666

I have a file called tester.py, located on /project.

/project has a subdirectory called lib, with a file called BoxTime.py:

/project/tester.py
/project/lib/BoxTime.py

I want to import BoxTime from tester. I have tried this:

import lib.BoxTime

Which resulted:

Traceback (most recent call last):
  File "./tester.py", line 3, in <module>
    import lib.BoxTime
ImportError: No module named lib.BoxTime

Any ideas how to import BoxTime from the subdirectory?

EDIT

The __init__.py was the problem, but don't forget to refer to BoxTime as lib.BoxTime, or use:

import lib.BoxTime as BT
...
BT.bt_function()
2

13 Answers 13

701

Take a look at the Packages documentation (Section 6.4).

In short, you need to put a blank file named

__init__.py

in the lib directory.

8
  • 84
    Why does it feel hacky? It's the way python marks safe/available import directories.
    – IAbstract
    Commented Aug 26, 2014 at 16:52
  • 12
    Not only it marks safe/available import directories, but also provides a way to run some initialization code when importing a directory name.
    – Sadjad
    Commented Nov 5, 2014 at 10:26
  • 56
    Yes this is hacky and even dirty, and in my opinion the language shouldn't impose its way of loading files across the filesystem. In PHP we solved the problem by letting the userland code register multiple autoloading functions that are called when a namespace/class is missing. Then the community has produced the PSR-4 standard and Composer implements it, and nowadays nobody has to worry about that. And no stupid hardcoded __init__ files (but if you want it, just register an autoloading hook ! This is the difference between hacky and hackable). Commented Oct 7, 2015 at 11:15
  • 4
    The necessary __init__.py file was at least not comfortable, may it be hacky or not. And this point got resolved in Python 3.3: A package can be made up without the file. Commented Oct 23, 2019 at 11:53
  • 2
    See also stackoverflow.com/questions/37139786/…
    – Étienne
    Commented Jan 28, 2020 at 10:48
227
  • Create a subdirectory named lib.
  • Create an empty file named lib\__init__.py.
  • In lib\BoxTime.py, write a function foo() like this:

    def foo():
        print "foo!"
    
  • In your client code in the directory above lib, write:

    from lib import BoxTime
    BoxTime.foo()
    
  • Run your client code. You will get:

    foo!
    

Much later -- in linux, it would look like this:

% cd ~/tmp
% mkdir lib
% touch lib/__init__.py
% cat > lib/BoxTime.py << EOF
heredoc> def foo():
heredoc>     print "foo!"
heredoc> EOF
% tree lib
lib
├── BoxTime.py
└── __init__.py

0 directories, 2 files
% python 
Python 2.7.6 (default, Mar 22 2014, 22:59:56) 
[GCC 4.8.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> from lib import BoxTime
>>> BoxTime.foo()
foo!
10
  • 2
    Could you provide a link to the Python documentation where this is explained? Thanks!
    – Zenon
    Commented Mar 13, 2012 at 0:19
  • 5
    Let's make that link clickable: docs.python.org/3/tutorial/modules.html#packages Commented Sep 3, 2018 at 5:42
  • Nice walkthrough for implementing a package lib Commented Oct 23, 2019 at 11:55
  • please note: subdirs shall not contain dashes or dots, but underscores are valid. for me that seems like the same restrictions as for other symbol names, but i have not yet digged it down to documentation level. Commented Mar 11, 2020 at 14:03
  • underscores => python3 (too late for editing the comment) Commented Mar 11, 2020 at 14:22
106

