How can I create imports that always work?

How can I create imports that always work? - python

I am struggling a bit to set up a working structure in one of my projects. The problem is, that I have main package and a subpackage in a structure like this (I left out all unnecessary files):
code.py
mypackage/__init__.py
mypackage/work.py
mypackage/utils.py
The utils.py has some utility code that is normally only used in the mypackage package.
I normally have some test code each module file, that calls some methods of the current module and prints some things to quickcheck if everything is working correctly. This code is placed in a if __name__ == "__main__" block at the end of the file. So I include the utils.py directly via import utils. E.g mypackage/work.py looks like:
import utils
def someMethod():
pass
if __name__ == "__main__":
print(someMethod())
But now when I use this module in the parent package (e.g. code.py) and I import it like this
import mypackage.work
I get the following error:
ImportError: No module named 'utils'
After some research I found out, that this can be fixed by adding the mypackage/ folder to the PYTHONPATH environment variable, but this feels strange for me. Isn't there any other way to fix this? I have heard about relative imports, but this is mentioned in the python docs about modules
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.
Any suggestions how I can have a if __name__ == "__main__" section in the submodule and also can use this file from the parent package without messing up the imports?
EDIT: If I use a relative import in work.py as suggested in a answer to import utils:
from . import utils
I get the following error:
SystemError: Parent module '' not loaded, cannot perform relative import

Unfortunately relative imports and direct running of submodules don't mix.
Add the parent directory of mypackage to your PYTHONPATH or always cd into the parent directory when you want to run a submodule.
Then you have two possibilities:
Use absolute (from mypackage import utils) instead of relative imports (from . import utils) and run them directly as before. The drawback with that solution is that you'll always need to write the fully qualified path, making it more work to rename mypackage later, among other things.
or
Run python3 -m mypackage.utils etc. to run your submodules instead of running python3 mypackage/utils.py.
This may take some time to adapt to, but it's the more correct way (a module in a package isn't the same as a standalone script) and you can continue to use relative imports.
There are more "magical" solutions involving __package__ and sys.path but they all require extra code at the top of every file with relative imports you want to run directly. I wouldn't recommend these.

You should create a structure like this:
flammi88
├── flammi88
│ ├── __init__.py
│   ├── code.py
│   └── mypackage
│   ├── __init__.py
│   ├── utils.py
│   └── work.py
└── setup.py
then put at least this in the setup.py:
import setuptools
from distutils.core import setup
setup(
name='flammi88',
packages=['flammi88'],
)
now, from the directory containing setup.py, run
pip install -e .
This will make the flammi88 package available in development mode. Now you can say:
from flammi88.mypackage import utils
everywhere. This is the normal way to develop packages in Python, and it solves all of your relative import problems. That being said, Guido frowns upon standalone scripts in sub-packages. With this structure I would move the tests inside flammi88/tests/... (and run them with e.g. py.test), but some people like to keep the tests next to the code.
Update:
an extended setup.py that describes external requirements and creates executables of the sub-packages you want to run can look like:
import setuptools
from distutils.core import setup
setup(
name='flammi88',
packages=['flammi88'],
install_requires=[
'markdown',
],
entry_points={
'console_scripts': [
'work = flammi88.mypackage.work:someMethod',
]
}
)
now, after pip installing your package, you can just type work at the command line.

Import utils inside the work.py as follows:
import mypackage.utils
or if you want to use shorter name:
from mypackage import utils
EDIT: If you need to run work.py outside of the package, then try:
try:
from mypackage import utils
except ImportError:
import utils

Use:
from . import utils
as suggested by Peter
In your code.py you should use:
from mypackage import work

Related

Can't Create Python Project Tests via Hitchhiker's Guide To Python Method

I'm trying to build a project via the method described in The Hitchhiker's Guide To Python. I'm running into problems with the test structure.
My file structure looks like this:
.
├── __init__.py
├── sample
│   ├── __init__.py
│   └── core.py
├── setup.py
└── tests
├── __init__.py
├── context.py
└── test_core.py
With:
# sample/core.py
class SampleClass:
def got_it(self):
return True
And:
# tests/context.py
import os
import sys
sys.path.insert(0, os.path.abspath(
os.path.join(os.path.dirname(__file__), '..')
))
import sample
And:
# tests/test_core.py
import unittest
from .context import sample
class SampleClassTest(unittest.TestCase):
def test_got_it(self):
# ...
pass
if __name__ == '__main__':
unittest.main()
(Note that I just thru the __init__.py files in the root and tests to see if that helped, but it didn't.)
When I try to run tests/test_core.py with Python 3.7. I get this error:
ImportError: attempted relative import with no known parent package
That happens if I run the test file from outside the tests directory or in it.
If I remote the . and do this in tests/test_core.py:
from context import sample
Everything loads, but I can't access SampleClass.
The things I've tried are:
sc = SampleClass()
NameError: name 'SampleClass' is not defined
sc = sample.SampleClass()
AttributeError: module 'sample' has no attribute 'SampleClass'
sc = sample.core.SampleClass()
AttributeError: module 'sample' has no attribute 'core'
sc = core.SampleClass()
NameError: name 'core' is not defined
I tried on Python 2 as well and had similar problems (with slightly different error methods). I also tried calling a function instead of a class and had similar problems there as well.
Can someone point me in the direction of what I'm doing wrong?

This once caused me some headaches, until I realized that it just depended on the way the test is started. If you use python tests/test_core.py, then you are starting a simple script outside any package. So relative imports are forbidden.
But if (still from the project root folder) you use:
python -m tests.test_core
then you are starting the module test_core from the tests package, and relative imports are allowed.
From that time, I always start my tests as modules and not as scripts. What is even better is that unittest discover knows about packages. So if you are consistent in naming the test folder and scripts with a initial test, you can forget about the number of tests and just use:
python -m unittest discover
and that is enough to run the whole test suite.
Happy testing!

Since you have a proper package complete with setup.py, you're better off using pip install -e . instead of using a sys.path hack.
Minimal setup.py
#setup.py
from setuptools import setup, find_packages
setup(name='sample', version='0.0.1', packages=find_packages(exclude=('tests')))
Then you should import everything from sample packages anywhere you want using the same syntax (i.e. absolute path). To import SampleClass for example:
from sample.core import SampleClass
Also, to answer your question on why you're getting on relative import:
ImportError: attempted relative import with no known parent package
You'd have to avoid invoking the file as a script using python ... because relative imports do not work on scripts. You should invoke the file as a module using python -m .... Here is a very nice explanation that's been viewed a billion times:

Importing from above path on python for windows

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.

How to import from a python package which is in a sibling directory?

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

Import Failing in Python3, but not 2 [duplicate]

I want to import a function from another file in the same directory.
Usually, one of the following works:
from .mymodule import myfunction
from mymodule import myfunction
...but the other one gives me one of these errors:
ImportError: attempted relative import with no known parent package
ModuleNotFoundError: No module named 'mymodule'
SystemError: Parent module '' not loaded, cannot perform relative import
Why is this?

unfortunately, this module needs to be inside the package, and it also
needs to be runnable as a script, sometimes. Any idea how I could
achieve that?
It's quite common to have a layout like this...
main.py
mypackage/
__init__.py
mymodule.py
myothermodule.py
...with a mymodule.py like this...
#!/usr/bin/env python3
# Exported function
def as_int(a):
return int(a)
# Test function for module
def _test():
assert as_int('1') == 1
if __name__ == '__main__':
_test()
...a myothermodule.py like this...
#!/usr/bin/env python3
from .mymodule import as_int
# Exported function
def add(a, b):
return as_int(a) + as_int(b)
# Test function for module
def _test():
assert add('1', '1') == 2
if __name__ == '__main__':
_test()
...and a main.py like this...
#!/usr/bin/env python3
from mypackage.myothermodule import add
def main():
print(add('1', '1'))
if __name__ == '__main__':
main()
...which works fine when you run main.py or mypackage/mymodule.py, but fails with mypackage/myothermodule.py, due to the relative import...
from .mymodule import as_int
The way you're supposed to run it is...
python3 -m mypackage.myothermodule
...but it's somewhat verbose, and doesn't mix well with a shebang line like #!/usr/bin/env python3.
The simplest fix for this case, assuming the name mymodule is globally unique, would be to avoid using relative imports, and just use...
from mymodule import as_int
...although, if it's not unique, or your package structure is more complex, you'll need to include the directory containing your package directory in PYTHONPATH, and do it like this...
from mypackage.mymodule import as_int
...or if you want it to work "out of the box", you can frob the PYTHONPATH in code first with this...
import sys
import os
SCRIPT_DIR = os.path.dirname(os.path.abspath(__file__))
sys.path.append(os.path.dirname(SCRIPT_DIR))
from mypackage.mymodule import as_int
It's kind of a pain, but there's a clue as to why in an email written by a certain Guido van Rossum...
I'm -1 on this and on any other proposed twiddlings of the __main__
machinery. The only use case seems to be running scripts that happen
to be living inside a module's directory, which I've always seen as an
antipattern. To make me change my mind you'd have to convince me that
it isn't.
Whether running scripts inside a package is an antipattern or not is subjective, but personally I find it really useful in a package I have which contains some custom wxPython widgets, so I can run the script for any of the source files to display a wx.Frame containing only that widget for testing purposes.

