Related
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?
EDIT: There are better/more coherent answers to this question in other questions:
Sibling package imports
Relative imports for the billionth time
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.
import sys
sys.path.append("..") # Adds higher directory to python modules path.
Try this.
Worked for me.
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.
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.
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:
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
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:
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.
from package.A import foo
I think it's clearer than
import sys
sys.path.append("..")
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())
This is actually a lot simpler than what other answers make it out to be.
TL;DR: Import A directly instead of attempting a relative import.
The current working directory is not a package, unless you import the folder package from a different folder. So the behavior of your package will work fine if you intend it to be imported by other applications. What's not working is the tests...
Without changing anything in your directory structure, all that needs to be changed is how test.py imports foo.py.
from A import foo
Now running python -m test_A.test from the package directory will run without an ImportError.
Why does that work?
Your current working directory is not a package, but it is added to the path. Therefore you can import folder A and its contents directly. It is the same reason you can import any other package that you have installed... they're all included in your path.
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.
Just remove .. in test.py
For me pytest works fine with that
Example:
from A import foo
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
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
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.
[2022-07-19] I think this "relative-import" limitation is quite an ugly design, totally against (one of) Python's motto "Simple is better than complex".
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.
Having
package/
__init__.py
A/
__init__.py
foo.py
test_A/
__init__.py
test.py
in A/__init__.py import foo:
from .foo import foo
when importing A/ from test_A/
import sys, os
sys.path.append(os.path.abspath('../A'))
# then import foo
import foo
I want to inherit from a class in a file that lies in a directory above the current one.
Is it possible to relatively import that file?
from ..subpkg2 import mod
Per the Python docs: When inside a package hierarchy, use two dots, as the import statement doc says:
When specifying what module to import you do not have to specify the absolute name of the module. When a module or package is contained within another package it is possible to make a relative import within the same top package without having to mention the package name. By using leading dots in the specified module or package after from you can specify how high to traverse up the current package hierarchy without specifying exact names. One leading dot means the current package where the module making the import exists. Two dots means up one package level. Three dots is up two levels, etc. So if you execute from . import mod from a module in the pkg package then you will end up importing pkg.mod. If you execute from ..subpkg2 import mod from within pkg.subpkg1 you will import pkg.subpkg2.mod. The specification for relative imports is contained within PEP 328.
PEP 328 deals with absolute/relative imports.
import sys
sys.path.append("..") # Adds higher directory to python modules path.
#gimel's answer is correct if you can guarantee the package hierarchy he mentions. If you can't -- if your real need is as you expressed it, exclusively tied to directories and without any necessary relationship to packaging -- then you need to work on __file__ to find out the parent directory (a couple of os.path.dirname calls will do;-), then (if that directory is not already on sys.path) prepend temporarily insert said dir at the very start of sys.path, __import__, remove said dir again -- messy work indeed, but, "when you must, you must" (and Pyhon strives to never stop the programmer from doing what must be done -- just like the ISO C standard says in the "Spirit of C" section in its preface!-).
Here is an example that may work for you:
import sys
import os.path
sys.path.append(
os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir)))
import module_in_parent_dir
Import module from a directory which is exactly one level above the current directory:
from .. import module
How to load a module that is a directory up
preface: I did a substantial rewrite of a previous answer with the hopes of helping ease people into python's ecosystem, and hopefully give everyone the best change of success with python's import system.
This will cover relative imports within a package, which I think is the most probable case to OP's question.
Python is a modular system
This is why we write import foo to load a module "foo" from the root namespace, instead of writing:
foo = dict(); # please avoid doing this
with open(os.path.join(os.path.dirname(__file__), '../foo.py') as foo_fh: # please avoid doing this
exec(compile(foo_fh.read(), 'foo.py', 'exec'), foo) # please avoid doing this
Python isn't coupled to a file-system
This is why we can embed python in environment where there isn't a defacto filesystem without providing a virtual one, such as Jython.
Being decoupled from a filesystem lets imports be flexible, this design allows for things like imports from archive/zip files, import singletons, bytecode caching, cffi extensions, even remote code definition loading.
So if imports are not coupled to a filesystem what does "one directory up" mean? We have to pick out some heuristics but we can do that, for example when working within a package, some heuristics have already been defined that makes relative imports like .foo and ..foo work within the same package. Cool!
