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Import function from another folder python [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('..')

Python can't import a module that I've created, but does it for the others [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('..')

Pythonic way to manage imports of objects from another directory [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('..')

Why can't I import class from parent directory? [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('..')

beyond top level package error in relative import

It seems there are already quite some questions here about relative import in python 3, but after going through many of them I still didn't find the answer for my issue.
so here is the question.
I have a package shown below
package/
__init__.py
A/
__init__.py
foo.py
test_A/
__init__.py
test.py
and I have a single line in test.py:
from ..A import foo
now, I am in the folder of package, and I run
python -m test_A.test
I got message
"ValueError: attempted relative import beyond top-level package"
but if I am in the parent folder of package, e.g., I run:
cd ..
python -m package.test_A.test
everything is fine.
Now my question is:
when I am in the folder of package, and I run the module inside the test_A sub-package as test_A.test, based on my understanding, ..A goes up only one level, which is still within the package folder, why it gives message saying beyond top-level package. What is exactly the reason that causes this error message?

EDIT: There are better/more coherent answers to this question in other questions:
Sibling package imports
Relative imports for the billionth time
Why doesn't it work? It's because python doesn't record where a package was loaded from. So when you do python -m test_A.test, it basically just discards the knowledge that test_A.test is actually stored in package (i.e. package is not considered a package). Attempting from ..A import foo is trying to access information it doesn't have any more (i.e. sibling directories of a loaded location). It's conceptually similar to allowing from ..os import path in a file in math. This would be bad because you want the packages to be distinct. If they need to use something from another package, then they should refer to them globally with from os import path and let python work out where that is with $PATH and $PYTHONPATH.
When you use python -m package.test_A.test, then using from ..A import foo resolves just fine because it kept track of what's in package and you're just accessing a child directory of a loaded location.
Why doesn't python consider the current working directory to be a package? NO CLUE, but gosh it would be useful.

import sys
sys.path.append("..") # Adds higher directory to python modules path.
Try this.
Worked for me.

Assumption:
If you are in the package directory, A and test_A are separate packages.
Conclusion:
..A imports are only allowed within a package.
Further notes:
Making the relative imports only available within packages is useful if you want to force that packages can be placed on any path located on sys.path.
EDIT:
Am I the only one who thinks that this is insane!? Why in the world is the current working directory not considered to be a package? – Multihunter
The current working directory is usually located in sys.path. So, all files there are importable. This is behavior since Python 2 when packages did not yet exist. Making the running directory a package would allow imports of modules as "import .A" and as "import A" which then would be two different modules. Maybe this is an inconsistency to consider.

None of these solutions worked for me in 3.6, with a folder structure like:
package1/
subpackage1/
module1.py
package2/
subpackage2/
module2.py
My goal was to import from module1 into module2. What finally worked for me was, oddly enough:
import sys
sys.path.append(".")
Note the single dot as opposed to the two-dot solutions mentioned so far.
Edit: The following helped clarify this for me:
import os
print (os.getcwd())
In my case, the working directory was (unexpectedly) the root of the project.

This is very tricky in Python.
I'll first comment on why you're having that problem and then I will mention two possible solutions.
What's going on?
You must take this paragraph from the Python documentation into consideration:
Note that relative imports are based on the name of the current
module. Since the name of the main module is always "main",
modules intended for use as the main module of a Python application
must always use absolute imports.
And also the following from PEP 328:
Relative imports use a module's name attribute to determine that
module's position in the package hierarchy. If the module's name does
not contain any package information (e.g. it is set to 'main')
then relative imports are resolved as if the module were a top level
module, regardless of where the module is actually located on the file
system.
Relative imports work from the filename (__name__ attribute), which can take two values:
It's the filename, preceded by the folder strucutre, separated by dots.
For eg: package.test_A.test
Here Python knows the parent directories: before test comes test_A and then package.
So you can use the dot notation for relative import.
# package.test_A/test.py
from ..A import foo
You can then have like a root file in the root directory which calls test.py:
# root.py
from package.test_A import test
When you run the module (test.py) directly, it becomes the entry point to the program , so __name__ == __main__. The filename has no indication of the directory structure, so Python doesn't know how to go up in the directory. For Python, test.py becomes the top-level script, there is nothing above it. That's why you cannot use relative import.
Possible Solutions
A) One way to solve this is to have a root file (in the root directory) which calls the modules/packages, like this:
root.py imports test.py. (entry point, __name__ == __main__).
test.py (relative) imports foo.py.
foo.py says the module has been imported.
The output is:
package.A.foo has been imported
Module's name is: package.test_A.test
B) If you want to execute the code as a module and not as a top-level script, you can try this from the command line:
python -m package.test_A.test
Any suggestions are welcomed.
You should also check: Relative imports for the billionth time , specially BrenBarn's answer.

from package.A import foo
I think it's clearer than
import sys
sys.path.append("..")