Explanation
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.
At some point PEP 338 conflicted with PEP 328:
... relative imports rely on __name__ to determine the current
module's position in the package hierarchy. In a main module, the
value of __name__ is always '__main__', so explicit relative imports
will always fail (as they only work for a module inside a package)
and to address the issue, PEP 366 introduced the top level variable __package__:
By adding a new module level attribute, this PEP allows relative
imports to work automatically if the module is executed using the -m
switch. A small amount of boilerplate in the module itself will allow
the relative imports to work when the file is executed by name. [...] When it [the attribute] is present, relative imports will be based on this attribute
rather than the module __name__ attribute. [...] When the main module is specified by its filename, then the __package__ attribute will be set to None. [...] When the import system encounters an explicit relative import in a
module without __package__ set (or with it set to None), it will
calculate and store the correct value (__name__.rpartition('.')[0]
for normal modules and __name__ for package initialisation modules)
(emphasis mine)
If the __name__ is '__main__', __name__.rpartition('.')[0] returns empty string. This is why there's empty string literal in the error description:
SystemError: Parent module '' not loaded, cannot perform relative import
The relevant part of the CPython's PyImport_ImportModuleLevelObject function:
if (PyDict_GetItem(interp->modules, package) == NULL) {
PyErr_Format(PyExc_SystemError,
"Parent module %R not loaded, cannot perform relative "
"import", package);
goto error;
}
CPython raises this exception if it was unable to find package (the name of the package) in interp->modules (accessible as sys.modules). Since sys.modules is "a dictionary that maps module names to modules which have already been loaded", it's now clear that the parent module must be explicitly absolute-imported before performing relative import.
Note: The patch from the issue 18018 has added another if block, which will be executed before the code above:
if (PyUnicode_CompareWithASCIIString(package, "") == 0) {
PyErr_SetString(PyExc_ImportError,
"attempted relative import with no known parent package");
goto error;
} /* else if (PyDict_GetItem(interp->modules, package) == NULL) {
...
*/
If package (same as above) is empty string, the error message will be
ImportError: attempted relative import with no known parent package
However, you will only see this in Python 3.6 or newer.
Solution #1: Run your script using -m
Consider a directory (which is a Python package):
.
├── package
│   ├── __init__.py
│   ├── module.py
│   └── standalone.py
All of the files in package begin with the same 2 lines of code:
from pathlib import Path
print('Running' if __name__ == '__main__' else 'Importing', Path(__file__).resolve())
I'm including these two lines only to make the order of operations obvious. We can ignore them completely, since they don't affect the execution.
__init__.py and module.py contain only those two lines (i.e., they are effectively empty).
standalone.py additionally attempts to import module.py via relative import:
from . import module # explicit relative import
We're well aware that /path/to/python/interpreter package/standalone.py will fail. However, we can run the module with the -m command line option that will "search sys.path for the named module and execute its contents as the __main__ module":
vaultah#base:~$ python3 -i -m package.standalone
Importing /home/vaultah/package/__init__.py
Running /home/vaultah/package/standalone.py
Importing /home/vaultah/package/module.py
>>> __file__
'/home/vaultah/package/standalone.py'
>>> __package__
'package'
>>> # The __package__ has been correctly set and module.py has been imported.
... # What's inside sys.modules?
... import sys
>>> sys.modules['__main__']
<module 'package.standalone' from '/home/vaultah/package/standalone.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/package/module.py'>
>>> sys.modules['package']
<module 'package' from '/home/vaultah/package/__init__.py'>
-m does all the importing stuff for you and automatically sets __package__, but you can do that yourself in the
Solution #2: Set __package__ manually
Please treat it as a proof of concept rather than an actual solution. It isn't well-suited for use in real-world code.
PEP 366 has a workaround to this problem, however, it's incomplete, because setting __package__ alone is not enough. You're going to need to import at least N preceding packages in the module hierarchy, where N is the number of parent directories (relative to the directory of the script) that will be searched for the module being imported.
Thus,
Add the parent directory of the Nth predecessor of the current module to sys.path
Remove the current file's directory from sys.path
Import the parent module of the current module using its fully-qualified name
Set __package__ to the fully-qualified name from 2
Perform the relative import
I'll borrow files from the Solution #1 and add some more subpackages:
package
├── __init__.py
├── module.py
└── subpackage
├── __init__.py
└── subsubpackage
├── __init__.py
└── standalone.py
This time standalone.py will import module.py from the package package using the following relative import
from ... import module # N = 3
We'll need to precede that line with the boilerplate code, to make it work.
import sys
from pathlib import Path
if __name__ == '__main__' and __package__ is None:
file = Path(__file__).resolve()
parent, top = file.parent, file.parents[3]
sys.path.append(str(top))
try:
sys.path.remove(str(parent))
except ValueError: # Already removed
pass
import package.subpackage.subsubpackage
__package__ = 'package.subpackage.subsubpackage'
from ... import module # N = 3
It allows us to execute standalone.py by filename:
vaultah#base:~$ python3 package/subpackage/subsubpackage/standalone.py
Running /home/vaultah/package/subpackage/subsubpackage/standalone.py
Importing /home/vaultah/package/__init__.py
Importing /home/vaultah/package/subpackage/__init__.py
Importing /home/vaultah/package/subpackage/subsubpackage/__init__.py
Importing /home/vaultah/package/module.py
A more general solution wrapped in a function can be found here. Example usage:
if __name__ == '__main__' and __package__ is None:
import_parents(level=3) # N = 3
from ... import module
from ...module.submodule import thing
Solution #3: Use absolute imports and setuptools
The steps are -
Replace explicit relative imports with equivalent absolute imports
Install package to make it importable
For instance, the directory structure may be as follows
.
├── project
│   ├── package
│   │   ├── __init__.py
│   │   ├── module.py
│   │   └── standalone.py
│   └── setup.py
where setup.py is
from setuptools import setup, find_packages
setup(
name = 'your_package_name',
packages = find_packages(),
)
The rest of the files were borrowed from the Solution #1.
Installation will allow you to import the package regardless of your working directory (assuming there'll be no naming issues).
We can modify standalone.py to use this advantage (step 1):
from package import module # absolute import
Change your working directory to project and run /path/to/python/interpreter setup.py install --user (--user installs the package in your site-packages directory) (step 2):
vaultah#base:~$ cd project
vaultah#base:~/project$ python3 setup.py install --user
Let's verify that it's now possible to run standalone.py as a script:
vaultah#base:~/project$ python3 -i package/standalone.py
Running /home/vaultah/project/package/standalone.py
Importing /home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/__init__.py
Importing /home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py
>>> module
<module 'package.module' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py'>
>>> import sys
>>> sys.modules['package']
<module 'package' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/__init__.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py'>
Note: If you decide to go down this route, you'd be better off using virtual environments to install packages in isolation.
Solution #4: Use absolute imports and some boilerplate code
Frankly, the installation is not necessary - you could add some boilerplate code to your script to make absolute imports work.
I'm going to borrow files from Solution #1 and change standalone.py:
Add the parent directory of package to sys.path before attempting to import anything from package using absolute imports:
import sys
from pathlib import Path # if you haven't already done so
file = Path(__file__).resolve()
parent, root = file.parent, file.parents[1]
sys.path.append(str(root))
# Additionally remove the current file's directory from sys.path
try:
sys.path.remove(str(parent))
except ValueError: # Already removed
pass
Replace the relative import by the absolute import:
from package import module # absolute import
standalone.py runs without problems:
vaultah#base:~$ python3 -i package/standalone.py
Running /home/vaultah/package/standalone.py
Importing /home/vaultah/package/__init__.py
Importing /home/vaultah/package/module.py
>>> module
<module 'package.module' from '/home/vaultah/package/module.py'>
>>> import sys
>>> sys.modules['package']
<module 'package' from '/home/vaultah/package/__init__.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/package/module.py'>
I feel that I should warn you: try not to do this, especially if your project has a complex structure.
As a side note, PEP 8 recommends the use of absolute imports, but states that in some scenarios explicit relative imports are acceptable:
Absolute imports are recommended, as they are usually more readable
and tend to be better behaved (or at least give better error
messages). [...] However, explicit relative imports are an acceptable
alternative to absolute imports, especially when dealing with complex
package layouts where using absolute imports would be unnecessarily
verbose.

Put this inside your package's __init__.py file:
# For relative imports to work in Python 3.6
import os, sys; sys.path.append(os.path.dirname(os.path.realpath(__file__)))
Assuming your package is like this:
├── project
│ ├── package
│ │ ├── __init__.py
│ │ ├── module1.py
│ │ └── module2.py
│ └── setup.py
Now use regular imports in you package, like:
# in module2.py
from module1 import class1
This works in both python 2 and 3.