If you sincerely want to couple your source code loading patterns to a filesystem, you can do that. You'll have to choose your own heuristics, and use some kind of importing machinery, I recommend importlib
Python's importlib example looks something like so:
import importlib.util
import sys
# For illustrative purposes.
file_path = os.path.join(os.path.dirname(__file__), '../foo.py')
module_name = 'foo'
foo_spec = importlib.util.spec_from_file_location(module_name, file_path)
# foo_spec is a ModuleSpec specifying a SourceFileLoader
foo_module = importlib.util.module_from_spec(foo_spec)
sys.modules[module_name] = foo_module
foo_spec.loader.exec_module(foo_module)
foo = sys.modules[module_name]
# foo is the sys.modules['foo'] singleton
Packaging
There is a great example project available officially here: https://github.com/pypa/sampleproject
A python package is a collection of information about your source code, that can inform other tools how to copy your source code to other computers, and how to integrate your source code into that system's path so that import foo works for other computers (regardless of interpreter, host operating system, etc)
Directory Structure
Lets have a package name foo, in some directory (preferably an empty directory).
some_directory/
foo.py # `if __name__ == "__main__":` lives here
My preference is to create setup.py as sibling to foo.py, because it makes writing the setup.py file simpler, however you can write configuration to change/redirect everything setuptools does by default if you like; for example putting foo.py under a "src/" directory is somewhat popular, not covered here.
some_directory/
foo.py
setup.py
.
#!/usr/bin/env python3
# setup.py
import setuptools
setuptools.setup(
name="foo",
...
py_modules=['foo'],
)
.
python3 -m pip install --editable ./ # or path/to/some_directory/
"editable" aka -e will yet-again redirect the importing machinery to load the source files in this directory, instead copying the current exact files to the installing-environment's library. This can also cause behavioral differences on a developer's machine, be sure to test your code!
There are tools other than pip, however I'd recommend pip be the introductory one :)
I also like to make foo a "package" (a directory containing __init__.py) instead of a module (a single ".py" file), both "packages" and "modules" can be loaded into the root namespace, modules allow for nested namespaces, which is helpful if we want to have a "relative one directory up" import.
some_directory/
foo/
__init__.py
setup.py
.
#!/usr/bin/env python3
# setup.py
import setuptools
setuptools.setup(
name="foo",
...
packages=['foo'],
)
I also like to make a foo/__main__.py, this allows python to execute the package as a module, eg python3 -m foo will execute foo/__main__.py as __main__.
some_directory/
foo/
__init__.py
__main__.py # `if __name__ == "__main__":` lives here, `def main():` too!
setup.py
.
#!/usr/bin/env python3
# setup.py
import setuptools
setuptools.setup(
name="foo",
...
packages=['foo'],
...
entry_points={
'console_scripts': [
# "foo" will be added to the installing-environment's text mode shell, eg `bash -c foo`
'foo=foo.__main__:main',
]
},
)
Lets flesh this out with some more modules:
Basically, you can have a directory structure like so:
some_directory/
bar.py # `import bar`
foo/
__init__.py # `import foo`
__main__.py
baz.py # `import foo.baz
spam/
__init__.py # `import foo.spam`
eggs.py # `import foo.spam.eggs`
setup.py
setup.py conventionally holds metadata information about the source code within, such as:
what dependencies are needed to install named "install_requires"
what name should be used for package management (install/uninstall "name"), I suggest this match your primary python package name in our case foo, though substituting underscores for hyphens is popular
licensing information
maturity tags (alpha/beta/etc),
audience tags (for developers, for machine learning, etc),
single-page documentation content (like a README),
shell names (names you type at user shell like bash, or names you find in a graphical user shell like a start menu),
a list of python modules this package will install (and uninstall)
a defacto "run tests" entry point python ./setup.py test
Its very expansive, it can even compile c extensions on the fly if a source module is being installed on a development machine. For a every-day example I recommend the PYPA Sample Repository's setup.py
If you are releasing a build artifact, eg a copy of the code that is meant to run nearly identical computers, a requirements.txt file is a popular way to snapshot exact dependency information, where "install_requires" is a good way to capture minimum and maximum compatible versions. However, given that the target machines are nearly identical anyway, I highly recommend creating a tarball of an entire python prefix. This can be tricky, too detailed to get into here. Check out pip install's --target option, or virtualenv aka venv for leads.
back to the example
how to import a file one directory up:
From foo/spam/eggs.py, if we wanted code from foo/baz we could ask for it by its absolute namespace:
import foo.baz
If we wanted to reserve capability to move eggs.py into some other directory in the future with some other relative baz implementation, we could use a relative import like:
import ..baz
Here's a three-step, somewhat minimalist version of ThorSummoner's answer for the sake of clarity. It doesn't quite do what I want (I'll explain at the bottom), but it works okay.