As the most popular answer suggests, basically its because your PYTHONPATH or sys.path includes . but not your path to your package. And the relative import is relative to your current working directory, not the file where the import happens; oddly.
You could fix this by first changing your relative import to absolute and then either starting it with:
PYTHONPATH=/path/to/package python -m test_A.test
OR forcing the python path when called this way, because:
With python -m test_A.test you're executing test_A/test.py with __name__ == '__main__' and __file__ == '/absolute/path/to/test_A/test.py'
That means that in test.py you could use your absolute import semi-protected in the main case condition and also do some one-time Python path manipulation:
from os import path
…
def main():
…
if __name__ == '__main__':
import sys
sys.path.append(path.join(path.dirname(__file__), '..'))
from A import foo
exit(main())

This is actually a lot simpler than what other answers make it out to be.
TL;DR: Import A directly instead of attempting a relative import.
The current working directory is not a package, unless you import the folder package from a different folder. So the behavior of your package will work fine if you intend it to be imported by other applications. What's not working is the tests...
Without changing anything in your directory structure, all that needs to be changed is how test.py imports foo.py.
from A import foo
Now running python -m test_A.test from the package directory will run without an ImportError.
Why does that work?
Your current working directory is not a package, but it is added to the path. Therefore you can import folder A and its contents directly. It is the same reason you can import any other package that you have installed... they're all included in your path.

Edit: 2020-05-08: Is seems the website I quoted is no longer controlled by the person who wrote the advice, so I'm removing the link to the site. Thanks for letting me know baxx.
If someone's still struggling a bit after the great answers already provided, I found advice on a website that no longer is available.
Essential quote from the site I mentioned:
"The same can be specified programmatically in this way:
import sys
sys.path.append('..')
Of course the code above must be written before the other import
statement.
It's pretty obvious that it has to be this way, thinking on it after the fact. I was trying to use the sys.path.append('..') in my tests, but ran into the issue posted by OP. By adding the import and sys.path defintion before my other imports, I was able to solve the problem.

Just remove .. in test.py
For me pytest works fine with that
Example:
from A import foo

if you have an __init__.py in an upper folder, you can initialize the import as
import file/path as alias in that init file. Then you can use it on lower scripts as:
import alias

In my case, I had to change to this:
Solution 1(more better which depend on current py file path. Easy to deploy)
Use pathlib.Path.parents make code cleaner
import sys
import os
import pathlib
target_path = pathlib.Path(os.path.abspath(__file__)).parents[3]
sys.path.append(target_path)
from utils import MultiFileAllowed
Solution 2
import sys
import os
sys.path.append(os.getcwd())
from utils import MultiFileAllowed

In my humble opinion, I understand this question in this way:
[CASE 1] When you start an absolute-import like
python -m test_A.test
or
import test_A.test
or
from test_A import test
you're actually setting the import-anchor to be test_A, in other word, top-level package is test_A . So, when we have test.py do from ..A import xxx, you are escaping from the anchor, and Python does not allow this.
[CASE 2] When you do
python -m package.test_A.test
or
from package.test_A import test
your anchor becomes package, so package/test_A/test.py doing from ..A import xxx does not escape the anchor(still inside package folder), and Python happily accepts this.
In short:
Absolute-import changes current anchor (=redefines what is the top-level package);
Relative-import does not change the anchor but confines to it.
Furthermore, we can use full-qualified module name(FQMN) to inspect this problem.
Check FQMN in each case:
[CASE2] test.__name__ = package.test_A.test
[CASE1] test.__name__ = test_A.test
So, for CASE2, an from .. import xxx will result in a new module with FQMN=package.xxx, which is acceptable.
While for CASE1, the .. from within from .. import xxx will jump out of the starting node(anchor) of test_A, and this is NOT allowed by Python.
[2022-07-19] I think this "relative-import" limitation is quite an ugly design, totally against (one of) Python's motto "Simple is better than complex".

Not sure in python 2.x but in python 3.6, assuming you are trying to run the whole suite, you just have to use -t
-t, --top-level-directory directory
Top level directory of project (defaults to start directory)
So, on a structure like
project_root
|
|----- my_module
| \
| \_____ my_class.py
|
\ tests
\___ test_my_func.py
One could for example use:
python3 unittest discover -s /full_path/project_root/tests -t /full_path/project_root/
And still import the my_module.my_class without major dramas.

Having
package/
__init__.py
A/
__init__.py
foo.py
test_A/
__init__.py
test.py
in A/__init__.py import foo:
from .foo import foo
when importing A/ from test_A/
import sys, os
sys.path.append(os.path.abspath('../A'))
# then import foo
import foo