I ran into this issue. A hack workaround is importing via an if/else block like follows:
#!/usr/bin/env python3
#myothermodule
if __name__ == '__main__':
from mymodule import as_int
else:
from .mymodule import as_int
# Exported function
def add(a, b):
return as_int(a) + as_int(b)
# Test function for module
def _test():
assert add('1', '1') == 2
if __name__ == '__main__':
_test()

SystemError: Parent module '' not loaded, cannot perform relative import
This means you are running a module inside the package as a script. Mixing scripts inside packages is tricky and should be avoided if at all possible. Use a wrapper script that imports the package and runs your scripty function instead.
If your top-level directory is called foo, which is on your PYTHONPATH module search path, and you have a package bar there (it is a directory you'd expect an __init__.py file in), scripts should not be placed inside bar, but should live on in foo at best.
Note that scripts differ from modules here in that they are used as a filename argument to the python command, either by using python <filename> or via a #! (shebang) line. It is loaded directly as the __main__ module (this is why if __name__ == "__main__": works in scripts), and there is no package context to build on for relative imports.
Your options
If you can, package your project with setuptools (or poetry or flit, which can help simplify packaging), and create console script entrypoints; installing your project with pip then creates scripts that know how to import your package properly. You can install your package locally with pip install -e ., so it can still be edited in-place.
Otherwise, never, ever, use python path/to/packagename/file.py, always use python path/to/script.py and script.py can use from packagename import ....
As a fallback, you could use the -m command-line switch to tell Python to import a module and use that as the __main__ file instead. This does not work with a shebang line, as there is no script file any more, however.
If you use python -m foo.bar and foo/bar.py is found in a sys.path directory, that is then imported and executed as __main__ with the right package context. If bar is also a package, inside foo/, it must have a __main__.py file (so foo/bar/__main__.py as the path from the sys.path directory).
In extreme circumstances, add the metadata Python uses to resolve relative imports by setting __package__ directly; the file foo/bar/spam.py, importable as foo.bar.spam, is given the global __package__ = "foo.bar". It is just another global, like __file__ and __name__, set by Python when imported.
On sys.path
The above all requires that your package can be imported, which means it needs to be found in one of the directories (or zipfiles) listed in sys.path. There are several options here too:
The directory where path/to/script.py was found (so path/to) is automatically added to sys.path. Executing python path/to/foo.py adds path/to to sys.path.
If you packaged your project (with setuptools, poetry, flit or another Python packaging tool), and installed it, the package has been added to the right place already.
As a last resort, add the right directory to sys.path yourself. If the package can be located relatively to the script file, use the __file__ variable in the script global namespace (e.g. using the pathlib.Path object, HERE = Path(__file__).resolve().parent is a reference to the directory the file lives in, as absolute path).

For PyCharm users:
I also was getting ImportError: attempted relative import with no known parent package because I was adding the . notation to silence a PyCharm parsing error. PyCharm innaccurately reports not being able to find:
lib.thing import function
If you change it to:
.lib.thing import function
it silences the error but then you get the aforementioned ImportError: attempted relative import with no known parent package. Just ignore PyCharm's parser. It's wrong and the code runs fine despite what it says.

To obviate this problem, I devised a solution with the repackage package, which has worked for me for some time. It adds the upper directory to the lib path:
import repackage
repackage.up()
from mypackage.mymodule import myfunction
Repackage can make relative imports that work in a wide range of cases, using an intelligent strategy (inspecting the call stack).

TL;DR: to #Aya's answer, updated with pathlib library, and working for Jupyter notebooks where __file__ is not defined:
You want to import my_function defined under ../my_Folder_where_the_package_lives/my_package.py
respect to where you are writing the code.
Then do:
import os
import sys
import pathlib
PACKAGE_PARENT = pathlib.Path(__file__).parent
#PACKAGE_PARENT = pathlib.Path.cwd().parent # if on jupyter notebook
SCRIPT_DIR = PACKAGE_PARENT / "my_Folder_where_the_package_lives"
sys.path.append(str(SCRIPT_DIR))
from my_package import my_function

Hopefully, this will be of value to someone out there - I went through half a dozen stackoverflow posts trying to figure out relative imports similar to whats posted above here. I set up everything as suggested but I was still hitting ModuleNotFoundError: No module named 'my_module_name'
Since I was just developing locally and playing around, I hadn't created/run a setup.py file. I also hadn't apparently set my PYTHONPATH.
I realized that when I ran my code as I had been when the tests were in the same directory as the module, I couldn't find my module:
$ python3 test/my_module/module_test.py 2.4.0
Traceback (most recent call last):
File "test/my_module/module_test.py", line 6, in <module>
from my_module.module import *
ModuleNotFoundError: No module named 'my_module'
However, when I explicitly specified the path things started to work:
$ PYTHONPATH=. python3 test/my_module/module_test.py 2.4.0
...........
----------------------------------------------------------------------
Ran 11 tests in 0.001s
OK
So, in the event that anyone has tried a few suggestions, believes their code is structured correctly and still finds themselves in a similar situation as myself try either of the following if you don't export the current directory to your PYTHONPATH:
Run your code and explicitly include the path like so:
$ PYTHONPATH=. python3 test/my_module/module_test.py
To avoid calling PYTHONPATH=., create a setup.py file with contents like the following and run python setup.py development to add packages to the path:
# setup.py
from setuptools import setup, find_packages
setup(
name='sample',
packages=find_packages()
)

TL;DR
You can only relatively import modules inside another module in the same package.
Concept Clarify
We see a lot of example code in books/docs/articles, they show us how to relatively import a module, but when we do so, it fails.
The reason is, put it in a simple sentence, we did not run the code as the python module mechanism expects, even though the code is written totally right. It's like some kind of runtime thing.
Module loading is depended on how you run the code. That is the source of confusion.
What is a module?
A module is a python file when and only when it is being imported by another file. Given a file mod.py, is it a module? Yes and No, if you run python mod.py, it is not a module, because it is not imported.
What is a package?
A package is a folder that includes Python module(s).
BTW, __init__.py is not necessary from python 3.3, if you don't need any package initialization or auto-load submodules. You don't need to place a blank __init__.py in a directory.
That proves a package is just a folder as long as there are files being imported.
Real Answer
Now, this description becomes clearer.
You can only relatively import modules inside another module in the same package.
Given a directory:
. CWD
|-- happy_maker.py # content: print('Sends Happy')
`-- me.py # content: from . import happy_maker
Run python me.py, we got attempted relative import with no known parent package
me.py is run directly, it is not a module, and we can't use relative import in it.
Solution 1
Use import happy_maker instead of from . import happy_maker
Solution 2
Switch our working directory to the parent folder.
. CWD
|-- happy
| |-- happy_maker.py
`-- me.py
Run python -m happy.me.
When we are in the directory that includes happy, happy is a package, me.py, happy_maker.py are modules, we can use relative import now, and we still want to run me.py, so we use -m which means run the module as a script.
Python Idiom
. CWD
|-- happy
| |-- happy_maker.py # content: print('Sends Happy')
| `-- me.py # content: from . import happy_maker
`-- main.py # content: import happy.me
This structure is the python idiom. main is our script, best practice in Python. Finally, we got there.
Siblings or Grandparents
Another common need:
.
|-- happy
| |-- happy_maker.py
| `-- me.py
`-- sad
`-- sad_maker.py
We want to import sad_maker in me.py, How to do that?
First, we need to make happy and sad in the same package, so we have to go up a directory level. And then from ..sad import sad_maker in the me.py.
That is all.

My boilerplate to make a module with relative imports in a package runnable standalone.
package/module.py
## Standalone boilerplate before relative imports
if __package__ is None:
DIR = Path(__file__).resolve().parent
sys.path.insert(0, str(DIR.parent))
__package__ = DIR.name
from . import variable_in__init__py
from . import other_module_in_package
...
Now you can use your module in any fashion:
Run module as usual: python -m package.module
Use it as a module: python -c 'from package import module'
Run it standalone: python package/module.py
or with shebang (#!/bin/env python) just: package/module.py
NB! Using sys.path.append instead of sys.path.insert will give you a hard to trace error if your module has the same name as your package. E.g. my_script/my_script.py
Of course if you have relative imports from higher levels in your package hierarchy, than this is not enough, but for most cases, it's just okay.

I needed to run python3 from the main project directory to make it work.
For example, if the project has the following structure:
project_demo/
├── main.py
├── some_package/
│ ├── __init__.py
│ └── project_configs.py
└── test/
└── test_project_configs.py
Solution
I would run python3 inside folder project_demo/ and then perform a
from some_package import project_configs

I was getting this ImportError: attempted relative import with no known parent package
In my program I was using the file from current path for importing its function.
from .filename import function
Then I modified the current path (Dot) with package name. Which resolved my issue.
from package_name.filename import function
I hope the above answer helps you.

