I am having trouble importing python packages only when running python from cmdline/console. However, when using pydev, everything seems to work fine.
I have the following filesystem...
---MarketData
---Parser
---Parser.py
---__init__.py
---IO
---__init__.py
---MarketSocket.py
Currently, Parser and IO are defined as python packages (they have init.py files, although there is no code in the Parser.init.py file.
I am trying to run the following line of code in MarketSocket.py
from Parser import Parser
Which should import the module 'Parser' within the package 'Parser' however, I get the following error.
ImportError: No Module Named Parser
Any help would be appreciated! This should work according to similar issues on stackOverflow, but for some odd reason it isn't.
MarketSocket.py is in the directory IO. Therefore it is not possible to find the package Parser.
The best way to resolve this, are relative imports: from ..Parser import Parser But they might not work, if you start the script like: python MarketSocket.py. To use this, you would also have to add an __init__.py to your MarketData directory.
If it doesn't work extend the sys.path like this:
import sys
sys.path.append('../')
With this addition, Python searches also the paths you want.
If I were you I would also think about restructuring your project. In my opinion executables should be (most of the time) at the top of your working tree, which is also like Python works.
the MarketSocket.py is one level below Parser and thus can't see it
do this:
import os
import sys
sys.path.insert(0, os.path.join(os.path.dirname(__file__), ".."))
sys.path.append(os.path.dirname(__file__))
Putting an (empty) __init__.py in the MarketData directory will make the whole thing a package (and avoids the ugly path hacks). That should just work then if you call the module from the top level of the package.
You encountered an issue with relative import. Only in the parent directory you may have the access to any child package/module. So in MarketSocket.py, you need
from ..Parser import Parser
Then when you run it with -m option, the trick is you have to run it in the top level directory. So in this case
1) you would go to the parent directory of MarketData
2) in that parent directory, run python -m MarketData.IO.marketSocket
Related
I have a layout like this:
src
__init__.py
main.py
examples
__init__.py
example_project.py
library
__init__.py
some_library_code.py
example_project.py uses code from some_library_code.py
I run the example_project.py like that:
***\src>: python examples\example_project.py and get ImportError: attempted relative import with no known parent package
I've read through some answers on SO and found that I need some construction
sys.path.append(os.path.normpath(os.path.join(os.path.dirname(os.path.abspath(__file__)), os.pardir)))
to be present in example_project.py
My example_project.py import section looks like that:
import os, sys
sys.path.append(os.path.normpath(os.path.join(os.path.dirname(os.path.abspath(__file__)), os.pardir)))
from ..library import some_library_code
but that doesn't work and shows the same ImportError
UPD:
if I change from ..library ***** to from library ***** then it works but the IDE doesn't recognize imported types and shows error around import clause
Revert the sys path hacks and instead run your code as:
***\src>: python -m examples.example_project
This way tells python to run the module example_project that lives in the package examples. Otherwise python has no way (when you run the script directly that is) to know that this script is part of a package - hence the error. The syspath hacks will fail in subtle ways (the IDE can't really follow those dynamic sys path additions - there are settings for this but then starts to become complicated, hence hack - but there are worst consequences even undefined behavior) while running your script with the -m switch from the parent dir of your root package is the recommended way of running scripts.
You forgot to add __init__.py in library folder. Here is the documentation https://docs.python.org/3/reference/import.html#regular-packages
Try from library.some_library_code import Something.
It seems there are already quite some questions here about relative import in python 3, but after going through many of them I still didn't find the answer for my issue.
so here is the question.
I have a package shown below
package/
__init__.py
A/
__init__.py
foo.py
test_A/
__init__.py
test.py
and I have a single line in test.py:
from ..A import foo
now, I am in the folder of package, and I run
python -m test_A.test
I got message
"ValueError: attempted relative import beyond top-level package"
but if I am in the parent folder of package, e.g., I run:
cd ..
python -m package.test_A.test
everything is fine.
Now my question is:
when I am in the folder of package, and I run the module inside the test_A sub-package as test_A.test, based on my understanding, ..A goes up only one level, which is still within the package folder, why it gives message saying beyond top-level package. What is exactly the reason that causes this error message?
