Importing files from different folder

Note: This answer was intended for a very specific question. For most programmers coming here from a search engine, this is not the answer you are looking for. Typically you would structure your files into packages (see other answers) instead of modifying the search path.

By default, you can't. When importing a file, Python only searches the directory that the entry-point script is running from and sys.path which includes locations such as the package installation directory (it's actually a little more complex than this, but this covers most cases).

However, you can add to the Python path at runtime:

# some_file.py import sys # caution: path[0] is reserved for script path (or '' in REPL) sys.path.insert(1, '/path/to/application/app/folder') import file 

There is nothing wrong with:

from application.app.folder.file import func_name 

Just make sure folder also contains an __init__.py. This allows it to be included as a package. I am not sure why the other answers talk about PYTHONPATH.

When modules are in parallel locations, as in the question:

application/app2/some_folder/some_file.py application/app2/another_folder/another_file.py 

This shorthand makes one module visible to the other:

import sys sys.path.append('../') 

First import sys in name-file.py

 import sys 

Second append the folder path in name-file.py

sys.path.insert(0, '/the/folder/path/name-package/') 

Third Make a blank file called __ init __.py in your subdirectory (this tells Python it is a package)

• name-file.py
• name-package
  • __ init __.py
  • name-module.py

Fourth import the module inside the folder in name-file.py

from name-package import name-module 

Try Python's relative imports:

from ...app.folder.file import func_name 

Every leading dot is another higher level in the hierarchy beginning with the current directory.

Problems? If this isn't working for you then you probably are getting bit by the many gotcha's relative imports has. Read answers and comments for more details: How to fix "Attempted relative import in non-package" even with __init__.py

Hint: have __init__.py at every directory level. You might need python -m application.app2.some_folder.some_file (leaving off .py) which you run from the top level directory or have that top level directory in your PYTHONPATH. Phew!

I think an ad hoc way would be to use the environment variable PYTHONPATH as described in the documentation: Python2, Python3

# Linux and OS X export PYTHONPATH=$HOME/dirWithScripts/:$PYTHONPATH # Windows set PYTHONPATH=C:\path\to\dirWithScripts\;%PYTHONPATH% 

The issue is that Python is looking in the wrong directory for the file. To solve this, try using relative import. Change

from application.app.folder.file import func_name 

to:

from .application.app.folder.file import func_name 

Adding the dot instructs Python to look for the application folder within the current folder, instead of in the Python install folder.

Given a folder structure like

├── main.py └── myfolder └── myfile.py 

Where myfile.py contains

def myfunc(): print('hello') 

To call myfunc from main.py, use:

from myfolder.myfile import myfunc myfunc() 

In Python 3.4 and later, you can import from a source file directly (link to documentation). This is not the simplest solution, but I'm including this answer for completeness.

Here is an example. First, the file to be imported, named foo.py:

def announce(): print("Imported!") 

The code that imports the file above, inspired heavily by the example in the documentation:

import importlib.util def module_from_file(module_name, file_path): spec = importlib.util.spec_from_file_location(module_name, file_path) module = importlib.util.module_from_spec(spec) spec.loader.exec_module(module) return module foo = module_from_file("foo", "/path/to/foo.py") if __name__ == "__main__": print(foo) print(dir(foo)) foo.announce() 

The output:

<module 'foo' from '/path/to/foo.py'> ['__builtins__', '__cached__', '__doc__', '__file__', '__loader__', '__name__', '__package__', '__spec__', 'announce'] Imported! 

Note that the variable name, the module name, and the filename need not match. This code still works:

import importlib.util def module_from_file(module_name, file_path): spec = importlib.util.spec_from_file_location(module_name, file_path) module = importlib.util.module_from_spec(spec) spec.loader.exec_module(module) return module baz = module_from_file("bar", "/path/to/foo.py") if __name__ == "__main__": print(baz) print(dir(baz)) baz.announce() 

The output:

<module 'bar' from '/path/to/foo.py'> ['__builtins__', '__cached__', '__doc__', '__file__', '__loader__', '__name__', '__package__', '__spec__', 'announce'] Imported! 