Importing from same directory
Firstly, you can import from the same directory.
Here is the file structure...
Folder
|
├─ Scripts
| ├─ module123.py
|
├─ main.py
├─ script123.py
Here is main.py
from . import script123
from Scripts import module123
As you can see, importing from . imports from current directory.
Note: if running using anything but IDLE, make sure that your terminal is navigated to the same directory as the main.py file before running.
Also, importing from a local folder also works.
Importing from parent directory
As seen in my GitHub gist here, there is the following method.
Take the following file tree...
ParentDirectory
├─ Folder
| |
| ├─ Scripts
| | ├─ module123.py
| |
| ├─ main.py
| ├─ script123.py
|
├─ parentModule.py
Then, just add this code to the top of your main.py file.
import inspect
import os
import sys
current_dir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
parent_dir = os.path.dirname(current_dir)
sys.path.insert(0, parent_dir)
from ParentDirectory import Stuff

I tried all of the above to no avail, only to realize I mistakenly had a - in my package name.
In short, don't have - in the directory where __init__.py is. I've never felt elated after finding out such inanity.

if both packages are in your import path (sys.path), and the module/class you want is in example/example.py, then to access the class without relative import try:
from example.example import fkt

If none of the above worked for you, you can specify the module explicitly.
Directory:
├── Project
│ ├── Dir
│ │ ├── __init__.py
│ │ ├── module.py
│ │ └── standalone.py
Solution:
#in standalone.py
from Project.Dir.module import ...
module - the module to be imported

Here is a three-liner for those who disagree with Guido:
import sys
from pathlib import Path
sys.path.append(str(Path(sys.argv[0]).absolute().parent.parent))
Hope it helps.

I think the best solution is to create a package for your module:
Here is more info on how to do it.
Once you have a package you don't need to worry about relative import, you can just do absolute imports.

I encounter this a lot when I am working with Django, since a lot of functionality is performed from the manage.py script but I also want to have some of my modules runnable directly as scripts as well (ideally you would make them manage.py directives but we're not there yet).
This is a mock up of what such a project might look like;
├── dj_app
│   ├── models.py
│   ├── ops
│   │   ├── bar.py
│   │   └── foo.py
│   ├── script.py
│   ├── tests.py
│   ├── utils.py
│   └── views.py
└── manage.py
The important parts here being manage.py, dj_app/script.py, and dj_app/tests.py. We also have submodules dj_app/ops/bar.py and dj_app/ops/foo.py which contain more items we want to use throughout the project.
The source of the issue commonly comes from wanting your dj_app/script.py script methods to have test cases in dj_app/tests.py which get invoked when you run manage.py test.
This is how I set up the project and its imports;
# dj_app/ops/foo.py
# Foo operation methods and classes
foo_val = "foo123"
.
# dj_app/ops/bar.py
# Bar operations methods and classes
bar_val = "bar123"
.
# dj_app/script.py
# script to run app methods from CLI
# if run directly from command line
if __name__ == '__main__':
from ops.bar import bar_val
from ops.foo import foo_val
# otherwise
else:
from .ops.bar import bar_val
from .ops.foo import foo_val
def script_method1():
print("this is script_method1")
print("bar_val: {}".format(bar_val))
print("foo_val: {}".format(foo_val))
if __name__ == '__main__':
print("running from the script")
script_method1()
.
# dj_app/tests.py
# test cases for the app
# do not run this directly from CLI or the imports will break
from .script import script_method1
from .ops.bar import bar_val
from .ops.foo import foo_val
def main():
print("Running the test case")
print("testing script method")
script_method1()
if __name__ == '__main__':
print("running tests from command line")
main()
.
# manage.py
# just run the test cases for this example
import dj_app.tests
dj_app.tests.main()
.
Running the test cases from manage.py;
$ python3 manage.py
Running the test case
testing script method
this is script_method1
bar_val: bar123
foo_val: foo123
Running the script on its own;
$ python3 dj_app/script.py
running from the script
this is script_method1
bar_val: bar123
foo_val: foo123
Note that you get an error if you try to run the test.py directly however, so don't do that;
$ python3 dj_app/tests.py
Traceback (most recent call last):
File "dj_app/tests.py", line 5, in <module>
from .script import script_method1
ModuleNotFoundError: No module named '__main__.script'; '__main__' is not a package
If I run into more complicated situations for imports, I usually end up implementing something like this to hack through it;
import os
import sys
THIS_DIR = os.path.dirname(os.path.realpath(__file__))
sys.path.insert(0, THIS_DIR)
from script import script_method1
sys.path.pop(0)

This my project structure
├── folder
| |
│ ├── moduleA.py
| | |
| | └--function1()
| | └~~ uses function2()
| |
│ └── moduleB.py
| |
| └--function2()
|
└── main.py
└~~ uses function1()
Here my moduleA imports moduleB and main imports moduleA
I added the snippet below in moduleA to import moduleB
try:
from .moduleB import function2
except:
from moduleB import function2
Now I can execute both main.py as well as moduleA.py individually
Is this a solution ?

The below solution is tested on Python3
├── classes
| |
| ├──__init__.py
| |
│ ├── userclass.py
| | |
| | └--viewDetails()
| |
| |
│ └── groupclass.py
| |
| └--viewGroupDetails()
|
└── start.py
└~~ uses function1()
Now, in order to use viewDetails of userclass or viewGroupDetails of groupclass define that in _ init _.py of classess directory first.
Ex: In _ init _.py
from .userclasss import viewDetails
from .groupclass import viewGroupDetails
Step2: Now, in start.py we can directly import viewDetails
Ex: In start.py
from classes import viewDetails
from classes import viewGroupDetails

I ran into a similar problem when trying to write a python file that can be loaded either as a module or an executable script.
Setup
/path/to/project/
├── __init__.py
└── main.py
└── mylib/
├── list_util.py
└── args_util.py
with:
main.py:
#!/usr/bin/env python3
import sys
import mylib.args_util
if __name__ == '__main__':
print(f'{mylib.args_util.parseargs(sys.argv[1:])=}')
mylib/list_util.py:
def to_int_list(args):
return [int(x) for x in args]
mylib/args_util.py:
#!/usr/bin/env python3
import sys
from . import list_util as lu
def parseargs(args):
return sum(lu.to_int_list(args))
if __name__ == '__main__':
print(f'{parseargs(sys.argv[1:])=}')
Output
$ ./main.py 1 2 3
mylib.args_util.parseargs(sys.argv[1:])=6
$ mylib/args_util.py 1 2 3
Traceback (most recent call last):
File "/path/to/project/mylib/args_util.py", line 10, in <module>
from . import list_util as lu
ImportError: attempted relative import with no known parent package
Solution
I settled for a Bash/Python polyglot solution. The Bash version of the program just calls python3 -m mylib.args_util then exits.
The Python version ignores the Bash code because it's contained in the docstring.
The Bash version ignores the Python code because it uses exec to stop parsing/running lines.
mylib/args_util.py:
#!/bin/bash
# -*- Mode: python -*-
''''true
exec /usr/bin/env python3 -m mylib.args_util "$#"
'''
import sys
from . import list_util as lu
def parseargs(args):
return sum(lu.to_int_list(args))
if __name__ == '__main__':
print(f'{parseargs(sys.argv[1:])=}')
Output
$ ./main.py 1 2 3
mylib.args_util.parseargs(sys.argv[1:])=6
$ mylib/args_util.py 1 2 3
parseargs(sys.argv[1:])=6
Explanation
Line 1: #!/bin/bash; this is the "shebang" line; it tells the interactive shell how run this script.
Python: ignored (comment)
Bash: ignored (comment)
Line 2: # -*- Mode: python -*- optional; this is called the "mode-line"; it tells Emacs to use Python syntax highlighting instead of guessing that the language is Bash when reading the file.
Python: ignored (comment)
Bash: ignored (comment)
Line 3: ''''true
Python: views this as an unassigned docstring starting with 'true\n
Bash: views this as three strings (of which the first two are empty strings) that expand to true (i.e. '' + '' + 'true' = 'true'); it then runs true (which does nothing) and continues to the next line
Line 4: exec /usr/bin/env python3 -m mylib.args_util "$#"
Python: still views this as part of the docstring from line 3.
Bash: runs python3 -m mylib.args_util then exits (it doesn't parse anything beyond this line)
Line 5: '''
Python: views this as the end of the docstring from line 3.
Bash: doesn't parse this line
Caveats
This doesn't work on Windows:
Workaround: Use WSL or a Batch wrapper script to call python -m mylib.args_util.
This only works if the current working directory is set to /path/to/project/.
Workaround: Set PYTHONPATH when calling /usr/bin/env
#!/bin/bash
# -*- Mode: python -*-
''''true
exec /usr/bin/env python3 \
PYTHONPATH="$(cd "$(dirname "$0")/.." ; pwd)" \
-m mylib.args_util "$#"
'''

I've created a new, experimental import library for Python: ultraimport
It gives the programmer more control over imports and makes them unambiguous. Also it gives better error messages when an import fails.
It allows you to do relative, file-system based imports that always work, no matter how you run your code and no matter what is your current working directory. It does not matter if you run a script or module. You also don't have to change sys.path which might have other side effects.
You would then change
from .mymodule import myfunction
to
import ultraimport
myfunction = ultraimport('__dir__/mymodule.py', 'myfunction')
This way the import will always work, even if you run the code as script.
One issue when importing scripts like this is that subsequent relative imports might fail. ultraimport has a builtin preprocessor to automatically rewrite relative imports.

