. Another way is to add a .pth to your site-packages location.
On my mac my python keeps site-packages in /Library/Python shown below
/Library/Python/2.7/site-packages
I created a file called awesome.pth at /Library/Python/2.7/site-packages/awesome.pth and in the file put the following path that references my awesome modules
/opt/awesome/custom_python_modules

You can try
from myapp.myapp import SomeObject
because your project name is the same as the myapp.py which makes it search the project document first

You need to have
__init__.py
in all the folders that have code you need to interact with.
You also need to specify the top folder name of your project in every import even if the file you tried to import is at the same level.

In your first myapp directory ,u can add a setup.py file and add two python code in setup.py
from setuptools import setup
setup(name='myapp')
in your first myapp directory in commandline , use pip install -e . to install the package

pip install on Windows 10 defaults to installing in 'Program Files/PythonXX/Lib/site-packages' which is a directory that requires administrative privileges. So I fixed my issue by running pip install as Administrator (you have to open command prompt as administrator even if you are logged in with an admin account). Also, it is safer to call pip from python.
e.g.
python -m pip install <package-name>
instead of
pip install <package-name>

In my case it was Windows vs Python surprise, despite Windows filenames are not case sensitive, Python import is. So if you have Stuff.py file you need to import this name as-is.

let's say i write a module
import os
my_home_dir=os.environ['HOME'] // in windows 'HOMEPATH'
file_abs_path=os.path.join(my_home_dir,"my_module.py")
with open(file_abs_path,"w") as f:
f.write("print('I am loaded successfully')")
import importlib
importlib.util.find_spec('my_module') ==> cannot find
we have to tell python where to look for the module. we have to add our path to the sys.path
import sys
sys.path.append(file_abs_path)
now importlib.util.find_spec('my_module') returns:
ModuleSpec(name='my_module', loader=<_frozen_importlib_external.SourceFileLoader object at 0x7fa40143e8e0>, origin='/Users/name/my_module.py')
we created our module, we informed python its path, now we should be able to import it
import my_module
//I am loaded successfully

This worked for me:
from .myapp import SomeObject
The . signifies that it will search any local modules from the parent module.

Short Answer:
python -m ParentPackage.Submodule
Executing the required file via module flag worked for me. Lets say we got a typical directory structure as below:
my_project:
| Core
->myScript.py
| Utils
->helpers.py
configs.py
Now if you want to run a file inside a directory, that has imports from other modules, all you need to do is like below:
python -m Core.myscript
PS: You gotta use dot notation to refer the submodules(Files/scripts you want to execute). Also I used python3.9+. So I didnt require neither any init.py nor any sys path append statements.
Hope that helps! Happy Coding!

If you use Anaconda you can do:
conda develop /Path/To/Your/Modules
from the Shell and it will write your path into a conda.pth file into the standard directory for 3rd party modules (site-packages in my case).

If you are using the IPython Console, make sure your IDE (e.g., spyder) is pointing to the right working directory (i.e., your project folder)

Besides the suggested solutions like the accepted answer, I had the same problem in Pycharm, and I didn't want to modify imports like the relative addressing suggested above.
I finally found out that if I mark my src/ (root directory of my python codes) as the source in Interpreter settings, the issue will be resolved.

How to import a module from a directory on level above the current script

For my Python application, I have the following directories structure:
\myapp
\myapp\utils\
\myapp\utils\GChartWrapper\
\myapp\model\
\myapp\view\
\myapp\controller\
One of my class in \myapp\view\ must import a class called GChartWrapper. However, I am getting an import error...
myview.py
from myapp.utils.GChartWrapper import *
Here is the error:
<type 'exceptions.ImportError'>: No module named GChartWrapper.GChart
args = ('No module named GChartWrapper.GChart',)
message = 'No module named GChartWrapper.GChart'
What am I doing wrong? I really have a hard time to import modules/classes in Python...

The __init__.py file of the GChartWrapper package expects the GChartWrapper package on PYTHONPATH. You can tell by the first line:
from GChartWrapper.GChart import *
Is it necessary to have the GChartWrapper included package in your package directory structure?
If so, then one thing you could do is adding the path where the package resides to sys.path at run time. I take it myview.py is in the myapp\view directory? Then you could do this before importing GChartWrapper:
import sys
import os
sys.path.append(os.path.abspath(os.path.join(os.path.dirname(__file__), '..', 'utils')))
If it is not necessary to have it in your directory structure, it could be easier to have it installed at the conventional location. You can do that by running the setup.py script that's included in the GChartWrapper source distribution.

You don't import modules and packages from arbritary paths. Instead, in python you use packages and absolute imports. That'll avoid all future problems.
Example:
create the following files:
MyApp\myapp\__init__.py
MyApp\myapp\utils\__init__.py
MyApp\myapp\utils\charts.py
MyApp\myapp\model\__init__.py
MyApp\myapp\view\__init__.py
MyApp\myapp\controller\__init__.py
MyApp\run.py
MyApp\setup.py
MyApp\README
The files should be empty except for those:
MyApp\myapp\utils\charts.py:
class GChartWrapper(object):
def __init__(self):
print "DEBUG: An instance of GChartWrapper is being created!"
MyApp\myapp\view\__init__.py:
from myapp.utils.charts import GChartWrapper
def start():
c = GChartWrapper() # creating instance of the class
MyApp\run.py:
from myapp.view import start
start()
That's all! When you run your entry point (run.py) it calls a function on the view, and that creates an instance of the GChartWrapper class. Using this structure you can import anything anywhere and use it.
To complement, in MyApp\setup.py you write an installation program for the MyApp\myapp package. Use distutils to write it:
from distutils.core import setup
setup(name='MyApp',
version='1.0',
description='My Beautiful Application',
author='Martin',
author_email='martin#xxxxxxx.com',
url='http://stackoverflow.com/questions/1003843/',
packages=['myapp'],
scripts=['run.py']
)
That is enough. Now when people download the MyApp folder, they can just install it using setup.py and run it using run.py. Distutils can generate packages in a number of formats including windows installable .EXE
It's the standard way of distributing python packages/applications.

You can change the path where python looks for files.
At the top of your source file, add:
import sys
sys.path.append("..")
Or alternatively change the environment variable:
export PYTHONPATH=..

Or starting in python 2.5 (again assuming myview is in myapp\view:
from __future__ import absolute_import
from ..utils.GChartWrapper import *
See: http://docs.python.org/whatsnew/2.5.html#pep-328-absolute-and-relative-imports

GChartWrapper is also available from PyPI so you can use easy_install or pip to install the module:
sudo pip install GChartWrapper==0.9
It will then be automatically added to your PYTHONPATH and then you can remove it from your /myapp/utils directory. If you can't use sudo, look at using virtualenv (and virtualenvwrapper).