python - Relative imports for the billionth time

Question

I've been here:

• http://www.python.org/dev/peps/pep-0328/
• http://docs.python.org/2/tutorial/modules.html#packages
• Python packages: relative imports
• python relative import example code does not work
• Relative imports in python 2.5
• Relative imports in Python
• Python: Disabling relative import

and plenty of URLs that I did not copy, some on SO, some on other sites, back when I thought I'd have the solution quickly.

The forever-recurring question is this: With Windows 7, 32-bit Python 2.7.3, how do I solve this "Attempted relative import in non-package" message? I built an exact replica of the package on pep-0328:

package/
    __init__.py
    subpackage1/
        __init__.py
        moduleX.py
        moduleY.py
    subpackage2/
        __init__.py
        moduleZ.py
    moduleA.py

The imports were done from the console.

I did make functions named spam and eggs in their appropriate modules. Naturally, it didn't work. The answer is apparently in the 4th URL I listed, but it's all alumni to me. There was this response on one of the URLs I visited:

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.

The above response looks promising, but it's all hieroglyphs to me. So my question, how do I make Python not return to me "Attempted relative import in non-package"? has an answer that involves -m, supposedly.

Can somebody please tell me why Python gives that error message, what it means by "non-package", why and how do you define a 'package', and the precise answer put in terms easy enough for a kindergartener to understand.

Answer 1

Script vs. Module

Here's an explanation. The short version is that there is a big difference between directly running a Python file, and importing that file from somewhere else. Just knowing what directory a file is in does not determine what package Python thinks it is in. That depends, additionally, on how you load the file into Python (by running or by importing).

There are two ways to load a Python file: as the top-level script, or as a module. A file is loaded as the top-level script if you execute it directly, for instance by typing python myfile.py on the command line. It is loaded as a module when an import statement is encountered inside some other file. There can only be one top-level script at a time; the top-level script is the Python file you ran to start things off.

Naming

When a file is loaded, it is given a name (which is stored in its __name__ attribute). If it was loaded as the top-level script, its name is __main__. If it was loaded as a module, its name is the filename, preceded by the names of any packages/subpackages of which it is a part, separated by dots.

So for instance in your example:

package/
    __init__.py
    subpackage1/
        __init__.py
        moduleX.py
    moduleA.py

if you imported moduleX (note: imported, not directly executed), its name would be package.subpackage1.moduleX. If you imported moduleA, its name would be package.moduleA. However, if you directly run moduleX from the command line, its name will instead be __main__, and if you directly run moduleA from the command line, its name will be __main__. When a module is run as the top-level script, it loses its normal name and its name is instead __main__.

Accessing a module NOT through its containing package

There is an additional wrinkle: the module's name depends on whether it was imported "directly" from the directory it is in, or imported via a package. This only makes a difference if you run Python in a directory, and try to import a file in that same directory (or a subdirectory of it). For instance, if you start the Python interpreter in the directory package/subpackage1 and then do import moduleX, the name of moduleX will just be moduleX, and not package.subpackage1.moduleX. This is because Python adds the current directory to its search path when the interpreter is entered interactively; if it finds the to-be-imported module in the current directory, it will not know that that directory is part of a package, and the package information will not become part of the module's name.

A special case is if you run the interpreter interactively (e.g., just type python and start entering Python code on the fly). In this case the name of that interactive session is __main__.

Now here is the crucial thing for your error message: if a module's name has no dots, it is not considered to be part of a package. It doesn't matter where the file actually is on disk. All that matters is what its name is, and its name depends on how you loaded it.

Now look at the quote you included in your question:

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...

Relative imports use the module's name to determine where it is in a package. When you use a relative import like from .. import foo, the dots indicate to step up some number of levels in the package hierarchy. For instance, if your current module's name is package.subpackage1.moduleX, then ..moduleA would mean package.moduleA. For a from .. import to work, the module's name must have at least as many dots as there are in the import statement.

... are only relative in a package

However, if your module's name is __main__, it is not considered to be in a package. Its name has no dots, and therefore you cannot use from .. import statements inside it. If you try to do so, you will get the "relative-import in non-package" error.

Scripts can't import relative

What you probably did is you tried to run moduleX or the like from the command line. When you did this, its name was set to __main__, which means that relative imports within it will fail, because its name does not reveal that it is in a package. Note that this will also happen if you run Python from the same directory where a module is, and then try to import that module, because, as described above, Python will find the module in the current directory "too early" without realizing it is part of a package.