I had a similar problem: I needed a Linux service and cgi plugin which use common constants to cooperate. The 'natural' way to do this is to place them in the init.py of the package, but I cannot start the cgi plugin with the -m parameter.
My final solution was similar to Solution #2 above:
import sys
import pathlib as p
import importlib
pp = p.Path(sys.argv[0])
pack = pp.resolve().parent
pkg = importlib.import_module('__init__', package=str(pack))
The disadvantage is that you must prefix the constants (or common functions) with pkg:
print(pkg.Glob)

TLDR; Append Script path to the System Path by adding following in the entry point of your python script.
import os.path
import sys
PACKAGE_PARENT = '..'
SCRIPT_DIR = os.path.dirname(os.path.realpath(os.path.join(os.getcwd(), os.path.expanduser(__file__))))
sys.path.append(os.path.normpath(os.path.join(SCRIPT_DIR, PACKAGE_PARENT)))
Thats it now you can run your project in PyCharma as well as from Terminal!!

Moving the file from which you are importing to an outside directory helps.
This is extra useful when your main file makes any other files in its own directory.
Ex:
Before:
Project
|---dir1
|-------main.py
|-------module1.py
After:
Project
|---module1.py
|---dir1
|-------main.py

I was getting the same error and my project structure was like
->project
->vendors
->vendors.py
->main.py
I was trying to call like this
from .vendors.Amazon import Amazom_Purchase
Here it was throwing an error so I fixed it simply by removing the first . from the statement
from vendors.Amazon import Amazom_Purchase
Hope this helps.

It's good to note that sometimes the cache causes of all it - I've tried different things after re-arranging classes into new directories and relative import started to work after I removed the __pycache__

If the following import:
from . import something
doesn't work for you it is because this is python-packaging import and will not work with your regular implementation, and here is an example to show how to use it:
Folder structure:
.
└── funniest
├── funniest
│ ├── __init__.py
│ └── text.py
├── main.py
└── setup.py
inside __init__.py add:
def available_module():
return "hello world"
text.py add:
from . import available_module
inside setup.py add
from setuptools import setup
setup(name='funniest',
version='0.1',
description='The funniest joke in the world',
url='http://github.com/storborg/funniest',
author='Flying Circus',
author_email='flyingcircus#example.com',
license='MIT',
packages=['funniest'],
zip_safe=False)
Now, this is the most important part you need to install your package:
pip install .
Anywhere else in our system using the same Python, we can do this now:
>> import funnies.text as fun
>> fun.available_module()
This should output 'hello world'
you can test this in main.py (this will not require any installation of the Package)
Here is main.py as well
import funniest.text as fun
print(fun.available_module())

Relative imports in Python 3

I want to import a function from another file in the same directory.
Usually, one of the following works:
from .mymodule import myfunction
from mymodule import myfunction
...but the other one gives me one of these errors:
ImportError: attempted relative import with no known parent package
ModuleNotFoundError: No module named 'mymodule'
SystemError: Parent module '' not loaded, cannot perform relative import
Why is this?

unfortunately, this module needs to be inside the package, and it also
needs to be runnable as a script, sometimes. Any idea how I could
achieve that?
It's quite common to have a layout like this...
main.py
mypackage/
__init__.py
mymodule.py
myothermodule.py
...with a mymodule.py like this...
#!/usr/bin/env python3
# Exported function
def as_int(a):
return int(a)
# Test function for module
def _test():
assert as_int('1') == 1
if __name__ == '__main__':
_test()
...a myothermodule.py like this...
#!/usr/bin/env python3
from .mymodule import as_int
# Exported function
def add(a, b):
return as_int(a) + as_int(b)
# Test function for module
def _test():
assert add('1', '1') == 2
if __name__ == '__main__':
_test()
...and a main.py like this...
#!/usr/bin/env python3
from mypackage.myothermodule import add
def main():
print(add('1', '1'))
if __name__ == '__main__':
main()
...which works fine when you run main.py or mypackage/mymodule.py, but fails with mypackage/myothermodule.py, due to the relative import...
from .mymodule import as_int
The way you're supposed to run it is...
python3 -m mypackage.myothermodule
...but it's somewhat verbose, and doesn't mix well with a shebang line like #!/usr/bin/env python3.
The simplest fix for this case, assuming the name mymodule is globally unique, would be to avoid using relative imports, and just use...
from mymodule import as_int
...although, if it's not unique, or your package structure is more complex, you'll need to include the directory containing your package directory in PYTHONPATH, and do it like this...
from mypackage.mymodule import as_int
...or if you want it to work "out of the box", you can frob the PYTHONPATH in code first with this...
import sys
import os
SCRIPT_DIR = os.path.dirname(os.path.abspath(__file__))
sys.path.append(os.path.dirname(SCRIPT_DIR))
from mypackage.mymodule import as_int
It's kind of a pain, but there's a clue as to why in an email written by a certain Guido van Rossum...
I'm -1 on this and on any other proposed twiddlings of the __main__
machinery. The only use case seems to be running scripts that happen
to be living inside a module's directory, which I've always seen as an
antipattern. To make me change my mind you'd have to convince me that
it isn't.
Whether running scripts inside a package is an antipattern or not is subjective, but personally I find it really useful in a package I have which contains some custom wxPython widgets, so I can run the script for any of the source files to display a wx.Frame containing only that widget for testing purposes.

