python – beyond top level package error in relative import

The Question :

381 people think this question is useful

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?

The Question Comments :
  • I have a thought here, so when run test_A.test as module, ‘..’ goes above test_A, which is already the highest level of the import test_A.test, I think the package level is not the directory level, but how many levels you import the package.
  • I promise you will understand everything about relative import after watching this answer stackoverflow.com/a/14132912/8682868.
  • Is there a way to avoid doing relative imports? Such as the way PyDev in Eclipse sees all packages within <PydevProject>/src?
  • Does your working dir also have an init.py?

The Answer 1

205 people think this answer is useful

EDIT: There are better/more coherent answers to this question in other questions:


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.

The Answer 2

185 people think this answer is useful
import sys
sys.path.append("..") # Adds higher directory to python modules path.

Try this. Worked for me.

The Answer 3

48 people think this answer is useful

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.

The Answer 4

31 people think this answer is useful

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.

The Answer 5

19 people think this answer is useful

from package.A import foo

I think it’s clearer than

import sys
sys.path.append("..")

The Answer 6

16 people think this answer is useful

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())

The Answer 7

10 people think this answer is useful

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.

The Answer 8

6 people think this answer is useful

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

The Answer 9

6 people think this answer is useful

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:

  1. 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

  1. 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:

enter image description here

  • 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.

The Answer 10

2 people think this answer is useful

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

The Answer 11

1 people think this answer is useful

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.

The Answer 12

1 people think this answer is useful

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.

The Answer 13

0 people think this answer is useful

This one didn’t work for me as I’m using Django 2.1.3:

import sys
sys.path.append("..") # Adds higher directory to python modules path.

I opted for a custom solution where I added a command to the server startup script to copy my shared script into the django ‘app’ that needed the shared python script. It’s not ideal but as I’m only developing a personal website, it fit the bill for me. I will post here again if I can find the django way of sharing code between Django Apps within a single website.

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