Programmatically importing modules was introduced in Python 3.1 and gives you more control over how modules are imported. Refer to the documentation for more information.

Using sys.path.append with an absolute path is not ideal when moving the application to other environments. Using a relative path won't always work because the current working directory depends on how the script was invoked.

Since the application folder structure is fixed, we can use os.path to get the full path of the module we wish to import. For example, if this is the structure:

/home/me/application/app2/some_folder/vanilla.py /home/me/application/app2/another_folder/mango.py 

And let's say that you want to import the mango module. You could do the following in vanilla.py:

import sys, os.path mango_dir = (os.path.abspath(os.path.join(os.path.dirname(__file__), '..')) + '/another_folder/') sys.path.append(mango_dir) import mango 

Of course, you don't need the mango_dir variable.

To understand how this works look at this interactive session example:

>>> import os >>> mydir = '/home/me/application/app2/some_folder' >>> newdir = os.path.abspath(os.path.join(mydir, '..')) >>> newdir '/home/me/application/app2' >>> newdir = os.path.abspath(os.path.join(mydir, '..')) + '/another_folder' >>> >>> newdir '/home/me/application/app2/another_folder' >>> 

And check the os.path documentation.

Also worth noting that dealing with multiple folders is made easier when using packages, as one can use dotted module names.

From what I know, add an __init__.py file directly in the folder of the functions you want to import will do the job.

I was faced with the same challenge, especially when importing multiple files, this is how I managed to overcome it.

import os, sys from os.path import dirname, join, abspath sys.path.insert(0, abspath(join(dirname(__file__), '..'))) from root_folder import file_name 

This worked for me in Python 3 on Linux:

import sys sys.path.append(pathToFolderContainingScripts) from scriptName import functionName #scriptName without .py extension 

Considering application as the root directory for your Python project, create an empty __init__.py file in the application, app and folder folders. Then in your some_file.py, make changes as follows to get the definition of func_name:

import sys sys.path.insert(0, r'/from/root/directory/application') from application.app.folder.file import func_name ## You can also use '*' wildcard to import all the functions in file.py file. func_name() 

One way is to create a package and use absolute import to access other modules from the package. Start the program from a script at the root of the package. This structure allows using and accessing sub-packages, parent package, and sibling packages and modules.

As an example, try creating the following folder structure:

package/ ├── __init__.py ├── main_module.py ├── module_0.py ├── subpackage_1/ | ├── __init__.py | ├── module_1.py | └── sub_subpackage_3/ | ├── __init__.py | └── module_3.py └── subpackage_2/ ├── __init__.py └── module_2.py 

Contents of main_module.py:

import subpackage_1.module_1 

Contents of module_0.py:

print('module_0 at parent directory, is imported') 

Contents of module_1.py:

print('importing other modules from module_1...') import module_0 import subpackage_2.module_2 import subpackage_1.sub_subpackage_3.module_3 

Contents of module_2.py:

print('module_2 at same level directory, is imported') 

Contents of module_3.py:

print('module_3 at sub directory, is imported') 

Leave all __init__.py files empty.

Now run main_module.py; the output will be

importing other modules from module_1... module_0 at parent directory, is imported module_2 at same level directory, is imported module_3 at sub directory, is imported 

├───root │ ├───dir_a │ │ ├───file_a.py │ │ └───file_xx.py │ ├───dir_b │ │ ├───file_b.py │ │ └───file_yy.py │ ├───dir_c │ └───dir_n 

You can add the parent directory to PYTHONPATH, in order to achieve that, you can use OS depending path in the "module search path" which is listed in sys.path. So you can easily add the parent directory like following:

# file_b.py import sys sys.path.insert(0, '..') from dir_a.file_a import func_name 

In my case I had a class to import. My file looked like this:

# /opt/path/to/code/log_helper.py class LogHelper: # stuff here 

In my main file I included the code via:

import sys sys.path.append("/opt/path/to/code/") from log_helper import LogHelper 

This works for me on Windows:

# some_file.py on mainApp/app2 import sys sys.path.insert(0, sys.path[0]+'\\app2') import some_file 

I bumped into the same question several times, so I would like to share my solution.