Explanation
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.
At some point PEP 338 conflicted with PEP 328:
... relative imports rely on __name__ to determine the current
module's position in the package hierarchy. In a main module, the
value of __name__ is always '__main__', so explicit relative imports
will always fail (as they only work for a module inside a package)
and to address the issue, PEP 366 introduced the top level variable __package__:
By adding a new module level attribute, this PEP allows relative
imports to work automatically if the module is executed using the -m
switch. A small amount of boilerplate in the module itself will allow
the relative imports to work when the file is executed by name. [...] When it [the attribute] is present, relative imports will be based on this attribute
rather than the module __name__ attribute. [...] When the main module is specified by its filename, then the __package__ attribute will be set to None. [...] When the import system encounters an explicit relative import in a
module without __package__ set (or with it set to None), it will
calculate and store the correct value (__name__.rpartition('.')[0]
for normal modules and __name__ for package initialisation modules)
(emphasis mine)
If the __name__ is '__main__', __name__.rpartition('.')[0] returns empty string. This is why there's empty string literal in the error description:
SystemError: Parent module '' not loaded, cannot perform relative import
The relevant part of the CPython's PyImport_ImportModuleLevelObject function:
if (PyDict_GetItem(interp->modules, package) == NULL) {
PyErr_Format(PyExc_SystemError,
"Parent module %R not loaded, cannot perform relative "
"import", package);
goto error;
}
CPython raises this exception if it was unable to find package (the name of the package) in interp->modules (accessible as sys.modules). Since sys.modules is "a dictionary that maps module names to modules which have already been loaded", it's now clear that the parent module must be explicitly absolute-imported before performing relative import.
Note: The patch from the issue 18018 has added another if block, which will be executed before the code above:
if (PyUnicode_CompareWithASCIIString(package, "") == 0) {
PyErr_SetString(PyExc_ImportError,
"attempted relative import with no known parent package");
goto error;
} /* else if (PyDict_GetItem(interp->modules, package) == NULL) {
...
*/
If package (same as above) is empty string, the error message will be
ImportError: attempted relative import with no known parent package
However, you will only see this in Python 3.6 or newer.
Solution #1: Run your script using -m
Consider a directory (which is a Python package):
.
├── package
│   ├── __init__.py
│   ├── module.py
│   └── standalone.py
All of the files in package begin with the same 2 lines of code:
from pathlib import Path
print('Running' if __name__ == '__main__' else 'Importing', Path(__file__).resolve())
I'm including these two lines only to make the order of operations obvious. We can ignore them completely, since they don't affect the execution.
__init__.py and module.py contain only those two lines (i.e., they are effectively empty).
standalone.py additionally attempts to import module.py via relative import:
from . import module # explicit relative import
We're well aware that /path/to/python/interpreter package/standalone.py will fail. However, we can run the module with the -m command line option that will "search sys.path for the named module and execute its contents as the __main__ module":
vaultah#base:~$ python3 -i -m package.standalone
Importing /home/vaultah/package/__init__.py
Running /home/vaultah/package/standalone.py
Importing /home/vaultah/package/module.py
>>> __file__
'/home/vaultah/package/standalone.py'
>>> __package__
'package'
>>> # The __package__ has been correctly set and module.py has been imported.
... # What's inside sys.modules?
... import sys
>>> sys.modules['__main__']
<module 'package.standalone' from '/home/vaultah/package/standalone.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/package/module.py'>
>>> sys.modules['package']
<module 'package' from '/home/vaultah/package/__init__.py'>
-m does all the importing stuff for you and automatically sets __package__, but you can do that yourself in the
Solution #2: Set __package__ manually
Please treat it as a proof of concept rather than an actual solution. It isn't well-suited for use in real-world code.
PEP 366 has a workaround to this problem, however, it's incomplete, because setting __package__ alone is not enough. You're going to need to import at least N preceding packages in the module hierarchy, where N is the number of parent directories (relative to the directory of the script) that will be searched for the module being imported.
Thus,
Add the parent directory of the Nth predecessor of the current module to sys.path
Remove the current file's directory from sys.path
Import the parent module of the current module using its fully-qualified name
Set __package__ to the fully-qualified name from 2
Perform the relative import
I'll borrow files from the Solution #1 and add some more subpackages:
package
├── __init__.py
├── module.py
└── subpackage
├── __init__.py
└── subsubpackage
├── __init__.py
└── standalone.py
This time standalone.py will import module.py from the package package using the following relative import
from ... import module # N = 3
We'll need to precede that line with the boilerplate code, to make it work.
import sys
from pathlib import Path
if __name__ == '__main__' and __package__ is None:
file = Path(__file__).resolve()
parent, top = file.parent, file.parents[3]
sys.path.append(str(top))
try:
sys.path.remove(str(parent))
except ValueError: # Already removed
pass
import package.subpackage.subsubpackage
__package__ = 'package.subpackage.subsubpackage'
from ... import module # N = 3
It allows us to execute standalone.py by filename:
vaultah#base:~$ python3 package/subpackage/subsubpackage/standalone.py
Running /home/vaultah/package/subpackage/subsubpackage/standalone.py
Importing /home/vaultah/package/__init__.py
Importing /home/vaultah/package/subpackage/__init__.py
Importing /home/vaultah/package/subpackage/subsubpackage/__init__.py
Importing /home/vaultah/package/module.py
A more general solution wrapped in a function can be found here. Example usage:
if __name__ == '__main__' and __package__ is None:
import_parents(level=3) # N = 3
from ... import module
from ...module.submodule import thing
Solution #3: Use absolute imports and setuptools
The steps are -
Replace explicit relative imports with equivalent absolute imports
Install package to make it importable
For instance, the directory structure may be as follows
.
├── project
│   ├── package
│   │   ├── __init__.py
│   │   ├── module.py
│   │   └── standalone.py
│   └── setup.py
where setup.py is
from setuptools import setup, find_packages
setup(
name = 'your_package_name',
packages = find_packages(),
)
The rest of the files were borrowed from the Solution #1.
Installation will allow you to import the package regardless of your working directory (assuming there'll be no naming issues).
We can modify standalone.py to use this advantage (step 1):
from package import module # absolute import
Change your working directory to project and run /path/to/python/interpreter setup.py install --user (--user installs the package in your site-packages directory) (step 2):
vaultah#base:~$ cd project
vaultah#base:~/project$ python3 setup.py install --user
Let's verify that it's now possible to run standalone.py as a script:
vaultah#base:~/project$ python3 -i package/standalone.py
Running /home/vaultah/project/package/standalone.py
Importing /home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/__init__.py
Importing /home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py
>>> module
<module 'package.module' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py'>
>>> import sys
>>> sys.modules['package']
<module 'package' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/__init__.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py'>
Note: If you decide to go down this route, you'd be better off using virtual environments to install packages in isolation.
Solution #4: Use absolute imports and some boilerplate code
Frankly, the installation is not necessary - you could add some boilerplate code to your script to make absolute imports work.
I'm going to borrow files from Solution #1 and change standalone.py:
Add the parent directory of package to sys.path before attempting to import anything from package using absolute imports:
import sys
from pathlib import Path # if you haven't already done so
file = Path(__file__).resolve()
parent, root = file.parent, file.parents[1]
sys.path.append(str(root))
# Additionally remove the current file's directory from sys.path
try:
sys.path.remove(str(parent))
except ValueError: # Already removed
pass
Replace the relative import by the absolute import:
from package import module # absolute import
standalone.py runs without problems:
vaultah#base:~$ python3 -i package/standalone.py
Running /home/vaultah/package/standalone.py
Importing /home/vaultah/package/__init__.py
Importing /home/vaultah/package/module.py
>>> module
<module 'package.module' from '/home/vaultah/package/module.py'>
>>> import sys
>>> sys.modules['package']
<module 'package' from '/home/vaultah/package/__init__.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/package/module.py'>
I feel that I should warn you: try not to do this, especially if your project has a complex structure.
As a side note, PEP 8 recommends the use of absolute imports, but states that in some scenarios explicit relative imports are acceptable:
Absolute imports are recommended, as they are usually more readable
and tend to be better behaved (or at least give better error
messages). [...] However, explicit relative imports are an acceptable
alternative to absolute imports, especially when dealing with complex
package layouts where using absolute imports would be unnecessarily
verbose.

Put this inside your package's __init__.py file:
# For relative imports to work in Python 3.6
import os, sys; sys.path.append(os.path.dirname(os.path.realpath(__file__)))
Assuming your package is like this:
├── project
│ ├── package
│ │ ├── __init__.py
│ │ ├── module1.py
│ │ └── module2.py
│ └── setup.py
Now use regular imports in you package, like:
# in module2.py
from module1 import class1
This works in both python 2 and 3.

I ran into this issue. A hack workaround is importing via an if/else block like follows:
#!/usr/bin/env python3
#myothermodule
if __name__ == '__main__':
from mymodule import as_int
else:
from .mymodule import as_int
# Exported function
def add(a, b):
return as_int(a) + as_int(b)
# Test function for module
def _test():
assert add('1', '1') == 2
if __name__ == '__main__':
_test()

SystemError: Parent module '' not loaded, cannot perform relative import
This means you are running a module inside the package as a script. Mixing scripts inside packages is tricky and should be avoided if at all possible. Use a wrapper script that imports the package and runs your scripty function instead.
If your top-level directory is called foo, which is on your PYTHONPATH module search path, and you have a package bar there (it is a directory you'd expect an __init__.py file in), scripts should not be placed inside bar, but should live on in foo at best.
Note that scripts differ from modules here in that they are used as a filename argument to the python command, either by using python <filename> or via a #! (shebang) line. It is loaded directly as the __main__ module (this is why if __name__ == "__main__": works in scripts), and there is no package context to build on for relative imports.
Your options
If you can, package your project with setuptools (or poetry or flit, which can help simplify packaging), and create console script entrypoints; installing your project with pip then creates scripts that know how to import your package properly. You can install your package locally with pip install -e ., so it can still be edited in-place.
Otherwise, never, ever, use python path/to/packagename/file.py, always use python path/to/script.py and script.py can use from packagename import ....
As a fallback, you could use the -m command-line switch to tell Python to import a module and use that as the __main__ file instead. This does not work with a shebang line, as there is no script file any more, however.
If you use python -m foo.bar and foo/bar.py is found in a sys.path directory, that is then imported and executed as __main__ with the right package context. If bar is also a package, inside foo/, it must have a __main__.py file (so foo/bar/__main__.py as the path from the sys.path directory).
In extreme circumstances, add the metadata Python uses to resolve relative imports by setting __package__ directly; the file foo/bar/spam.py, importable as foo.bar.spam, is given the global __package__ = "foo.bar". It is just another global, like __file__ and __name__, set by Python when imported.
On sys.path
The above all requires that your package can be imported, which means it needs to be found in one of the directories (or zipfiles) listed in sys.path. There are several options here too:
The directory where path/to/script.py was found (so path/to) is automatically added to sys.path. Executing python path/to/foo.py adds path/to to sys.path.
If you packaged your project (with setuptools, poetry, flit or another Python packaging tool), and installed it, the package has been added to the right place already.
As a last resort, add the right directory to sys.path yourself. If the package can be located relatively to the script file, use the __file__ variable in the script global namespace (e.g. using the pathlib.Path object, HERE = Path(__file__).resolve().parent is a reference to the directory the file lives in, as absolute path).

For PyCharm users:
I also was getting ImportError: attempted relative import with no known parent package because I was adding the . notation to silence a PyCharm parsing error. PyCharm innaccurately reports not being able to find:
lib.thing import function
If you change it to:
.lib.thing import function
it silences the error but then you get the aforementioned ImportError: attempted relative import with no known parent package. Just ignore PyCharm's parser. It's wrong and the code runs fine despite what it says.

To obviate this problem, I devised a solution with the repackage package, which has worked for me for some time. It adds the upper directory to the lib path:
import repackage
repackage.up()
from mypackage.mymodule import myfunction
Repackage can make relative imports that work in a wide range of cases, using an intelligent strategy (inspecting the call stack).