Step 1: Make directory and setup.py
filepath_to/project_name/
setup.py
In setup.py, write:
import setuptools
setuptools.setup(name='project_name')
Step 2: Install this directory as a package
Run this code in console:
python -m pip install --editable filepath_to/project_name
Instead of python, you may need to use python3 or something, depending on how your python is installed. Also, you can use -e instead of --editable.
Now, your directory will look more or less like this. I don't know what the egg stuff is.
filepath_to/project_name/
setup.py
test_3.egg-info/
dependency_links.txt
PKG-INFO
SOURCES.txt
top_level.txt
This folder is considered a python package and you can import from files in this parent directory even if you're writing a script anywhere else on your computer.
Step 3. Import from above
Let's say you make two files, one in your project's main directory and another in a sub directory. It'll look like this:
filepath_to/project_name/
top_level_file.py
subdirectory/
subfile.py
setup.py |
test_3.egg-info/ |----- Ignore these guys
... |
Now, if top_level_file.py looks like this:
x = 1
Then I can import it from subfile.py, or really any other file anywhere else on your computer.
# subfile.py OR some_other_python_file_somewhere_else.py
import random # This is a standard package that can be imported anywhere.
import top_level_file # Now, top_level_file.py works similarly.
print(top_level_file.x)
This is different than what I was looking for: I hoped python had a one-line way to import from a file above. Instead, I have to treat the script like a module, do a bunch of boilerplate, and install it globally for the entire python installation to have access to it. It's overkill. If anyone has a simpler method than doesn't involve the above process or importlib shenanigans, please let me know.
Polished answer of #alex-martelli with pathlib:
import pathlib
import sys
_parentdir = pathlib.Path(__file__).parent.parent.resolve()
sys.path.insert(0, str(_parentdir))
import module_in_parent_dir
sys.path.remove(str(_parentdir))
To run python /myprogram/submodule/mymodule.py which imports /myprogram/mainmodule.py, e.g., via
from mainmodule import *
on Linux (e.g., in the python Docker image), I had to add the program root directory to PYTHONPATH:
export PYTHONPATH=/myprogram
It is 2022 and none of the answers really worked for me. Here is what worked in the end
import sys
sys.path.append('../my_class')
import my_class
My directory structure:
src
--my_class.py
notebooks
-- mynotebook.ipynb
I imported my_class from mynotebook.ipynb.
You can use the sys.path.append() method to add the directory containing the package to the list of paths searched for modules. For example, if the package is located two directories above the current directory, you can use the following code:
import sys
sys.path.append("../../")
if the package is location one directory above the current directory, you can use below code:
import sys
sys.path.append("..")
Python is a modular system
Python doesn't rely on a file system
To load python code reliably, have that code in a module, and that module installed in python's library.
Installed modules can always be loaded from the top level namespace with import <name>
There is a great sample project available officially here: https://github.com/pypa/sampleproject
Basically, you can have a directory structure like so:
the_foo_project/
setup.py
bar.py # `import bar`
foo/
__init__.py # `import foo`
baz.py # `import foo.baz`
faz/ # `import foo.faz`
__init__.py
daz.py # `import foo.faz.daz` ... etc.
.
Be sure to declare your setuptools.setup() in setup.py,
official example: https://github.com/pypa/sampleproject/blob/master/setup.py
In our case we probably want to export bar.py and foo/__init__.py, my brief example:
setup.py
#!/usr/bin/env python3
import setuptools
setuptools.setup(
...
py_modules=['bar'],
packages=['foo'],
...
entry_points={},
# Note, any changes to your setup.py, like adding to `packages`, or
# changing `entry_points` will require the module to be reinstalled;
# `python3 -m pip install --upgrade --editable ./the_foo_project
)
.
Now we can install our module into the python library;
with pip, you can install the_foo_project into your python library in edit mode,
so we can work on it in real time
python3 -m pip install --editable=./the_foo_project
# if you get a permission error, you can always use
# `pip ... --user` to install in your user python library
.