EDIT: There are better/more coherent answers to this question in other questions:
Sibling package imports
Relative imports for the billionth time
Why doesn't it work? It's because python doesn't record where a package was loaded from. So when you do python -m test_A.test, it basically just discards the knowledge that test_A.test is actually stored in package (i.e. package is not considered a package). Attempting from ..A import foo is trying to access information it doesn't have any more (i.e. sibling directories of a loaded location). It's conceptually similar to allowing from ..os import path in a file in math. This would be bad because you want the packages to be distinct. If they need to use something from another package, then they should refer to them globally with from os import path and let python work out where that is with $PATH and $PYTHONPATH.
When you use python -m package.test_A.test, then using from ..A import foo resolves just fine because it kept track of what's in package and you're just accessing a child directory of a loaded location.
Why doesn't python consider the current working directory to be a package? NO CLUE, but gosh it would be useful.
import sys
sys.path.append("..") # Adds higher directory to python modules path.
Try this.
Worked for me.
Assumption:
If you are in the package directory, A and test_A are separate packages.
Conclusion:
..A imports are only allowed within a package.
Further notes:
Making the relative imports only available within packages is useful if you want to force that packages can be placed on any path located on sys.path.
EDIT:
Am I the only one who thinks that this is insane!? Why in the world is the current working directory not considered to be a package? – Multihunter
The current working directory is usually located in sys.path. So, all files there are importable. This is behavior since Python 2 when packages did not yet exist. Making the running directory a package would allow imports of modules as "import .A" and as "import A" which then would be two different modules. Maybe this is an inconsistency to consider.
None of these solutions worked for me in 3.6, with a folder structure like:
package1/
subpackage1/
module1.py
package2/
subpackage2/
module2.py
My goal was to import from module1 into module2. What finally worked for me was, oddly enough:
import sys
sys.path.append(".")
Note the single dot as opposed to the two-dot solutions mentioned so far.
Edit: The following helped clarify this for me:
import os
print (os.getcwd())
In my case, the working directory was (unexpectedly) the root of the project.
This is very tricky in Python.
I'll first comment on why you're having that problem and then I will mention two possible solutions.
What's going on?
You must take this paragraph from the Python documentation into consideration:
Note that relative imports are based on the name of the current
module. Since the name of the main module is always "main",
modules intended for use as the main module of a Python application
must always use absolute imports.
And also the following from PEP 328:
Relative imports use a module's name attribute to determine that
module's position in the package hierarchy. If the module's name does
not contain any package information (e.g. it is set to 'main')
then relative imports are resolved as if the module were a top level
module, regardless of where the module is actually located on the file
system.
Relative imports work from the filename (__name__ attribute), which can take two values:
It's the filename, preceded by the folder strucutre, separated by dots.
For eg: package.test_A.test
Here Python knows the parent directories: before test comes test_A and then package.
So you can use the dot notation for relative import.
# package.test_A/test.py
from ..A import foo
You can then have like a root file in the root directory which calls test.py:
# root.py
from package.test_A import test
When you run the module (test.py) directly, it becomes the entry point to the program , so __name__ == __main__. The filename has no indication of the directory structure, so Python doesn't know how to go up in the directory. For Python, test.py becomes the top-level script, there is nothing above it. That's why you cannot use relative import.
Possible Solutions
A) One way to solve this is to have a root file (in the root directory) which calls the modules/packages, like this:
root.py imports test.py. (entry point, __name__ == __main__).
test.py (relative) imports foo.py.
foo.py says the module has been imported.
The output is:
package.A.foo has been imported
Module's name is: package.test_A.test
B) If you want to execute the code as a module and not as a top-level script, you can try this from the command line:
python -m package.test_A.test
Any suggestions are welcomed.
You should also check: Relative imports for the billionth time , specially BrenBarn's answer.
from package.A import foo
I think it's clearer than
import sys
sys.path.append("..")
As the most popular answer suggests, basically its because your PYTHONPATH or sys.path includes . but not your path to your package. And the relative import is relative to your current working directory, not the file where the import happens; oddly.
You could fix this by first changing your relative import to absolute and then either starting it with:
PYTHONPATH=/path/to/package python -m test_A.test
OR forcing the python path when called this way, because:
With python -m test_A.test you're executing test_A/test.py with __name__ == '__main__' and __file__ == '/absolute/path/to/test_A/test.py'
That means that in test.py you could use your absolute import semi-protected in the main case condition and also do some one-time Python path manipulation:
from os import path
…
def main():
…
if __name__ == '__main__':
import sys
sys.path.append(path.join(path.dirname(__file__), '..'))
from A import foo
exit(main())
This is actually a lot simpler than what other answers make it out to be.