TL;DR: to #Aya's answer, updated with pathlib library, and working for Jupyter notebooks where __file__ is not defined:
You want to import my_function defined under ../my_Folder_where_the_package_lives/my_package.py
respect to where you are writing the code.
Then do:
import os
import sys
import pathlib
PACKAGE_PARENT = pathlib.Path(__file__).parent
#PACKAGE_PARENT = pathlib.Path.cwd().parent # if on jupyter notebook
SCRIPT_DIR = PACKAGE_PARENT / "my_Folder_where_the_package_lives"
sys.path.append(str(SCRIPT_DIR))
from my_package import my_function

Hopefully, this will be of value to someone out there - I went through half a dozen stackoverflow posts trying to figure out relative imports similar to whats posted above here. I set up everything as suggested but I was still hitting ModuleNotFoundError: No module named 'my_module_name'
Since I was just developing locally and playing around, I hadn't created/run a setup.py file. I also hadn't apparently set my PYTHONPATH.
I realized that when I ran my code as I had been when the tests were in the same directory as the module, I couldn't find my module:
$ python3 test/my_module/module_test.py 2.4.0
Traceback (most recent call last):
File "test/my_module/module_test.py", line 6, in <module>
from my_module.module import *
ModuleNotFoundError: No module named 'my_module'
However, when I explicitly specified the path things started to work:
$ PYTHONPATH=. python3 test/my_module/module_test.py 2.4.0
...........
----------------------------------------------------------------------
Ran 11 tests in 0.001s
OK
So, in the event that anyone has tried a few suggestions, believes their code is structured correctly and still finds themselves in a similar situation as myself try either of the following if you don't export the current directory to your PYTHONPATH:
Run your code and explicitly include the path like so:
$ PYTHONPATH=. python3 test/my_module/module_test.py
To avoid calling PYTHONPATH=., create a setup.py file with contents like the following and run python setup.py development to add packages to the path:
# setup.py
from setuptools import setup, find_packages
setup(
name='sample',
packages=find_packages()
)

TL;DR
You can only relatively import modules inside another module in the same package.
Concept Clarify
We see a lot of example code in books/docs/articles, they show us how to relatively import a module, but when we do so, it fails.
The reason is, put it in a simple sentence, we did not run the code as the python module mechanism expects, even though the code is written totally right. It's like some kind of runtime thing.
Module loading is depended on how you run the code. That is the source of confusion.
What is a module?
A module is a python file when and only when it is being imported by another file. Given a file mod.py, is it a module? Yes and No, if you run python mod.py, it is not a module, because it is not imported.
What is a package?
A package is a folder that includes Python module(s).
BTW, __init__.py is not necessary from python 3.3, if you don't need any package initialization or auto-load submodules. You don't need to place a blank __init__.py in a directory.
That proves a package is just a folder as long as there are files being imported.
Real Answer
Now, this description becomes clearer.
You can only relatively import modules inside another module in the same package.
Given a directory:
. CWD
|-- happy_maker.py # content: print('Sends Happy')
`-- me.py # content: from . import happy_maker
Run python me.py, we got attempted relative import with no known parent package
me.py is run directly, it is not a module, and we can't use relative import in it.
Solution 1
Use import happy_maker instead of from . import happy_maker
Solution 2
Switch our working directory to the parent folder.
. CWD
|-- happy
| |-- happy_maker.py
`-- me.py
Run python -m happy.me.
When we are in the directory that includes happy, happy is a package, me.py, happy_maker.py are modules, we can use relative import now, and we still want to run me.py, so we use -m which means run the module as a script.
Python Idiom
. CWD
|-- happy
| |-- happy_maker.py # content: print('Sends Happy')
| `-- me.py # content: from . import happy_maker
`-- main.py # content: import happy.me
This structure is the python idiom. main is our script, best practice in Python. Finally, we got there.
Siblings or Grandparents
Another common need:
.
|-- happy
| |-- happy_maker.py
| `-- me.py
`-- sad
`-- sad_maker.py
We want to import sad_maker in me.py, How to do that?
First, we need to make happy and sad in the same package, so we have to go up a directory level. And then from ..sad import sad_maker in the me.py.
That is all.

My boilerplate to make a module with relative imports in a package runnable standalone.
package/module.py
## Standalone boilerplate before relative imports
if __package__ is None:
DIR = Path(__file__).resolve().parent
sys.path.insert(0, str(DIR.parent))
__package__ = DIR.name
from . import variable_in__init__py
from . import other_module_in_package
...
Now you can use your module in any fashion:
Run module as usual: python -m package.module
Use it as a module: python -c 'from package import module'
Run it standalone: python package/module.py
or with shebang (#!/bin/env python) just: package/module.py
NB! Using sys.path.append instead of sys.path.insert will give you a hard to trace error if your module has the same name as your package. E.g. my_script/my_script.py
Of course if you have relative imports from higher levels in your package hierarchy, than this is not enough, but for most cases, it's just okay.

I needed to run python3 from the main project directory to make it work.
For example, if the project has the following structure:
project_demo/
├── main.py
├── some_package/
│ ├── __init__.py
│ └── project_configs.py
└── test/
└── test_project_configs.py
Solution
I would run python3 inside folder project_demo/ and then perform a
from some_package import project_configs

I was getting this ImportError: attempted relative import with no known parent package
In my program I was using the file from current path for importing its function.
from .filename import function
Then I modified the current path (Dot) with package name. Which resolved my issue.
from package_name.filename import function
I hope the above answer helps you.

Importing from same directory
Firstly, you can import from the same directory.
Here is the file structure...
Folder
|
├─ Scripts
| ├─ module123.py
|
├─ main.py
├─ script123.py
Here is main.py
from . import script123
from Scripts import module123
As you can see, importing from . imports from current directory.
Note: if running using anything but IDLE, make sure that your terminal is navigated to the same directory as the main.py file before running.
Also, importing from a local folder also works.
Importing from parent directory
As seen in my GitHub gist here, there is the following method.
Take the following file tree...
ParentDirectory
├─ Folder
| |
| ├─ Scripts
| | ├─ module123.py
| |
| ├─ main.py
| ├─ script123.py
|
├─ parentModule.py
Then, just add this code to the top of your main.py file.
import inspect
import os
import sys
current_dir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
parent_dir = os.path.dirname(current_dir)
sys.path.insert(0, parent_dir)
from ParentDirectory import Stuff

I tried all of the above to no avail, only to realize I mistakenly had a - in my package name.
In short, don't have - in the directory where __init__.py is. I've never felt elated after finding out such inanity.

if both packages are in your import path (sys.path), and the module/class you want is in example/example.py, then to access the class without relative import try:
from example.example import fkt

If none of the above worked for you, you can specify the module explicitly.
Directory:
├── Project
│ ├── Dir
│ │ ├── __init__.py
│ │ ├── module.py
│ │ └── standalone.py
Solution:
#in standalone.py
from Project.Dir.module import ...
module - the module to be imported

Here is a three-liner for those who disagree with Guido:
import sys
from pathlib import Path
sys.path.append(str(Path(sys.argv[0]).absolute().parent.parent))
Hope it helps.

I think the best solution is to create a package for your module:
Here is more info on how to do it.
Once you have a package you don't need to worry about relative import, you can just do absolute imports.

I encounter this a lot when I am working with Django, since a lot of functionality is performed from the manage.py script but I also want to have some of my modules runnable directly as scripts as well (ideally you would make them manage.py directives but we're not there yet).
This is a mock up of what such a project might look like;
├── dj_app
│   ├── models.py
│   ├── ops
│   │   ├── bar.py
│   │   └── foo.py
│   ├── script.py
│   ├── tests.py
│   ├── utils.py
│   └── views.py
└── manage.py
The important parts here being manage.py, dj_app/script.py, and dj_app/tests.py. We also have submodules dj_app/ops/bar.py and dj_app/ops/foo.py which contain more items we want to use throughout the project.
The source of the issue commonly comes from wanting your dj_app/script.py script methods to have test cases in dj_app/tests.py which get invoked when you run manage.py test.
This is how I set up the project and its imports;
# dj_app/ops/foo.py
# Foo operation methods and classes
foo_val = "foo123"
.
# dj_app/ops/bar.py
# Bar operations methods and classes
bar_val = "bar123"
.
# dj_app/script.py
# script to run app methods from CLI
# if run directly from command line
if __name__ == '__main__':
from ops.bar import bar_val
from ops.foo import foo_val
# otherwise
else:
from .ops.bar import bar_val
from .ops.foo import foo_val
def script_method1():
print("this is script_method1")
print("bar_val: {}".format(bar_val))
print("foo_val: {}".format(foo_val))
if __name__ == '__main__':
print("running from the script")
script_method1()
.
# dj_app/tests.py
# test cases for the app
# do not run this directly from CLI or the imports will break
from .script import script_method1
from .ops.bar import bar_val
from .ops.foo import foo_val
def main():
print("Running the test case")
print("testing script method")
script_method1()
if __name__ == '__main__':
print("running tests from command line")
main()
.
# manage.py
# just run the test cases for this example
import dj_app.tests
dj_app.tests.main()
.
Running the test cases from manage.py;
$ python3 manage.py
Running the test case
testing script method
this is script_method1
bar_val: bar123
foo_val: foo123
Running the script on its own;
$ python3 dj_app/script.py
running from the script
this is script_method1
bar_val: bar123
foo_val: foo123
Note that you get an error if you try to run the test.py directly however, so don't do that;
$ python3 dj_app/tests.py
Traceback (most recent call last):
File "dj_app/tests.py", line 5, in <module>
from .script import script_method1
ModuleNotFoundError: No module named '__main__.script'; '__main__' is not a package
If I run into more complicated situations for imports, I usually end up implementing something like this to hack through it;
import os
import sys
THIS_DIR = os.path.dirname(os.path.realpath(__file__))
sys.path.insert(0, THIS_DIR)
from script import script_method1
sys.path.pop(0)

