552

Imagine this directory structure:

app/
   __init__.py
   sub1/
      __init__.py
      mod1.py
   sub2/
      __init__.py
      mod2.py

I'm coding mod1, and I need to import something from mod2. How should I do it?

I tried from ..sub2 import mod2 but I'm getting an "Attempted relative import in non-package".

I googled around but found only "sys.path manipulation" hacks. Isn't there a clean way?


Edit: all my __init__.py's are currently empty

Edit2: I'm trying to do this because sub2 contains classes that are shared across sub packages (sub1, subX, etc.).

Edit3: The behaviour I'm looking for is the same as described in PEP 366 (thanks John B)

rypel
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Joril
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    I recommend updating your question to make it more clear that you're describing the issue addressed in PEP 366. – John B Sep 16 '08 at 19:36
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    It's a long winded explanation but check here: http://stackoverflow.com/a/10713254/1267156 I answered a very similar question. I had this same problem until last night. – Sevvy325 May 23 '12 at 04:05
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    For those who wish to load a module located at an arbitrary path, see this: http://stackoverflow.com/questions/67631/how-to-import-a-module-given-the-full-path – Evgeni Sergeev Jun 08 '14 at 06:29
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    On a related note, Python 3 will change the default handling of imports to be absolute by default; relative imports will have to be explicitly specified. – Ross Mar 30 '15 at 23:28

15 Answers15

354

Everyone seems to want to tell you what you should be doing rather than just answering the question.

The problem is that you're running the module as '__main__' by passing the mod1.py as an argument to the interpreter.

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.

In Python 2.6, they're adding the ability to reference modules relative to the main module. PEP 366 describes the change.

Update: According to Nick Coghlan, the recommended alternative is to run the module inside the package using the -m switch.

John B
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    The answer here involves messing with sys.path at every entry point to your program. I guess that's the only way to do it. – Nick Retallack May 11 '10 at 04:27
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    The recommended alternative is to run modules inside packages using the `-m` switch, rather than by specifying their filename directly. – ncoghlan Feb 23 '11 at 04:25
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    I don't understand: where is the answer here? How can one import modules in such a directory structure? – Tom Sep 29 '12 at 16:34
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    @Tom: In this instance, mod1 would `from sub2 import mod2`. Then, to run mod1, from within app, do `python -m sub1.mod1`. – Xiong Chiamiov Nov 20 '12 at 06:06
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    @XiongChiamiov: does this mean you can't do it if your python is embedded in an application, so you don't have access to python command line switches? – LarsH Jan 29 '13 at 16:58
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    @MattJoiner: It works if you run mod1.py as `python -m app.sub1.mod1` (from the parent dir of `app`) as [Pankaj wrote](http://stackoverflow.com/a/15458607/229381). – Alexey Kuzminich Apr 25 '13 at 05:12
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    To make it clear, when you do `from .m import whatever` in a script you pass as an argument to interpreter, it basically [resolves](https://github.com/python/cpython/blob/v3.6.4/Python/import.c#L1471) to `from __main__.m import whatever`. Which makes it look for `__main__/m.py`. – x-yuri Dec 29 '17 at 14:08
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    "Everyone seems to want to tell you what you should be doing rather than just answering the question." Well, *good*. That's the way getting help writing software *should* be! When I fly in the face of good practice, *I want to know*. – jpmc26 Apr 03 '19 at 03:52
  • @Tom I suppose the answer is in the PEP 366 and boils down to something like `if(__name__ == "__main__" and __package__ is None): __package__ = "app"` – AstroFloyd Apr 10 '21 at 05:57
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    @jpmc26 Agreed, but I think it should be a comment then, not an answer. – AstroFloyd Apr 10 '21 at 06:01
142

Here is the solution which works for me:

I do the relative imports as from ..sub2 import mod2 and then, if I want to run mod1.py then I go to the parent directory of app and run the module using the python -m switch as python -m app.sub1.mod1.

The real reason why this problem occurs with relative imports, is that relative imports works by taking the __name__ property of the module. If the module is being directly run, then __name__ is set to __main__ and it doesn't contain any information about package structure. And, thats why python complains about the relative import in non-package error.

So, by using the -m switch you provide the package structure information to python, through which it can resolve the relative imports successfully.

I have encountered this problem many times while doing relative imports. And, after reading all the previous answers, I was still not able to figure out how to solve it, in a clean way, without needing to put boilerplate code in all files. (Though some of the comments were really helpful, thanks to @ncoghlan and @XiongChiamiov)

Hope this helps someone who is fighting with relative imports problem, because going through PEP is really not fun.