TL;DR: Import A directly instead of attempting a relative import.
The current working directory is not a package, unless you import the folder package from a different folder. So the behavior of your package will work fine if you intend it to be imported by other applications. What's not working is the tests...
Without changing anything in your directory structure, all that needs to be changed is how test.py imports foo.py.
from A import foo
Now running python -m test_A.test from the package directory will run without an ImportError.
Why does that work?
Your current working directory is not a package, but it is added to the path. Therefore you can import folder A and its contents directly. It is the same reason you can import any other package that you have installed... they're all included in your path.
Edit: 2020-05-08: Is seems the website I quoted is no longer controlled by the person who wrote the advice, so I'm removing the link to the site. Thanks for letting me know baxx.
If someone's still struggling a bit after the great answers already provided, I found advice on a website that no longer is available.
Essential quote from the site I mentioned:
"The same can be specified programmatically in this way:
import sys
sys.path.append('..')
Of course the code above must be written before the other import
statement.
It's pretty obvious that it has to be this way, thinking on it after the fact. I was trying to use the sys.path.append('..') in my tests, but ran into the issue posted by OP. By adding the import and sys.path defintion before my other imports, I was able to solve the problem.
Just remove .. in test.py
For me pytest works fine with that
Example:
from A import foo
if you have an __init__.py in an upper folder, you can initialize the import as
import file/path as alias in that init file. Then you can use it on lower scripts as:
import alias
In my case, I had to change to this:
Solution 1(more better which depend on current py file path. Easy to deploy)
Use pathlib.Path.parents make code cleaner
import sys
import os
import pathlib
target_path = pathlib.Path(os.path.abspath(__file__)).parents[3]
sys.path.append(target_path)
from utils import MultiFileAllowed
Solution 2
import sys
import os
sys.path.append(os.getcwd())
from utils import MultiFileAllowed
In my humble opinion, I understand this question in this way:
[CASE 1] When you start an absolute-import like
python -m test_A.test
or
import test_A.test
or
from test_A import test
you're actually setting the import-anchor to be test_A, in other word, top-level package is test_A . So, when we have test.py do from ..A import xxx, you are escaping from the anchor, and Python does not allow this.
[CASE 2] When you do
python -m package.test_A.test
or
from package.test_A import test
your anchor becomes package, so package/test_A/test.py doing from ..A import xxx does not escape the anchor(still inside package folder), and Python happily accepts this.
In short:
Absolute-import changes current anchor (=redefines what is the top-level package);
Relative-import does not change the anchor but confines to it.
Furthermore, we can use full-qualified module name(FQMN) to inspect this problem.
Check FQMN in each case:
[CASE2] test.__name__ = package.test_A.test
[CASE1] test.__name__ = test_A.test
So, for CASE2, an from .. import xxx will result in a new module with FQMN=package.xxx, which is acceptable.
While for CASE1, the .. from within from .. import xxx will jump out of the starting node(anchor) of test_A, and this is NOT allowed by Python.
[2022-07-19] I think this "relative-import" limitation is quite an ugly design, totally against (one of) Python's motto "Simple is better than complex".
Not sure in python 2.x but in python 3.6, assuming you are trying to run the whole suite, you just have to use -t
-t, --top-level-directory directory
Top level directory of project (defaults to start directory)
So, on a structure like
project_root
|
|----- my_module
| \
| \_____ my_class.py
|
\ tests
\___ test_my_func.py
One could for example use:
python3 unittest discover -s /full_path/project_root/tests -t /full_path/project_root/
And still import the my_module.my_class without major dramas.
Having
package/
__init__.py
A/
__init__.py
foo.py
test_A/
__init__.py
test.py
in A/__init__.py import foo:
from .foo import foo
when importing A/ from test_A/
import sys, os
sys.path.append(os.path.abspath('../A'))
# then import foo
import foo
I have a python project structured like this:
repo_dir/
----project_package/
--------__init__.py
--------process.py
--------config.py
----tests/
--------test_process.py
__init__.py is empty
config.py looks like this:
name = 'brian'
USAGE
I use the library by running python process.py from the project/project/ directory, or by specifying the python file path absolutely. I'm running Python 2.7 on Amazon EC2 Linux.