Python Version: 3.X

The following solution is for someone who develops your application in Python version 3.X because Python 2 is not supported since Jan/1/2020.

Project Structure

In python 3, you don't need __init__.py in your project subdirectory due to the Implicit Namespace Packages. See Is init.py not required for packages in Python 3.3+

Project ├── main.py ├── .gitignore | ├── a | └── file_a.py | └── b └── file_b.py 

Problem Statement

In file_b.py, I would like to import a class A in file_a.py under the folder a.

Solutions

#1 A quick but dirty way

Without installing the package like you are currently developing a new project

Using the try catch to check if the errors. Code example:

import sys try: # The insertion index should be 1 because index 0 is this file sys.path.insert(1, '/absolute/path/to/folder/a') # the type of path is string # because the system path already have the absolute path to folder a # so it can recognize file_a.py while searching from file_a import A except (ModuleNotFoundError, ImportError) as e: print("{} fileure".format(type(e))) else: print("Import succeeded") 

#2 Install your package

Once you installed your application (in this post, the tutorial of installation is not included)

You can simply

try: from __future__ import absolute_import # now it can reach class A of file_a.py in folder a # by relative import from ..a.file_a import A except (ModuleNotFoundError, ImportError) as e: print("{} fileure".format(type(e))) else: print("Import succeeded") 

Happy coding!

I'm quite special: I use Python with Windows!

I just complete information: for both Windows and Linux, both relative and absolute paths work into sys.path (I need relative paths because I use my scripts on the several PCs and under different main directories).

And when using Windows, both \ and / can be used as a separator for file names and of course you must double \ into Python strings. Here are some valid examples:

sys.path.append('c:\\tools\\mydir') sys.path.append('..\\mytools') sys.path.append('c:/tools/mydir') sys.path.append('../mytools') 

(Note: I think that / is more convenient than \, even if it is less 'Windows-native', because it is Linux-compatible and simpler to write and copy to Windows Explorer)

The following worked for me:

OS: Windows 10

Python: v3.10.0

Note: Since I am Python v3.10.0, I am not using __init__.py files, which did not work for me anyway.

application ├── app │ └── folder │ └── file.py └── app2 └── some_folder └── some_file.py 

WY Hsu's first solution worked for me. I have reposted it with an absolute file reference for clarity:

import sys sys.path.insert(1, 'C:\\Users\\<Your Username>\\application') import app2.some_folder.some_file some_file.hello_world() 

Alternative Solution: However, this also worked for me:

import sys sys.path.append( '.' ) import app2.some_folder.some_file some_file.hello_world() 

Although, I do not understand why it works. I thought the dot is a reference to the current directory. However, when printing out the paths to the current folder, the current directory is already listed at the top:

for path in sys.path: print(path) 

There are plenty of other solutions already but here is my two cents. Let's say you don't want to do any of these:

• add __init__.py files
• run with python -m mymodule
• edit __package__
• add if check in __main__
• edit sys.path by hand
• edit PYTHONPATH
• restructure the project

You can instead use a tool that will add a given absolute/relative path to sys.path while making sure the path is valid and in the correct format.

$ pip install importmonkey [github] [pip]

# Example structure ├─ src │ └─ project │ ├─ __init__.py │ └─ module.py └─ test └─ test.py 

# Example solution using the tool, in test.py from importmonkey import add_path add_path("../src") # relative to current __file__ import project # You can add as many paths as needed, absolute or relative, in any file. # Relative paths start from the current __file__ directory. # Normal unix path conventions work so you can use '..' and '.' and so on. # The paths you try to add are checked for validity etc. help(add_path) for details. 

Disclosure of affiliation: I made importmonkey.

