This question already has answers here:
How to do relative imports in Python?
(18 answers)
Closed 4 years ago.
I have a simple project structure like this:
➜ (venv:evernote) evernote_bear_project git:(master) ✗ tree | grep -v pyc
.
├── README.md
...
(snip)
...
├── manage.py
├── sample
│ ├── EDAMTest.py <==== here is an import that won't work
│ └── enlogo.png
└── util
├── __init__.py
├── files.py <====== This is being imported
└── test_files.py
Now I have a relative import in sample/EDAMTest.py:
from ..util.files import *
When I try to run python sample/EDAMTest.py from project root folder in command line, I get an error saying:
ValueError: attempted relative import beyond top-level package
I know this has been asked many times, but I still don't get it.
Since I'm running the script from the project root, in my understanding Python should be able to "know" that when I try to import from ..util.files import *, that it should go up one directory, no?
EDIT
Thanks for all the answers.
So what I understand from the link above is this:
I was running the module sample/EDAMTest.py directly via python sample/EDAMTest.py, and that meant
that the __name__ of the module was __main__
and now the path (sample/) was my "root" so to speak.
So now Python searches only this path and any path that's below it for modules / packages. Hence the error message attempted relative import _beyond top-level package_, so it cannot go one more level up, because it is at the root already.
Also Python cannot look one level "up", since this syntax from ..util.files import * does not go up a level in directory, but in a list of modules / packages it keeps on the "import search path" (sys.path)?
Is that correct?
sys.path.append() is a tweak, if your directory structure is fixed and there is nothing you can do about it.
Otherwise, you can try rearranging the folders. The easiest is moving util under sample, another option is making both of the folders psrt of a larger package.
Also import * is not ensorsed.
The relative import syntax is for importing other modules from the same package, not from the file system.
Depending on what you want, you could...
Fix the package so that the relative import works
Put __init__.py files in the project root and sample directory and run the script from one level up. This doesn't seem like what you want.
Tell python where to find the package
Set the PYTHONPATH environment variable so that python can find the package.
PYTHONPATH='.':$PYTHONPATH python samples/EDAMTest.py
Install util so that python can find it
Add a setup script and use it to install the util package and avoid setting PYTHONPATH.
"The relative import syntax is for importing other modules from the same package, not from the file system.", This is right as stated by George G.
Put __init__.py in your subfolders, which will make them package.
__init__.py can be an empty file but it is often used to perform
setup needed for the package(import things, load things into path, etc).
However you can import File into your __init__.py to make it
available at the package level:
# in your __init__.py
from util import files
# now import File from util package
from util import files
of if you want to import some specific method or class, you can do
from util.files import some_function
Another thing to do is at the package level make util/modules
available with the __all__ variable. When the interpeter sees
an __all__ variable defined in an __init__.py it imports the
modules listed in the __all__ variable when you do:
from package import *
__all__ is a list containing the names of modules that you want to be
imported with import * so looking at our above example again if we
wanted to import the submodules in util the all variable
in util/init.py would be:
__all__ = ['files', 'test_files']
With the __all__ variable populated like that, when you perform
from util import *
it would import files and test_files.
Related
I have a python project where I use grpc.
I create the files with python -m grpc_tools.protoc -I "pathToMyProtoFile" --python_out=. --grpc_python_out=. "pathToMyProtoFile\module.proto"
I want all the grpc-stuff to be in a python package. So I created a sub folder "my_package_folder" and added an empty __init__.py in it.
My Problem: How to access and where to place the generated module_pb2.py and module_pb2_grpc.py.
If I place them into the root folder of my application I cannot access them from my package with from .. import module_pb2_grpc "attempted relative import beyond top-level package"
If I place them into my "my_package_folder" the 2 generated files do not find each other.
(import module_pb2 as module__pb2 in "module_pb2_grpc.py")
This import mechanism in python is so extremely confusing... I have no idea where to start to solve this problem.
My folder structure is just the main project folder and a sub folder "my_package_folder" for all the grpc stuff.
Let's say you have a folder structure like this. I'm just taking the example of one file.