Now from any python context, we can load our shared py_modules and packages
foo_script.py
#!/usr/bin/env python3
import bar
import foo
print(dir(bar))
print(dir(foo))
Currently trying to work in Python3 and use absolute imports to import one module into another but I get the error ModuleNotFoundError: No module named '__main__.moduleB'; '__main__' is not a package. Consider this project structure:
proj
__init__.py3 (empty)
moduleA.py3
moduleB.py3
moduleA.py3
from .moduleB import ModuleB
ModuleB.hello()
moduleB.py3
class ModuleB:
def hello():
print("hello world")
Then running python3 moduleA.py3 gives the error. What needs to be changed here?
.moduleB is a relative import. Relative only works when the parent module is imported or loaded first. That means you need to have proj imported somewhere in your current runtime environment. When you are are using command python3 moduleA.py3, it is getting no chance to import parent module. You can:
from proj.moduleB import moduleB OR
You can create another script, let's say run.py, to invoke from proj import moduleA
Good luck with your journey to the awesome land of Python.
Foreword
I'm developing a project which in fact is a Python package that can be installed through pip, but it also exposes a command line interface. I don't have problems running my project after installing it with pip install ., but hey, who does this every time after changing something in one of the project files? I needed to run the whole thing through simple python mypackage/main.py.
/my-project
- README.md
- setup.py
/mypackage
- __init__.py
- main.py
- common.py
The different faces of the same problem
I tried importing a few functions in main.py from my common.py module. I tried different configurations that gave different errors, and I want to share with you with my observations and leave a quick note for future me as well.
Relative import
The first what I tried was a relative import:
from .common import my_func
I ran my application with simple: python mypackage/main.py. Unfortunately this gave the following error:
ModuleNotFoundError: No module named '__main__.common'; '__main__' is not a package
The cause of this problem is that the main.py was executed directly by python command, thus becoming the main module named __main__. If we connect this information with the relative import we used, we get what we have in the error message: __main__.common. This is explained in the Python documentation:
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.
When I installed my package with pip install . and then ran it, it worked perfectly fine. I was also able to import mypackage.main module in a Python console. So it looks like there's a problem only with running it directly.
Absolute import
Let's follow the advise from the documentation and change the import statement to something different:
from common import my_func
If we now try to run this as before: python mypackage/main.py, then it works as expected! But, there's a caveat when you, like me, develop something that need to work as a standalone command line tool after installing it with pip. I installed my package with pip install . and then tried to run it...
ModuleNotFoundError: No module named 'common'
What's worse, when I opened a Python console, and tried to import the main module manually (import mypackage.main), then I got the same error as above. The reason for that is simple: common is no longer a relative import, so Python tries to find it in installed packages. We don't have such package, that's why it fails.
The solution with an absolute import works well only when you create a typical Python app that is executed with a python command.
Import with a package name
There is also a third possibility to import the common module:
from mypackage.common import my_func
This is not very different from the relative import approach, as long as we do it from the context of mypackage. And again, trying to run this with python mypackage/main.py ends similar:
ModuleNotFoundError: No module named 'mypackage'
How irritating that could be, the interpreter is right, you don't have such package installed.
The solution
For simple Python apps
Just use absolute imports (without the dot), and everything will be fine.
For installable Python apps in development
Use relative imports, or imports with a package name on the beginning, because you need them like this when your app is installed. When it comes to running such module in development, Python can be executed with the -m option:
-m mod : run library module as a script (terminates option list)
So instead of python mypackage/main.py, do it like this: python -m mypackage.main.
In addition to md-sabuj-sarker's answer, there is a really good example in the Python modules documentation.
This is what the docs say about intra-package-references:
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.
If you run python3 moduleA.py3, moduleA is used as the main module, so using the absolute import looks like the right thing to do.
However, beware that this absolute import (from package.module import something) fails if, for some reason, the package contains a module file with the same name as the package (at least, on my Python 3.7). So, for example, it would fail if you have (using the OP's example):
proj/
__init__.py (empty)
proj.py (same name as package)
moduleA.py
moduleB.py
in which case you would get:
ModuleNotFoundError: No module named 'proj.moduleB'; 'proj' is not a package
Alternatively, you could remove the . in from .moduleB import, as suggested here and here, which seems to work, although my PyCharm (2018.2.4) marks this as an "Unresolved reference" and fails to autocomplete.
Maybe you can do this before importing the module：
moduleA.py3
import os
import sys
sys.path.append(os.path.dirname(os.path.abspath(__file__)))
from moduleB import ModuleB
ModuleB.hello()
Add the current directory to the environment directory
Just rename the file from where you run the app to main.py:
from app import app
if __name__ == '__main__':
app.run()
import os
import sys
sys.path.append(os.path.dirname(os.path.abspath(__file__)))
will solve the issue of import path issue.