Also remember that when you run the interactive interpreter, the "name" of that interactive session is always __main__. Thus you cannot do relative imports directly from an interactive session. Relative imports are only for use within module files.

Two solutions:

1. If you really do want to run moduleX directly, but you still want it to be considered part of a package, you can do python -m package.subpackage1.moduleX. The -m tells Python to load it as a module, not as the top-level script.

2. Or perhaps you don't actually want to run moduleX, you just want to run some other script, say myfile.py, that uses functions inside moduleX. If that is the case, put myfile.py somewhere else – not inside the package directory – and run it. If inside myfile.py you do things like from package.moduleA import spam, it will work fine.

Notes

• For either of these solutions, the package directory (package in your example) must be accessible from the Python module search path (sys.path). If it is not, you will not be able to use anything in the package reliably at all.

• Since Python 2.6, the module's "name" for package-resolution purposes is determined not just by its __name__ attributes but also by the __package__ attribute. That's why I'm avoiding using the explicit symbol __name__ to refer to the module's "name". Since Python 2.6 a module's "name" is effectively __package__ + '.' + __name__, or just __name__ if __package__ is None.)

Answer 2

This is really a problem within python. The origin of confusion is that people mistakenly takes the relative import as path relative which is not.

For example when you write in faa.py:

from .. import foo

This has a meaning only if faa.py was identified and loaded by python, during execution, as a part of a package. In that case,the module's name for faa.py would be for example some_packagename.faa. If the file was loaded just because it is in the current directory, when python is run, then its name would not refer to any package and eventually relative import would fail.

A simple solution to refer modules in the current directory, is to use this:

if __package__ is None or __package__ == '':
    # uses current directory visibility
    import foo
else:
    # uses current package visibility
    from . import foo

Answer 3

Here's a general recipe, modified to fit as an example, that I am using right now for dealing with Python libraries written as packages, that contain interdependent files, where I want to be able to test parts of them piecemeal. Let's call this lib.foo and say that it needs access to lib.fileA for functions f1 and f2, and lib.fileB for class Class3.

I have included a few print calls to help illustrate how this works. In practice you would want to remove them (and maybe also the from __future__ import print_function line).

This particular example is too simple to show when we really need to insert an entry into sys.path. (See Lars' answer for a case where we do need it, when we have two or more levels of package directories, and then we use os.path.dirname(os.path.dirname(__file__))—but it doesn't really hurt here either.) It's also safe enough to do this without the if _i in sys.path test. However, if each imported file inserts the same path—for instance, if both fileA and fileB want to import utilities from the package—this clutters up sys.path with the same path many times, so it's nice to have the if _i not in sys.path in the boilerplate.

from __future__ import print_function # only when showing how this works

if __package__:
    print('Package named {!r}; __name__ is {!r}'.format(__package__, __name__))
    from .fileA import f1, f2
    from .fileB import Class3
else:
    print('Not a package; __name__ is {!r}'.format(__name__))
    # these next steps should be used only with care and if needed
    # (remove the sys.path manipulation for simple cases!)
    import os, sys
    _i = os.path.dirname(os.path.abspath(__file__))
    if _i not in sys.path:
        print('inserting {!r} into sys.path'.format(_i))
        sys.path.insert(0, _i)
    else:
        print('{!r} is already in sys.path'.format(_i))
    del _i # clean up global name space

    from fileA import f1, f2
    from fileB import Class3

... all the code as usual ...

if __name__ == '__main__':
    import doctest, sys
    ret = doctest.testmod()
    sys.exit(0 if ret.failed == 0 else 1)

The idea here is this (and note that these all function the same across python2.7 and python 3.x):

1. If run as import lib or from lib import foo as a regular package import from ordinary code, __package is lib and __name__ is lib.foo. We take the first code path, importing from .fileA, etc.

2. If run as python lib/foo.py, __package__ will be None and __name__ will be __main__.

   We take the second code path. The lib directory will already be in sys.path so there is no need to add it. We import from fileA, etc.

3. If run within the lib directory as python foo.py, the behavior is the same as for case 2.

4. If run within the lib directory as python -m foo, the behavior is similar to cases 2 and 3. However, the path to the lib directory is not in sys.path, so we add it before importing. The same applies if we run Python and then import foo.