Pankaj
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    Best answer IMHO: not only explains why OP had the issue, but also finds a way to solve it *without changing the way his modules do imports*. Afterall, OP's relative imports were fine. The culprit was the lack of access to outer packages when directly running as script, something `-m` was designed to solve. – MestreLion Nov 07 '13 at 03:40
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    Also take note: this answer was 5 years after the question. These features were not available at the time. – JeremyKun Apr 01 '14 at 21:05
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    If you want to import a module from the same directory you can do `from . import some_module`. – Rotareti Sep 16 '16 at 23:47
  • This is the answer that helped me and it also helped me condense my thought down to this: **In order to run a Python script which contains relative imports, I must run the script as a module while `$ PWD` is its parent directory like `$ python -m app.main`**. For clarity, `$ python -m .` – Jesse H. May 15 '21 at 15:33
126
main.py
setup.py
app/ ->
    __init__.py
    package_a/ ->
       __init__.py
       module_a.py
    package_b/ ->
       __init__.py
       module_b.py
  1. You run python main.py.
  2. main.py does: import app.package_a.module_a
  3. module_a.py does import app.package_b.module_b

Alternatively 2 or 3 could use: from app.package_a import module_a

That will work as long as you have app in your PYTHONPATH. main.py could be anywhere then.

So you write a setup.py to copy (install) the whole app package and subpackages to the target system's python folders, and main.py to target system's script folders.

nosklo
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    Excellent answer. Is there some way to import that way without install the package in PYTHONPATH? – auraham Jul 27 '12 at 16:34
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    Suggested additional reading: http://blog.habnab.it/blog/2013/07/21/python-packages-and-you/ – nosklo Oct 17 '13 at 11:40
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    then, one day, need to change name of app to test_app. what would happen? You will need to change all the source codes, import app.package_b.module_b --> test_app.package_b.module_b. this is absolutely BAD practice... And we should try to use relative import within the package. – Spybdai Dec 16 '16 at 11:30
50

"Guido views running scripts within a package as an anti-pattern" (rejected PEP-3122)

I have spent so much time trying to find a solution, reading related posts here on Stack Overflow and saying to myself "there must be a better way!". Looks like there is not.

lesnik
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    Note: Already mentioned [pep-366](http://python.org/dev/peps/pep-0366/) (created around the same time as [pep-3122](http://www.python.org/dev/peps/pep-3122/)) provides the same capabilities but uses a different backward-compatible implementation i.e., if you want to run a module inside a package as a script *and* use explicit relative imports in it then you could run it using `-m` switch: `python -m app.sub1.mod1` or invoke `app.sub1.mod1.main()` from a top-level script (e.g., generated from setuptools' entry_points defined in setup.py). – jfs Apr 29 '13 at 00:14
  • +1 for using setuptools and entry points - it is a proper way to set up scripts that will be run from the outside, in a well-defined location, as opposed to endlessly hacking PYTHONPATH – RecencyEffect Mar 11 '20 at 11:51
  • Didn't found the definition of 'run' on the peps. For me, it doesn't look like 'running' is the best definition (for the ant pattern) cause at the end the 'interpretation' will link the dependencies and not actually 'run' it at the sense of executing immediately. [Reference 1](http://people.duke.edu/~ccc14/sta-663-2016/17_C_InOneLecture.html) and [reference 2](https://www.c-sharpcorner.com/article/why-learn-python-an-introduction-to-python/) –  Dec 29 '20 at 23:48
38

This is solved 100%:

  • app/
    • main.py
  • settings/
    • local_setings.py

Import settings/local_setting.py in app/main.py:

main.py:

import sys
sys.path.insert(0, "../settings")


try:
    from local_settings import *
except ImportError:
    print('No Import')
Vit Bernatik
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    thank you! all ppl were forcing me to run my script differently instead of telling me how to solve it within script. But I had to change the code to use `sys.path.insert(0, "../settings")` and then `from local_settings import *` – Vit Bernatik Nov 04 '16 at 20:22
24
def import_path(fullpath):
    """ 
    Import a file with full path specification. Allows one to
    import from anywhere, something __import__ does not do. 
    """
    path, filename = os.path.split(fullpath)
    filename, ext = os.path.splitext(filename)
    sys.path.append(path)
    module = __import__(filename)
    reload(module) # Might be out of date
    del sys.path[-1]
    return module

I'm using this snippet to import modules from paths, hope that helps

new123456
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iElectric
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    I'm using this snippet, combined with the imp module (as explained here [1]) to great effect. [1]: http://stackoverflow.com/questions/1096216/override-namespace-in-python/1096247#1096247 – Xiong Chiamiov Jul 16 '09 at 21:20
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    Probably, sys.path.append(path) should be replaced with sys.path.insert(0, path), and sys.path[-1] should be replaced with sys.path[0]. Otherwise the function will import the wrong module, if there is already a module with the same name in search path. E.g., if there is "some.py" in current dir, import_path("/imports/some.py") will import the wrong file. – Alex Che Jun 16 '10 at 08:24
  • I agree! Sometimes other relative imports will make precedance. Use sys.path.insert – iElectric Jun 19 '10 at 07:13
  • How would you replicate the behavior of from x import y (or *)? – levesque Dec 07 '10 at 19:26
  • It's not clear, please specify full usage of this script to solve OP problem. – mrgloom Sep 24 '18 at 13:12
23

explanation of nosklo's answer with examples

note: all __init__.py files are empty.