I've been coding in Python for long, but never actually tried to pack a piece of code so that I can share it. I started reading
https://python-packaging.readthedocs.io/en/latest/.
I started with the simplest possible case, say I want to share a module named 'clipper', and the only important thing is a class called Clipper. It seems in case I use setuptools I should create somewhere folders
clipper/clipper
and inside the inner clipper, place a file __init__.py
with the definition of class Clipper. So far so good. Theoretically, after installing the package, the way to use the class would be:
import clipper
cl = clipper.Clipper()
My problem is, I am assuming that while I am developing and before any installation, the same code should work. I mean, the previous code should create an instance of the object. But how would that work? How should I set PYTHONPATH so that the previous import would actually work?
Maybe I got something really wrong, I thought packing would be easier compared to coding, but I've spent some time and I don't get it. Any help, please?
Rather than modifying your Python path, install the packaged module as an editable version and your environment will handle this for you. When you have it running as an editable version you'll be able to make changes to the code on your local development instance.
For example, assuming you have Pip you can run the following command in the first 'clipper' folder (the same folder as the setup.py file you created during packaging):
pip install -e .
-e means editable
the . means install the package located in the current folder.
More detail here in an SO answer from 2015: "pip install --editable ./" vs "python setup.py develop"
Your directory tree would look a bit like this:
~/clipper/
setup.py
clipper/
__init__.py
clipper.py
The setup.py file contains information telling Python how to 'parse' your package. Things like the name of your project, the version and what packages to include are defined here. For your example, setup.py may look like this:
from distutils.core import setup
setup(
name="Clipper",
# A name for your package, typically your project name
description="My first package",
# A short description
version="1.0.0",
# A version specification
packages=["clipper"]
# A list of packages to include
)
Within clipper, clipper.py contains the actual Clipper class:
class Clipper(object):
def __init__(self):
pass
def foo(self):
print("Invoked foo!")
__init__.py is a special type of file. It defines the public interface for interacting with your package. Typically, it imports all public functions and classes:
from .clipper import Clipper
Finally, to turn this into a proper package, run python3 setup.py sdist. This creates the source distribution for your package and allows you to import it 1. Let's try that now. Navigate back to ~/clipper/ and start Python:
>>> from clipper import Clipper
>>> c = Clipper()
>>> c.foo()
Invoked foo!
>>>
And here's 'real' example of what a package directory would look like:
~/calculator/
setup.py
calculator/
__init__.py
add.py
substract.py
setup.py
from distutils.core import setup
setup(
name="Calculator",
description="Calculate stuff!",
version="1.0.0",
packages=["calculator"]
)
__init__.py
from .add import *
from .substract import *
add.py
def add(a, b):
"""Return `a` + `b`."""
return a + b
substract.py
def substract(a, b):
"""Return `a` - `b`."""
return a - b
For more information, see the Python tutorial on packaging.
1 You may get some warnings about missing information, but you can ignore that for now.
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?
EDIT: There are better/more coherent answers to this question in other questions:
Sibling package imports
Relative imports for the billionth time
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.
import sys
sys.path.append("..") # Adds higher directory to python modules path.
Try this.
Worked for me.
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.
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.
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:
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
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:
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.
from package.A import foo
I think it's clearer than
import sys
sys.path.append("..")
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())
This is actually a lot simpler than what other answers make it out to be.
TL;DR: Import A directly instead of attempting a relative import.
The current working directory is not a package, unless you import the folder package from a different folder. So the behavior of your package will work fine if you intend it to be imported by other applications. What's not working is the tests...
Without changing anything in your directory structure, all that needs to be changed is how test.py imports foo.py.
from A import foo
Now running python -m test_A.test from the package directory will run without an ImportError.
Why does that work?
Your current working directory is not a package, but it is added to the path. Therefore you can import folder A and its contents directly. It is the same reason you can import any other package that you have installed... they're all included in your path.
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.
Just remove .. in test.py
For me pytest works fine with that
Example:
from A import foo
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
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
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.
[2022-07-19] I think this "relative-import" limitation is quite an ugly design, totally against (one of) Python's motto "Simple is better than complex".
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.
Having
package/
__init__.py
A/
__init__.py
foo.py
test_A/
__init__.py
test.py
in A/__init__.py import foo:
from .foo import foo
when importing A/ from test_A/
import sys, os
sys.path.append(os.path.abspath('../A'))
# then import foo
import foo