You can try inserting it in sys.path:

sys.path.insert(0, './lib')
import BoxTime
9
  • 16
    This is great if you for some reason can't or won't create the init.py file.
    – jpihl
    Commented Mar 19, 2014 at 8:29
  • 5
    It works if you run python from the "project" directory. The "." is interpreted relative to your current working directory, not relative to the directory where the file you're executing lives. Say you cd /data, python ../project/tester.py. Then it won't work. Commented Dec 27, 2014 at 20:56
  • 2
    This worked for me. I prefer this over an init.py file, it makes for cleaner import statements. Commented Mar 19, 2015 at 21:42
  • 9
    This works MUCH better and is the "correct" solution. init.py messes up packages like boto that have their own child folders with modules. Commented Jul 3, 2015 at 0:47
  • 1
    @jpihl You have to create (at least) a empy file named __init__.py to allow python import modules from that folder. I've tried that solution and works perfectly (v2.7.6).
    – m3nda
    Commented Aug 28, 2015 at 0:44
55

I am writing this down because everyone seems to suggest that you have to create a lib directory.

You don't need to name your sub-directory lib. You can name it anything provided you put an __init__.py into it.

You can do that by entering the following command in a linux shell:

$ touch anything/__init__.py 

So now you have this structure:

$ ls anything/
__init__.py
mylib.py

$ ls
main.py

Then you can import mylib into main.py like this:

from anything import mylib 

mylib.myfun()

You can also import functions and classes like this:

from anything.mylib import MyClass
from anything.mylib import myfun

instance = MyClass()
result = myfun()

Any variable function or class you place inside __init__.py can also be accessed:

import anything

print(anything.myvar)

Or like this:

from anything import myvar

print(myvar)
3
  • 3
    My folder structure is utils\__init__.py and utils\myfile.py. (Utils contain both files) This is how I am trying to import from utils.myfile import myMethod. But I get ModuleNotFoundError: No module named 'utils'. What could be wrong? P.S: I am using Django and trying to import in views.py which is at the same level as utils folder
    – Jagruti
    Commented Mar 18, 2019 at 8:16
  • It is possible to use absolute paths when importing modules and run your program with PYTHONPATH=. python path/to/program.py
    – nurettin
    Commented Dec 10, 2019 at 19:35
  • clearest answer! thanks Commented Oct 29, 2021 at 23:51
31

Try import .lib.BoxTime. For more information read about relative import in PEP 328.

5
  • 2
    I don't think I've ever seen that syntax used before. Is there strong reason (not) to use this method?
    – tgray
    Commented Aug 11, 2009 at 18:53
  • 2
    Why wasn't this the answer. Sure, if you want to do the whole packages thing, you should do that. But that's not what the original question was. Commented Jan 29, 2014 at 17:53
  • This gives me: ValueError: Attempted relative import in non-package
    – Alex
    Commented Mar 7, 2014 at 7:51
  • 6
    This only works if the file you're importing from is itself part of a package. If not, you'll receive the error that @Alex pointed out. Commented Apr 22, 2015 at 23:58
  • (Tried in Python 3) This results in SyntaxError: invalid syntax. PyLance tells me Relative imports cannot be used with "import .a" form; use "from . import a" instead. In this case this would be from .lib import BoxTime
    – qrsngky
    Commented Feb 17, 2023 at 4:26
20

Does your lib directory contain a __init__.py file?

Python uses __init__.py to determine if a directory is a module.

20

Full example included

This basically covers all cases (make sure you have __init__.py in relative/path/to/your/lib/folder):

import sys, os
sys.path.append(os.path.dirname(os.path.realpath(__file__)) + "/relative/path/to/your/lib/folder")
import someFileNameWhichIsInTheFolder
...
somefile.foo()

Example:

You have in your project folder:

/root/myproject/app.py

You have in another project folder:

/root/anotherproject/utils.py
/root/anotherproject/__init__.py

You want to use /root/anotherproject/utils.py and call foo function which is in it.