├── module_pb2_grpc.py
├── my_package_folder
│ ├── __init__.py
Then to resolve the attempted relative import beyond top-level package, you can add this.
init.py
import os
import sys
SCRIPT_DIR = os.path.dirname(os.path.abspath(__file__))
sys.path.append(os.path.dirname(SCRIPT_DIR))
from module_pb2_grpc import *
I have the script I want to run in the following structure
scripts/
project/
main.py
libraries/
a.py
In main.py I need to import things from a.py. How can I import things in subfolders that are two or more folders above main.py?
The proper way to handle this would be putting everything that needs to know about each other under a shared package, then the individual sub-packages and sub-modules can be accessed through that package. But this will also require moving the application's entrypoint to either the package, or a module that's a sibling of the package in the directory and can import it. If moving the entrypoint is an issue, or something quick and dirty is required for prototyping, Python also implements a couple other methods for affecting where imports search for modules which can be found near the end of the answer.
For the package approach, let's say you have this structure and want to import something between the two modules:
.
├── bar_dir
│ └── bar.py
└── foo_dir
└── foo.py
Currently, the two packages do not know about each other because Python only adds the entrypoint file's parent (either bar_dir or foo_dir depending on which file you run) to the import search path, so we have to tell them about each other in some way, this is done through the top level package they'll both share.
.
└── top_level
├── __init__.py
├── bar_dir
│ ├── __init__.py
│ └── bar.py
└── foo_dir
├── __init__.py
└── foo.py
This is the package layout we need, but to be able to use the package in imports, the top packagehas to be initialized.
If you only need to run the one file, you can do for example python -m top_level.bar_dir.bar but a hidden entry point like that could be confusing to work with.
To avoid that, you can define the package as a runnable module by implementing a __main__.py file inside of it, next to __init__.py, which is ran when doing python -m top_level. The new __main__.py entrypoint would then contain the actual code that runs the app (e.g. the main function) while the other modules would only have definitions.
The __init__.py files are used to mark the directories as proper packages and are ran when the package is imported to initialize its namespace.
With this done the packages now can see each other and can be accessed through either absolute or relative imports, an absolute import would being with the top_level package and use the whole dotted path to the module/package we need to import, e.g. from top_level.bar_dir import bar can be used to import bar.
Packages also allow relative imports which are a special form of a from-style import that begins with one or more dots, where each dot means the import goes up one package - from the foo module from . import module would attempt to import module from the foo_dir package, from .. import module would search for it in the top_level package etc.
One thing to note is that importing a package doesn't initialize the modules under it unless it's an explicit import of that module, for example only importing top_level won't make foo_dir and bar_dir available in its namespace unless they're imported directly through import top_level.foo_dir/top_level.bar_dir or the package's __init__.py added them to the package's namespace through its own import.
If this doesn't work in your structure, an another way is to let Python know where to search for your modules by adding to its module search path, this can be done either at runtime by inserting path strings into the sys.path list, or through the PYTHONPATH environment variable.
Continuing with the above example with a scenario and importing bar from foo, an entry for the bar_dir directory (or the directory above it) can be added to the sys.path list or the aforementioned environment variable. After that import bar (or from bar_dir import bar if the parent was added) can be used to import the module, just as if they were next to each other. The inserted path can also be relative, but that is prone to breakage with a changing cwd.
Although I think this should be pretty simple I still can't make it run.
I have following folder structure:
├── apartment
│ ├── src
│ ├── train_model
│ ├── __init__.py
│ ├── train_model.py
│ ├── utils.py
│ ├── interference.py
│ └── __init__.py
In utils.py I tried:
from src.interference import create_sample
Error: ModuleNotFoundError: No module named 'src'
from .interference import create_sample
Error: ImportError: attempted relative import with no known parent package
from interference import create_features_sample
ModuleNotFoundError: No module named 'interference'
What is the way to make it work? I am not a huge fan of non-pythonic ways as it looks dirty.
The structure starting with a src/ is aimed explicitly at not enabling import by from src.intereference import ..., and you should not put an __init__.py file in src/ folder.
Instead, following nice explanation and examples here: https://blog.ionelmc.ro/2014/05/25/python-packaging/, here is what I recommend:
install the package:
add a setup.py file at root of your folder (this is clearly not as hard as it seems)
maybe create a virtual environment
using pip install -e . (with trailing dot!) command
then simply import your package by from interference import ...
To answer to your primary request, you could update src/__init__.py with from intereference import create_sample, to expose this function at a higher level, then the chained import would work. However, I do not recommend this, as it renders everything very rigid.
