I’ve been here:
- http://www.python.org/dev/peps/pep-0328/
- http://docs.python.org/2/tutorial/modules.html#packages
- 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.
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Answer
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.
Naming
When a file is loaded, it is given a name (which is stored in its __name__
attribute).
- If it was loaded as the top-level script, its name is
__main__
. - 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,
package.subpackage1.moduleX
.
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 __main__
.
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 __main__
.
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 __main__
.
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…
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 package.subpackage1.moduleX
, then ..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 import
statement.
… 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.
Two solutions:
If you really do want to run
moduleX
directly, but you still want it to be considered part of a package, you can dopython -m package.subpackage1.moduleX
. The-m
tells 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, saymyfile.py
, that uses functions insidemoduleX
. If that is the case, putmyfile.py
somewhere else – not inside thepackage
directory – and run it. If insidemyfile.py
you do things likefrom package.moduleA import spam
, it will work fine.
Notes
For either of these solutions, the package directory (
package
in 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__name__
if__package__
isNone
.)