When process.py looks like below, everything works fine and process.py prints brian.
import config
print config.name
When process.py looks like below, I get the error ImportError: No module named project.config.
import project.config
print config.name
When process.py looks like below, I get the error ImportError: No module named project. This makes sense as the same behavior from the previous example should be expected.
from project import config
print config.name
If I add these lines to process.py to include the library root in sys.path, all configurations above, work fine.
import os
import sys
sys.path.append(os.path.abspath(os.path.join(os.path.dirname(__file__), '..')))
MY CONFUSION
Many resources suggest setting up python libraries to import modules using project.module_name, but it doesn't seem like sys.path appending is standard, and seems weird that I need it. I can see that the sys.path append added my library root as a path in sys, but I thought that's what the __init__.py in my library root was supposed to do. What gives? What am I missing? I know Python importing creates lots of headaches so I've tried to simplify this as much as possible to wrap my head around it. I'm going crazy and it's Friday before a holiday. I'm bummed. Please help!!
QUESTIONS
How should I set up my libraries? How should I import packages? Where should I have __init__.py files? Do I need to append my library root to sys.path in every project? Why is this so confusing?
Your project setup is alright. I renamed the directories just for clarity
in this example, but the structure is the same as yours:
repo_dir/
project_package/
__init__.py
process.py
config.py
# Declare your project in a setup.py file, so that
# it will be installable, both by users and by you.
setup.py
When you have a module that wants to import from another module in
the same project, the best approach is to use relative imports. For example:
# In process.py
from .config import name
...
While working on the code on your dev box, do your work in a Python virtualenv,
and pip install your project in "editable" mode.
# From the root of your repo:
pip install -e .
With that approach, you'll never need to muck around with sys.path -- which
is almost always the wrong approach.
I think the problem is how you're running your script. If you want the script to be living in a package (the inner project folder), you should run it with python -m project.process, rather than by filename. Then you can make absolute or explicit relative imports to get config from process.
An absolute import would be from project import config or import project.config.
An explicit relative import would be from . import config.
Python 2 also allows implicit relative imports, but they're a really bad misfeature that you should never use. With implicit relative imports, internal package modules can shadow top-level modules. For instance, a project/json.py file would hide the standard library's json module from all the other modules in the package. You can tell Python you want to forbid implicit relative imports by putting from __future__ import absolute_import at the top of the file. It's the standard behavior in Python 3.
It seems there are already quite some questions here about relative import in python 3, but after going through many of them I still didn't find the answer for my issue.
so here is the question.
I have a package shown below
package/
__init__.py
A/
__init__.py
foo.py
test_A/
__init__.py
test.py
and I have a single line in test.py:
from ..A import foo
now, I am in the folder of package, and I run
python -m test_A.test
I got message
"ValueError: attempted relative import beyond top-level package"
but if I am in the parent folder of package, e.g., I run:
cd ..
python -m package.test_A.test
everything is fine.
Now my question is:
when I am in the folder of package, and I run the module inside the test_A sub-package as test_A.test, based on my understanding, ..A goes up only one level, which is still within the package folder, why it gives message saying beyond top-level package. What is exactly the reason that causes this error message?
EDIT: There are better/more coherent answers to this question in other questions:
Sibling package imports
Relative imports for the billionth time
Why doesn't it work? It's because python doesn't record where a package was loaded from. So when you do python -m test_A.test, it basically just discards the knowledge that test_A.test is actually stored in package (i.e. package is not considered a package). Attempting from ..A import foo is trying to access information it doesn't have any more (i.e. sibling directories of a loaded location). It's conceptually similar to allowing from ..os import path in a file in math. This would be bad because you want the packages to be distinct. If they need to use something from another package, then they should refer to them globally with from os import path and let python work out where that is with $PATH and $PYTHONPATH.
When you use python -m package.test_A.test, then using from ..A import foo resolves just fine because it kept track of what's in package and you're just accessing a child directory of a loaded location.
Why doesn't python consider the current working directory to be a package? NO CLUE, but gosh it would be useful.
import sys
sys.path.append("..") # Adds higher directory to python modules path.