   (Since . is in sys.path, we don't really need to add the absolute version of the path here. This is where a deeper package nesting structure, where we want to do from ..otherlib.fileC import ..., makes a difference. If you're not doing this, you can omit all the sys.path manipulation entirely.)

Notes

There is still a quirk. If you run this whole thing from outside:

$ python2 lib.foo

or:

$ python3 lib.foo

the behavior depends on the contents of lib/__init__.py. If that exists and is empty, all is well:

Package named 'lib'; __name__ is '__main__'

But if lib/__init__.py itself imports routine so that it can export routine.name directly as lib.name, you get:

$ python2 lib.foo
Package named 'lib'; __name__ is 'lib.foo'
Package named 'lib'; __name__ is '__main__'

That is, the module gets imported twice, once via the package and then again as __main__ so that it runs your main code. Python 3.6 and later warn about this:

$ python3 lib.routine
Package named 'lib'; __name__ is 'lib.foo'
[...]/runpy.py:125: RuntimeWarning: 'lib.foo' found in sys.modules
after import of package 'lib', but prior to execution of 'lib.foo';
this may result in unpredictable behaviour
  warn(RuntimeWarning(msg))
Package named 'lib'; __name__ is '__main__'

The warning is new, but the warned-about behavior is not. It is part of what some call the double import trap. (For additional details see issue 27487.) Nick Coghlan says:

This next trap exists in all current versions of Python, including 3.3, and can be summed up in the following general guideline: "Never add a package directory, or any directory inside a package, directly to the Python path".

Note that while we violate that rule here, we do it only when the file being loaded is not being loaded as part of a package, and our modification is specifically designed to allow us to access other files in that package. (And, as I noted, we probably shouldn't do this at all for single level packages.) If we wanted to be extra-clean, we might rewrite this as, e.g.:

    import os, sys
    _i = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
    if _i not in sys.path:
        sys.path.insert(0, _i)
    else:
        _i = None

    from sub.fileA import f1, f2
    from sub.fileB import Class3

    if _i:
        sys.path.remove(_i)
    del _i

That is, we modify sys.path long enough to achieve our imports, then put it back the way it was (deleting one copy of _i if and only if we added one copy of _i).

Answer 4

So after carping about this along with many others, I came across a note posted by Dorian B in this article that solved the specific problem I was having where I would develop modules and classes for use with a web service, but I also want to be able to test them as I'm coding, using the debugger facilities in PyCharm. To run tests in a self-contained class, I would include the following at the end of my class file:

if __name__ == '__main__':
   # run test code here...

but if I wanted to import other classes or modules in the same folder, I would then have to change all my import statements from relative notation to local references (i.e. remove the dot (.)) But after reading Dorian's suggestion, I tried his 'one-liner' and it worked! I can now test in PyCharm and leave my test code in place when I use the class in another class under test, or when I use it in my web service!

# import any site-lib modules first, then...
import sys
parent_module = sys.modules['.'.join(__name__.split('.')[:-1]) or '__main__']
if __name__ == '__main__' or parent_module.__name__ == '__main__':
    from codex import Codex # these are in same folder as module under test!
    from dblogger import DbLogger
else:
    from .codex import Codex
    from .dblogger import DbLogger

The if statement checks to see if we're running this module as main or if it's being used in another module that's being tested as main. Perhaps this is obvious, but I offer this note here in case anyone else frustrated by the relative import issues above can make use of it.

Answer 5

Here is one solution that I would not recommend, but might be useful in some situations where modules were simply not generated:

import os
import sys
parent_dir_name = os.path.dirname(os.path.dirname(os.path.realpath(__file__)))
sys.path.append(parent_dir_name + "/your_dir")
import your_script
your_script.a_function()

Answer 6

@BrenBarn's answer says it all, but if you're like me it might take a while to understand. Here's my case and how @BrenBarn's answer applies to it, perhaps it will help you.

The case

package/
    __init__.py
    subpackage1/
        __init__.py
        moduleX.py
    moduleA.py

Using our familiar example, and add to it that moduleX.py has a relative import to ..moduleA. Given that I tried writing a test script in the subpackage1 directory that imported moduleX, but then got the dreaded error described by the OP.