This my project structure
├── folder
| |
│ ├── moduleA.py
| | |
| | └--function1()
| | └~~ uses function2()
| |
│ └── moduleB.py
| |
| └--function2()
|
└── main.py
└~~ uses function1()
Here my moduleA imports moduleB and main imports moduleA
I added the snippet below in moduleA to import moduleB
try:
from .moduleB import function2
except:
from moduleB import function2
Now I can execute both main.py as well as moduleA.py individually
Is this a solution ?

The below solution is tested on Python3
├── classes
| |
| ├──__init__.py
| |
│ ├── userclass.py
| | |
| | └--viewDetails()
| |
| |
│ └── groupclass.py
| |
| └--viewGroupDetails()
|
└── start.py
└~~ uses function1()
Now, in order to use viewDetails of userclass or viewGroupDetails of groupclass define that in _ init _.py of classess directory first.
Ex: In _ init _.py
from .userclasss import viewDetails
from .groupclass import viewGroupDetails
Step2: Now, in start.py we can directly import viewDetails
Ex: In start.py
from classes import viewDetails
from classes import viewGroupDetails

I ran into a similar problem when trying to write a python file that can be loaded either as a module or an executable script.
Setup
/path/to/project/
├── __init__.py
└── main.py
└── mylib/
├── list_util.py
└── args_util.py
with:
main.py:
#!/usr/bin/env python3
import sys
import mylib.args_util
if __name__ == '__main__':
print(f'{mylib.args_util.parseargs(sys.argv[1:])=}')
mylib/list_util.py:
def to_int_list(args):
return [int(x) for x in args]
mylib/args_util.py:
#!/usr/bin/env python3
import sys
from . import list_util as lu
def parseargs(args):
return sum(lu.to_int_list(args))
if __name__ == '__main__':
print(f'{parseargs(sys.argv[1:])=}')
Output
$ ./main.py 1 2 3
mylib.args_util.parseargs(sys.argv[1:])=6
$ mylib/args_util.py 1 2 3
Traceback (most recent call last):
File "/path/to/project/mylib/args_util.py", line 10, in <module>
from . import list_util as lu
ImportError: attempted relative import with no known parent package
Solution
I settled for a Bash/Python polyglot solution. The Bash version of the program just calls python3 -m mylib.args_util then exits.
The Python version ignores the Bash code because it's contained in the docstring.
The Bash version ignores the Python code because it uses exec to stop parsing/running lines.
mylib/args_util.py:
#!/bin/bash
# -*- Mode: python -*-
''''true
exec /usr/bin/env python3 -m mylib.args_util "$#"
'''
import sys
from . import list_util as lu
def parseargs(args):
return sum(lu.to_int_list(args))
if __name__ == '__main__':
print(f'{parseargs(sys.argv[1:])=}')
Output
$ ./main.py 1 2 3
mylib.args_util.parseargs(sys.argv[1:])=6
$ mylib/args_util.py 1 2 3
parseargs(sys.argv[1:])=6
Explanation
Line 1: #!/bin/bash; this is the "shebang" line; it tells the interactive shell how run this script.
Python: ignored (comment)
Bash: ignored (comment)
Line 2: # -*- Mode: python -*- optional; this is called the "mode-line"; it tells Emacs to use Python syntax highlighting instead of guessing that the language is Bash when reading the file.
Python: ignored (comment)
Bash: ignored (comment)
Line 3: ''''true
Python: views this as an unassigned docstring starting with 'true\n
Bash: views this as three strings (of which the first two are empty strings) that expand to true (i.e. '' + '' + 'true' = 'true'); it then runs true (which does nothing) and continues to the next line
Line 4: exec /usr/bin/env python3 -m mylib.args_util "$#"
Python: still views this as part of the docstring from line 3.
Bash: runs python3 -m mylib.args_util then exits (it doesn't parse anything beyond this line)
Line 5: '''
Python: views this as the end of the docstring from line 3.
Bash: doesn't parse this line
Caveats
This doesn't work on Windows:
Workaround: Use WSL or a Batch wrapper script to call python -m mylib.args_util.
This only works if the current working directory is set to /path/to/project/.
Workaround: Set PYTHONPATH when calling /usr/bin/env
#!/bin/bash
# -*- Mode: python -*-
''''true
exec /usr/bin/env python3 \
PYTHONPATH="$(cd "$(dirname "$0")/.." ; pwd)" \
-m mylib.args_util "$#"
'''

I've created a new, experimental import library for Python: ultraimport
It gives the programmer more control over imports and makes them unambiguous. Also it gives better error messages when an import fails.
It allows you to do relative, file-system based imports that always work, no matter how you run your code and no matter what is your current working directory. It does not matter if you run a script or module. You also don't have to change sys.path which might have other side effects.
You would then change
from .mymodule import myfunction
to
import ultraimport
myfunction = ultraimport('__dir__/mymodule.py', 'myfunction')
This way the import will always work, even if you run the code as script.
One issue when importing scripts like this is that subsequent relative imports might fail. ultraimport has a builtin preprocessor to automatically rewrite relative imports.

I had a similar problem: I needed a Linux service and cgi plugin which use common constants to cooperate. The 'natural' way to do this is to place them in the init.py of the package, but I cannot start the cgi plugin with the -m parameter.
My final solution was similar to Solution #2 above:
import sys
import pathlib as p
import importlib
pp = p.Path(sys.argv[0])
pack = pp.resolve().parent
pkg = importlib.import_module('__init__', package=str(pack))
The disadvantage is that you must prefix the constants (or common functions) with pkg:
print(pkg.Glob)

TLDR; Append Script path to the System Path by adding following in the entry point of your python script.
import os.path
import sys
PACKAGE_PARENT = '..'
SCRIPT_DIR = os.path.dirname(os.path.realpath(os.path.join(os.getcwd(), os.path.expanduser(__file__))))
sys.path.append(os.path.normpath(os.path.join(SCRIPT_DIR, PACKAGE_PARENT)))
Thats it now you can run your project in PyCharma as well as from Terminal!!

Moving the file from which you are importing to an outside directory helps.
This is extra useful when your main file makes any other files in its own directory.
Ex:
Before:
Project
|---dir1
|-------main.py
|-------module1.py
After:
Project
|---module1.py
|---dir1
|-------main.py

I was getting the same error and my project structure was like
->project
->vendors
->vendors.py
->main.py
I was trying to call like this
from .vendors.Amazon import Amazom_Purchase
Here it was throwing an error so I fixed it simply by removing the first . from the statement
from vendors.Amazon import Amazom_Purchase
Hope this helps.

It's good to note that sometimes the cache causes of all it - I've tried different things after re-arranging classes into new directories and relative import started to work after I removed the __pycache__

If the following import:
from . import something
doesn't work for you it is because this is python-packaging import and will not work with your regular implementation, and here is an example to show how to use it:
Folder structure:
.
└── funniest
├── funniest
│ ├── __init__.py
│ └── text.py
├── main.py
└── setup.py
inside __init__.py add:
def available_module():
return "hello world"
text.py add:
from . import available_module
inside setup.py add
from setuptools import setup
setup(name='funniest',
version='0.1',
description='The funniest joke in the world',
url='http://github.com/storborg/funniest',
author='Flying Circus',
author_email='flyingcircus#example.com',
license='MIT',
packages=['funniest'],
zip_safe=False)
Now, this is the most important part you need to install your package:
pip install .
Anywhere else in our system using the same Python, we can do this now:
>> import funnies.text as fun
>> fun.available_module()
This should output 'hello world'
you can test this in main.py (this will not require any installation of the Package)
Here is main.py as well
import funniest.text as fun
print(fun.available_module())

We Keep Coding

Python is a programming language that lets you work quickly and integrate systems more effectively.

How can I create imports that always work? - python

Import utils inside the work.py as follows: import mypackage.utils or if you want to use shorter name: from mypackage import utils EDIT: If you need to run work.py outside of the package, then try: try: from mypackage import utils except ImportError: import utils

Use: from . import utils as suggested by Peter In your code.py you should use: from mypackage import work

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