So you write in app.py:

import sys, os
sys.path.append(os.path.dirname(os.path.realpath(__file__)) + "/../anotherproject")
import utils

utils.foo()
5
  • 2
    if you're using os.path you probably want to use os.path.join((os.path.dirname(os.path.realpath(__file__)),'..','anotherproject') instead of hardcoding the '/' in your path concatenation.
    – cowbert
    Commented Nov 15, 2017 at 2:58
  • Why can't you just do "../anotherproject" without the os.path.dirname()?
    – abc
    Commented Jul 30, 2018 at 22:46
  • @MosheRabaev - It is good practice to use os.path functions. In case of wrting "../anotherproject" and moving the code to Windows OS, the code will break! os.path utils knows how to return correct path considering the OS the code running on. for more info docs.python.org/2/library/os.path.html
    – Mercury
    Commented Jul 31, 2018 at 8:30
  • 1
    @MosheRabaev and if you use ".." without the dirname(realpath(__file__)), then it will compute the path relative to your current working directory when you run the script, not relative to where the script lives.
    – Tara
    Commented Dec 20, 2018 at 19:20
  • os.path.realpath(os.path.dirname(os.path.realpath(file)) + "/../anotherproject") It will change "/" if you are in windows
    – Tirsvad
    Commented May 20 at 11:48
8

Create an empty file __init__.py in subdirectory /lib. And add at the begin of main code

from __future__ import absolute_import 

then

import lib.BoxTime as BT
...
BT.bt_function()

or better

from lib.BoxTime import bt_function
...
bt_function()
1
  • This is the only right fix that worked for me in Pycharm, thanks Commented Aug 27, 2021 at 20:35
6

Just an addition to these answers.

If you want to import all files from all subdirectories, you can add this to the root of your file.

import sys, os
sys.path.extend([f'./{name}' for name in os.listdir(".") if os.path.isdir(name)])

And then you can simply import files from the subdirectories just as if these files are inside the current directory.

Working example

If I have the following directory with subdirectories in my project...

.
├── a.py
├── b.py
├── c.py
├── subdirectory_a
│   ├── d.py
│   └── e.py
├── subdirectory_b
│   └── f.py
├── subdirectory_c
│   └── g.py
└── subdirectory_d
    └── h.py

I can put the following code inside my a.py file

import sys, os
sys.path.extend([f'./{name}' for name in os.listdir(".") if os.path.isdir(name)])

# And then you can import files just as if these files are inside the current directory

import b
import c
import d
import e
import f
import g
import h

In other words, this code will abstract from which directory the file is coming from.

2
  • 1
    Hey! I modified your one liner to sys.path.extend([f'{item[0]}' for item in os.walk(".") if os.path.isdir(item[0])]) and now it scrapes recursively from the project's directory. Thank you for your answer! Super useful.
    – Shrout1
    Commented Sep 1, 2021 at 17:27
  • 1
    @Shrout1 , Good job! Commented Nov 2, 2021 at 2:20
2

For this folder hierarchy diagram example:

/project/tester.py    
/project/lib/BoxTime.py

1- Create a blank py file __init__.py inside lib folder

2- In the caller py file tester.py add theses code lines

import os, sys
sys.path.insert(0,'lib')# insert the folder lib in system path
from BoxTime import Function_name # from the py file import the needed function

Easy explanation can be found in here.

Notice: This is refered to as creating/importing modules in/from different folder.

Personel experience: I tried to create module from jupyter notebook, it did not not work (maybe I done it improperly using .ipynb), I needed to do it manually outside the juypyter notebook, or using other IDE (e.g. pycharm).

1
  • sys.path.insert was key for me. Thanks!
    – Greg Biles
    Commented Oct 6, 2022 at 20:41
1

create_card.py

 init():
   print('Hello world!')

app.py

import create_card

create_card.init()

if you want to import only required functions

from create_card import init

If you have nested directories (Ex: modules/aadhaar/create-card.py)

import modules.aadhaar.create_card as create_card or from modules.aadhaar.create_card import init

-1

/project/tester.py

/project/lib/BoxTime.py

create blank file __init__.py down the line till you reach the file

/project/lib/somefolder/BoxTime.py

#lib -- needs has two items one __init__.py and a directory named somefolder #somefolder has two items boxtime.py and __init__.py

-3

try this:

from lib import BoxTime

1
  • 9
    without any explanation this isn't very useful. Commented Jan 9, 2018 at 21:13

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