main.py
app/ ->
    __init__.py
    package_a/ ->
       __init__.py
       fun_a.py
    package_b/ ->
       __init__.py
       fun_b.py

app/package_a/fun_a.py

def print_a():
    print 'This is a function in dir package_a'

app/package_b/fun_b.py

from app.package_a.fun_a import print_a
def print_b():
    print 'This is a function in dir package_b'
    print 'going to call a function in dir package_a'
    print '-'*30
    print_a()

main.py

from app.package_b import fun_b
fun_b.print_b()

if you run $ python main.py it returns:

This is a function in dir package_b
going to call a function in dir package_a
------------------------------
This is a function in dir package_a
  • main.py does: from app.package_b import fun_b
  • fun_b.py does from app.package_a.fun_a import print_a

so file in folder package_b used file in folder package_a, which is what you want. Right??

suhailvs
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12

This is unfortunately a sys.path hack, but it works quite well.

I encountered this problem with another layer: I already had a module of the specified name, but it was the wrong module.

what I wanted to do was the following (the module I was working from was module3):

mymodule\
   __init__.py
   mymodule1\
      __init__.py
      mymodule1_1
   mymodule2\
      __init__.py
      mymodule2_1


import mymodule.mymodule1.mymodule1_1  

Note that I have already installed mymodule, but in my installation I do not have "mymodule1"

and I would get an ImportError because it was trying to import from my installed modules.

I tried to do a sys.path.append, and that didn't work. What did work was a sys.path.insert

if __name__ == '__main__':
    sys.path.insert(0, '../..')

So kind of a hack, but got it all to work! So keep in mind, if you want your decision to override other paths then you need to use sys.path.insert(0, pathname) to get it to work! This was a very frustrating sticking point for me, allot of people say to use the "append" function to sys.path, but that doesn't work if you already have a module defined (I find it very strange behavior)

Garrett Berg
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Let me just put this here for my own reference. I know that it is not good Python code, but I needed a script for a project I was working on and I wanted to put the script in a scripts directory.

import os.path
import sys
sys.path.append(os.path.abspath(os.path.join(os.path.dirname(__file__), "..")))
milkypostman
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9

As @EvgeniSergeev says in the comments to the OP, you can import code from a .py file at an arbitrary location with:

import imp

foo = imp.load_source('module.name', '/path/to/file.py')
foo.MyClass()

This is taken from this SO answer.

Community
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LondonRob
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6

Take a look at http://docs.python.org/whatsnew/2.5.html#pep-328-absolute-and-relative-imports. You could do

from .mod1 import stuff
Yi Jiang
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mossplix
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2

From Python doc,

In Python 2.5, you can switch import‘s behaviour to absolute imports using a from __future__ import absolute_import directive. This absolute- import behaviour will become the default in a future version (probably Python 2.7). Once absolute imports are the default, import string will always find the standard library’s version. It’s suggested that users should begin using absolute imports as much as possible, so it’s preferable to begin writing from pkg import string in your code

jung rhew
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1

On top of what John B said, it seems like setting the __package__ variable should help, instead of changing __main__ which could screw up other things. But as far as I could test, it doesn't completely work as it should.

I have the same problem and neither PEP 328 or 366 solve the problem completely, as both, by the end of the day, need the head of the package to be included in sys.path, as far as I could understand.

I should also mention that I did not find how to format the string that should go into those variables. Is it "package_head.subfolder.module_name" or what?

the Tin Man
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Gabriel
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1

I found it's more easy to set "PYTHONPATH" enviroment variable to the top folder:

bash$ export PYTHONPATH=/PATH/TO/APP

then:

import sub1.func1
#...more import

of course, PYTHONPATH is "global", but it didn't raise trouble for me yet.

Andrew_1510
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0

You have to append the module’s path to PYTHONPATH:

export PYTHONPATH="${PYTHONPATH}:/path/to/your/module/"
Giorgos Myrianthous
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    This is roughly the same as manipulating `sys.path`, since `sys.path` gets initialized from `PYTHONPATH` – Joril Mar 25 '20 at 14:50
  • @Joril That's correct but `sys.path` needs to be hardcoded in the source code in contrast to `PYTHONPATH` which is an environment variable and can be exported. – Giorgos Myrianthous Mar 25 '20 at 14:51