Solution

Move test script to the same level as package and import package.subpackage1.moduleX

Explanation

As explained, relative imports are made relative to the current name. When my test script imports moduleX from the same directory, then module name inside moduleX is moduleX. When it encounters a relative import the interpreter can't back up the package hierarchy because it's already at the top

When I import moduleX from above, then name inside moduleX is package.subpackage1.moduleX and the relative import can be found

Answer 7

There are too much too long anwers in a foreign language. So I'll try to make it short.

If you write from . import module, opposite to what you think, module will not be imported from current directory, but from the top level of your package! If you run .py file as a script, it simply doesn't know where the top level is and thus refuses to work.

If you start it like this py -m package.module from the directory above package, then python knows where the top level is. That's very similar to java: java -cp bin_directory package.class

---

Next question: how to import from current level?

Answer 8

I had a similar problem where I didn't want to change the Python module search path and needed to load a module relatively from a script (in spite of "scripts can't import relative with all" as BrenBarn explained nicely above).

So I used the following hack. Unfortunately, it relies on the imp module that became deprecated since version 3.4 to be dropped in favour of importlib. (Is this possible with importlib, too? I don't know.) Still, the hack works for now.

Example for accessing members of moduleX in subpackage1 from a script residing in the subpackage2 folder:

#!/usr/bin/env python3

import inspect
import imp
import os

def get_script_dir(follow_symlinks=True):
    """
    Return directory of code defining this very function.
    Should work from a module as well as from a script.
    """
    script_path = inspect.getabsfile(get_script_dir)
    if follow_symlinks:
        script_path = os.path.realpath(script_path)
    return os.path.dirname(script_path)

# loading the module (hack, relying on deprecated imp-module)
PARENT_PATH = os.path.dirname(get_script_dir())
(x_file, x_path, x_desc) = imp.find_module('moduleX', [PARENT_PATH+'/'+'subpackage1'])
module_x = imp.load_module('subpackage1.moduleX', x_file, x_path, x_desc)

# importing a function and a value
function = module_x.my_function
VALUE = module_x.MY_CONST

A cleaner approach seems to be to modify the sys.path used for loading modules as mentioned by Federico.

#!/usr/bin/env python3

if __name__ == '__main__' and __package__ is None:
    from os import sys, path
    # __file__ should be defined in this case
    PARENT_DIR = path.dirname(path.dirname(path.abspath(__file__)))
   sys.path.append(PARENT_DIR)
from subpackage1.moduleX import *

Answer 9

__name__ changes depending on whether the code in question is run in the global namespace or as part of an imported module.

If the code is not running in the global space, __name__ will be the name of the module. If it is running in global namespace -- for example, if you type it into a console, or run the module as a script using python.exe yourscriptnamehere.py then __name__ becomes "__main__".

You'll see a lot of python code with if __name__ == '__main__' is used to test whether the code is being run from the global namespace – that allows you to have a module that doubles as a script.

Did you try to do these imports from the console?

Answer 10

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.

Wrote a little python package to PyPi that might help viewers of this question. The package acts as workaround if one wishes to be able to run python files containing imports containing upper level packages from within a package / project without being directly in the importing file's directory. https://pypi.org/project/import-anywhere/

Answer 11

Another dirty but working workaround. Assumes you are on top level of your package.

import sys
from os.path import dirname, basename

if __package__ is None:
    sys.path.insert(0, '..')
    __package__ = basename(dirname(sys.argv[0]))

from . import your_module

The advantage vs another answer here is that you don't need to change imports which are autogenerated by IDE.

Edit: Attention! Pitfall!!

• If you use sys.path.append('..') instead of sys.path.insert(0, '..'), this would fail on this file structure. In that case, your_module.py will be imported instead of your_module.

    your_module
      your_module.py

• You need to have __init__.py in your directory.

Answer 12

To make Python not return to me "Attempted relative import in non-package". package/

init.py subpackage1/ init.py moduleX.py moduleY.py subpackage2/ init.py moduleZ.py moduleA.py

This error occurs only if you are applying relative import to the parent file. For example parent file already returns main after you code "print(name)" in moduleA.py .so THIS file is already main it cannot return any parent package further on. relative imports are required in files of packages subpackage1 and subpackage2 you can use ".." to refer to the parent directory or module .But parent is if already top level package it cannot go further above that parent directory(package). Such files where you are applying relative importing to parents can only work with the application of absolute import. If you will use ABSOLUTE IMPORT IN PARENT PACKAGE NO ERROR will come as python knows who is at the top level of package even if your file is in subpackages because of the concept of PYTHON PATH which defines the top level of the project