Instead of just doing an import ..., do this :

from <MySubFolder> import <MyFile>

MyFile is inside the MySubFolder.

I was working on project a that I wanted users to install via pip install a with the following file list:

. ├── setup.py ├── MANIFEST.in └── a ├── __init__.py ├── a.py └── b ├── __init__.py └── b.py 

setup.py

from setuptools import setup setup ( name='a', version='0.0.1', packages=['a'], package_data={ 'a': ['b/*'], }, ) 

MANIFEST.in

recursive-include b *.* 

a/init.py

from __future__ import absolute_import from a.a import cats import a.b 

a/a.py

cats = 0 

a/b/init.py

from __future__ import absolute_import from a.b.b import dogs 

a/b/b.py

dogs = 1 

I installed the module by running the following from the directory with MANIFEST.in:

python setup.py install 

Then, from a totally different location on my filesystem /moustache/armwrestle I was able to run:

import a dir(a) 

Which confirmed that a.cats indeed equalled 0 and a.b.dogs indeed equalled 1, as intended.

My solution for people who have all the necessary __init__.py in the package, but import still doesn't work.

import sys import os sys.path.insert(0, os.getcwd()) import application.app.folder.file as file 

The code below imports the Python script given by its path, no matter where it is located, in a Python version-safe way:

def import_module_by_path(path): name = os.path.splitext(os.path.basename(path))[0] if sys.version_info[0] == 2: # Python 2 import imp return imp.load_source(name, path) elif sys.version_info[:2] <= (3, 4): # Python 3, version <= 3.4 from importlib.machinery import SourceFileLoader return SourceFileLoader(name, path).load_module() else: # Python 3, after 3.4 import importlib.util spec = importlib.util.spec_from_file_location(name, path) mod = importlib.util.module_from_spec(spec) spec.loader.exec_module(mod) return mod 

I found this in the codebase of psutils, at line 1042 in psutils.test.__init__.py (most recent commit as of 09.10.2020).

Usage example:

script = "/home/username/Documents/some_script.py" some_module = import_module_by_path(script) print(some_module.foo()) 

Important caveat: The module will be treated as top-level; any relative imports from parent packages in it will fail.

This worked for me.

Python adds the folder containing the script you launch to the PYTHONPATH, so if you run

python application/app2/some_folder/some_file.py 

Only the folder application/app2/some_folder is added to the path (not the base directory that you're executing the command in). Instead, run your file as a module and add a __init__.py in your some_folder directory.

python -m application.app2.some_folder.some_file 

This will add the base directory to the path to executable python, and then classes will be accessible via a non-relative import.

If the purpose of loading a module from a specific path is to assist you during the development of a custom module, you can create a symbolic link in the same folder of the test script that points to the root of the custom module. This module reference will take precedence over any other modules installed of the same name for any script run in that folder.

I tested this on Linux but it should work in any modern OS that supports symbolic links.

One advantage to this approach is that you can you can point to a module that's sitting in your own local software version control branch working copy which can greatly simplify the development cycle time and reduce failure modes of managing different versions of the module.

You can use pip's pip install -e . command. You must create a file called setup.py in the root of the project's directory which contains the following:

from setuptools import find_packages, setup setup( name='src', packages=find_packages(), version='0.1.0', description='my_project', author='author', license='MIT', ) 

Afterwards, enter pip install -e . while in your project's root directory. This will enable all directories to be called with their name as a module. For example, if your root directory contains the subdirectories module1 and module2, each with scripts inside them, you will be able to access them from any subdirectories with the following command, for module1:

import module1.script1 as script1

This problem may be due to PyCharm

I had the same problem while using PyCharm. I had this project structure

skylake\ backend\ apps\ example.py configuration\ settings.py frontend\ ...some_stuff 

and code from configuration import settings in example.py raised an import error.

The problem was that when I opened PyCharm, it considered that skylake is the root path and ran this code.

sys.path.extend(['D:\\projects\\skylake', 'D:/projects/skylake']) 

To fix this I just marked backend directory as the source root.

And it's fixed my problem.