I’ve been here:
- Python packages: relative imports
- python relative import example code does not work
- Relative imports in python 2.5
- Relative imports in Python
- Python: Disabling relative import
and plenty of URLs that I did not copy, some on SO, some on other sites, back when I thought I’d have the solution quickly.
The forever-recurring question is this: how do I solve this “Attempted relative import in non-package” message?
ImportError: attempted relative import with no known parent package
I built an exact replica of the package on pep-0328:
package/ __init__.py subpackage1/ __init__.py moduleX.py moduleY.py subpackage2/ __init__.py moduleZ.py moduleA.py
The imports were done from the console.
I did make functions named spam and eggs in their appropriate modules. Naturally, it didn’t work. The answer is apparently in the 4th URL I listed, but it’s all alumni to me. There was this response on one of the URLs I visited:
Relative imports use a module’s name attribute to determine that module’s position in the package hierarchy. If the module’s name does not contain any package information (e.g. it is set to ‘main’) then relative imports are resolved as if the module were a top level module, regardless of where the module is actually located on the file system.
The above response looks promising, but it’s all hieroglyphs to me. So my question, how do I make Python not return to me “Attempted relative import in non-package”? has an answer that involves -m, supposedly.
Can somebody please tell me why Python gives that error message, what it means by “non-package”, why and how do you define a ‘package’, and the precise answer put in terms easy enough for a kindergartener to understand.
Script vs. Module
Here’s an explanation. The short version is that there is a big difference between directly running a Python file, and importing that file from somewhere else. Just knowing what directory a file is in does not determine what package Python thinks it is in. That depends, additionally, on how you load the file into Python (by running or by importing).
There are two ways to load a Python file: as the top-level script, or as a
module. A file is loaded as the top-level script if you execute it directly, for instance by typing
python myfile.py on the command line. It is loaded as a module when an
import statement is encountered inside some other file. There can only be one top-level script at a time; the top-level script is the Python file you ran to start things off.
When a file is loaded, it is given a name (which is stored in its
- If it was loaded as the top-level script, its name is
- If it was loaded as a module, its name is [ the filename, preceded by the names of any packages/subpackages of which it is a part, separated by dots ], for example,
But be aware, if you load
moduleX as a module from shell command line using something like
python -m package.subpackage1.moduleX, the
__name__ will still be
So for instance in your example:
package/ __init__.py subpackage1/ __init__.py moduleX.py moduleA.py
if you imported
moduleX (note: imported, not directly executed), its name would be
package.subpackage1.moduleX. If you imported
moduleA, its name would be
package.moduleA. However, if you directly run
moduleX from the command line, its name will instead be
__main__, and if you directly run
moduleA from the command line, its name will be
__main__. When a module is run as the top-level script, it loses its normal name and its name is instead
Accessing a module NOT through its containing package
There is an additional wrinkle: the module’s name depends on whether it was imported “directly” from the directory it is in or imported via a package. This only makes a difference if you run Python in a directory, and try to import a file in that same directory (or a subdirectory of it). For instance, if you start the Python interpreter in the directory
package/subpackage1 and then do
import moduleX, the name of
moduleX will just be
moduleX, and not
package.subpackage1.moduleX. This is because Python adds the current directory to its search path when the interpreter is entered interactively; if it finds the to-be-imported module in the current directory, it will not know that that directory is part of a package, and the package information will not become part of the module’s name.
A special case is if you run the interpreter interactively (e.g., just type
python and start entering Python code on the fly). In this case, the name of that interactive session is
Now here is the crucial thing for your error message: if a module’s name has no dots, it is not considered to be part of a package. It doesn’t matter where the file actually is on disk. All that matters is what its name is, and its name depends on how you loaded it.
Now look at the quote you included in your question:
Relative imports use a module’s name attribute to determine that module’s position in the package hierarchy. If the module’s name does not contain any package information (e.g. it is set to ‘main’) then relative imports are resolved as if the module were a top-level module, regardless of where the module is actually located on the file system.
Relative imports use the module’s name to determine where it is in a package. When you use a relative import like
from .. import foo, the dots indicate to step up some number of levels in the package hierarchy. For instance, if your current module’s name is
..moduleA would mean
package.moduleA. For a
from .. import to work, the module’s name must have at least as many dots as there are in the
… are only relative in a package
However, if your module’s name is
__main__, it is not considered to be in a package. Its name has no dots, and therefore you cannot use
from .. import statements inside it. If you try to do so, you will get the “relative-import in non-package” error.
Scripts can’t import relative
What you probably did is you tried to run
moduleX or the like from the command line. When you did this, its name was set to
__main__, which means that relative imports within it will fail, because its name does not reveal that it is in a package. Note that this will also happen if you run Python from the same directory where a module is, and then try to import that module, because, as described above, Python will find the module in the current directory “too early” without realizing it is part of a package.
Also remember that when you run the interactive interpreter, the “name” of that interactive session is always
__main__. Thus you cannot do relative imports directly from an interactive session. Relative imports are only for use within module files.
If you really do want to run
moduleXdirectly, but you still want it to be considered part of a package, you can do
python -m package.subpackage1.moduleX. The
-mtells Python to load it as a module, not as the top-level script.
Or perhaps you don’t actually want to run
moduleX, you just want to run some other script, say
myfile.py, that uses functions inside
moduleX. If that is the case, put
myfile.pysomewhere else – not inside the
packagedirectory – and run it. If inside
myfile.pyyou do things like
from package.moduleA import spam, it will work fine.
For either of these solutions, the package directory (
packagein your example) must be accessible from the Python module search path (
sys.path). If it is not, you will not be able to use anything in the package reliably at all.
Since Python 2.6, the module’s “name” for package-resolution purposes is determined not just by its
__name__attributes but also by the
__package__attribute. That’s why I’m avoiding using the explicit symbol
__name__to refer to the module’s “name”. Since Python 2.6 a module’s “name” is effectively
__package__ + '.' + __name__, or just
This is really a problem within python. The origin of confusion is that people mistakenly takes the relative import as path relative which is not.
For example when you write in faa.py:
from .. import foo
This has a meaning only if faa.py was identified and loaded by python, during execution, as a part of a package. In that case,the module’s name
for faa.py would be for example some_packagename.faa. If the file was loaded just because it is in the current directory, when python is run, then its name would not refer to any package and eventually relative import would fail.
A simple solution to refer modules in the current directory, is to use this:
if __package__ is None or __package__ == '': # uses current directory visibility import foo else: # uses current package visibility from . import foo
There are too much too long anwers in a foreign language. So I’ll try to make it short.
If you write
from . import module, opposite to what you think,
module will not be imported from current directory, but from the top level of your package! If you run .py file as a script, it simply doesn’t know where the top level is and thus refuses to work.
If you start it like this
py -m package.module from the directory above
package, then python knows where the top level is. That’s very similar to java:
java -cp bin_directory package.class