You need to add the directory that contains interference to PYTHONPATH.
You can use OS depending path in "module search path" which is listed in sys.path. So you can easily add parent directory like following:
import sys
sys.path.insert(0, '..')
from interference import create_features_sample
Note that the previous code uses a relative path, so you must launch the file inside the same location or it will likely not work.
To launch from anywhere, you can use Path from the pathlib module.
from pathlib import Path
import sys
path = str(Path(Path(__file__).parent.absolute()).parent.absolute())
sys.path.insert(0, path)
from interference import create_features_sample
If you want add a custom path to your PYTHONPATH permanently, go to the 'site-packages' folder of your current Python environment and add the file 'custompaths.pth' in which each line should consist of a directory which will then be checked if you try to import the module.
Assuming 'src' being the module you want to import you should add the following line to the .pth file:
your_preceding_path/apartment
Error because of python can't able to find the specific file or package. Python usually find only for the child packages or files, otherwise you need to specify the absolute path.
import sys, os
sys.path.append(os.path.abspath(os.path.join('../..', 'src')))
Please refer the "How to access a module from outside your file folder in Python?"
Have you tried from ..interference import create_sample ?
Or for the whole module from .. import interference.
I've checked here, that I'm using another command, something that is missing from the question at the time of writing this. The command that I'm using is python -m src.train_model.utils from apartment folder.
Thanks to RMPR for helping me.
This problem is similar to this one.
This question already has answers here:
Attempted relative import with no known parent package [duplicate]
(4 answers)
Relative imports in Python 3
(31 answers)
Closed 1 year ago.
I have the following directory structure:
py_test
├── __init__.py
├── dir1
│ ├── __init__.py
│ └── script1.py
└── dir2
├── __init__.py
└── script2.py
In script2 I want to "import ..\script1".
What I tried in script2:
Does not work
from ..dir1 import script1
ImportError: attempted relative import with no known parent package`
Works
import sys, os
path2add = os.path.normpath(os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir, 'dir1')))
if (not (path2add in sys.path)) :
sys.path.append(path2add)
If I want to go with option 1, what is the simplest (i.e., with the least files) file/dir structure that makes it work?
I am aware of this, but I wonder if creating that directory structure can be avoided, and still use type-1 import.
I am currently using this workaround, which uses type-2 import.
Related:
How to import a Python class that is in a directory above?
Import a module from a directory (package) one level up
Getting "ImportError: attempted relative import with no known parent package" when running from Python Interpreter
Using importlib to dynamically import module(s) containing relative imports
How to import variables in a different python file
As mentioned in the comments, attempting to import modules a directory up will not work if script2.py is your entry point.
As mentioned in this link you included:
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 module's __name__ is set to __main__ if it is the entry point, or the one that you pass to the interpreter with something like python script2.py.
Any python module run as such no longer has the information needed to import files from higher directories.
Therefore you have two options:
Option 1: Keep using the workaround with sys.path.append
This will work how you want it to, but it is rather cumbersome.
Option 2: Make your entry point at the top level
Assuming your package has more than one script that needs to be run, you could create a new file that imports both script1 and script2 and then calls the functionality you want based on a command line argument. Then you will be able to keep your current directory structure and have your relative imports work just fine, without any kind of fiddling with sys.path.
I am new to python and i am having some issues with packages and imports.
My structure is as follows:
src/
base/
packcore/
__init__.py
utils/
__init__.py
util_1.py
util_2.py
helpers/
__init__.py
helper_1.py
helper_2.py
some_script.py
app.py
packcore is an external package that has been installed using pip and put into the --target=base.
some of the helpers in the packcore uses some of the utils.
From app.py i want to be able to import a helper/util.
But when i use from base.packcore.utils import some_util i get an error saying that no module named packcore from inside the helper/util
and if i do from packcore.utils import some_util i get an error no module named packcorefrom theapp.py`
help would be much appreciated :)
If you add an __init__.py to base/, you can make it a Python package to import from. You also need to make the parent a package (Which is currently called src) so it is actually a sibling module, rather than many isolated modules.
From there, you can either do an absolute import from the main package:
from src.base.packcore.helpers import helper_1
Or relative (Assuming you are in some_script.py or app.py):
from .base.packcore.helpers import helper_1