Try this.
Worked for me.
Assumption:
If you are in the package directory, A and test_A are separate packages.
Conclusion:
..A imports are only allowed within a package.
Further notes:
Making the relative imports only available within packages is useful if you want to force that packages can be placed on any path located on sys.path.
EDIT:
Am I the only one who thinks that this is insane!? Why in the world is the current working directory not considered to be a package? – Multihunter
The current working directory is usually located in sys.path. So, all files there are importable. This is behavior since Python 2 when packages did not yet exist. Making the running directory a package would allow imports of modules as "import .A" and as "import A" which then would be two different modules. Maybe this is an inconsistency to consider.
None of these solutions worked for me in 3.6, with a folder structure like:
package1/
subpackage1/
module1.py
package2/
subpackage2/
module2.py
My goal was to import from module1 into module2. What finally worked for me was, oddly enough:
import sys
sys.path.append(".")
Note the single dot as opposed to the two-dot solutions mentioned so far.
Edit: The following helped clarify this for me:
import os
print (os.getcwd())
In my case, the working directory was (unexpectedly) the root of the project.
This is very tricky in Python.
I'll first comment on why you're having that problem and then I will mention two possible solutions.
What's going on?
You must take this paragraph from the Python documentation into consideration:
Note that relative imports are based on the name of the current
module. Since the name of the main module is always "main",
modules intended for use as the main module of a Python application
must always use absolute imports.
And also the following from PEP 328:
Relative imports use a module's name attribute to determine that
module's position in the package hierarchy. If the module's name does
not contain any package information (e.g. it is set to 'main')
then relative imports are resolved as if the module were a top level
module, regardless of where the module is actually located on the file
system.
Relative imports work from the filename (__name__ attribute), which can take two values:
It's the filename, preceded by the folder strucutre, separated by dots.
For eg: package.test_A.test
Here Python knows the parent directories: before test comes test_A and then package.
So you can use the dot notation for relative import.
# package.test_A/test.py
from ..A import foo
You can then have like a root file in the root directory which calls test.py:
# root.py
from package.test_A import test
When you run the module (test.py) directly, it becomes the entry point to the program , so __name__ == __main__. The filename has no indication of the directory structure, so Python doesn't know how to go up in the directory. For Python, test.py becomes the top-level script, there is nothing above it. That's why you cannot use relative import.
Possible Solutions
A) One way to solve this is to have a root file (in the root directory) which calls the modules/packages, like this:
root.py imports test.py. (entry point, __name__ == __main__).
test.py (relative) imports foo.py.
foo.py says the module has been imported.
The output is:
package.A.foo has been imported
Module's name is: package.test_A.test
B) If you want to execute the code as a module and not as a top-level script, you can try this from the command line:
python -m package.test_A.test
Any suggestions are welcomed.
You should also check: Relative imports for the billionth time , specially BrenBarn's answer.
from package.A import foo
I think it's clearer than
import sys
sys.path.append("..")
As the most popular answer suggests, basically its because your PYTHONPATH or sys.path includes . but not your path to your package. And the relative import is relative to your current working directory, not the file where the import happens; oddly.
You could fix this by first changing your relative import to absolute and then either starting it with:
PYTHONPATH=/path/to/package python -m test_A.test
OR forcing the python path when called this way, because:
With python -m test_A.test you're executing test_A/test.py with __name__ == '__main__' and __file__ == '/absolute/path/to/test_A/test.py'
That means that in test.py you could use your absolute import semi-protected in the main case condition and also do some one-time Python path manipulation:
from os import path
…
def main():
…
if __name__ == '__main__':
import sys
sys.path.append(path.join(path.dirname(__file__), '..'))
from A import foo
exit(main())
This is actually a lot simpler than what other answers make it out to be.
TL;DR: Import A directly instead of attempting a relative import.
The current working directory is not a package, unless you import the folder package from a different folder. So the behavior of your package will work fine if you intend it to be imported by other applications. What's not working is the tests...
Without changing anything in your directory structure, all that needs to be changed is how test.py imports foo.py.
from A import foo
Now running python -m test_A.test from the package directory will run without an ImportError.
Why does that work?
Your current working directory is not a package, but it is added to the path. Therefore you can import folder A and its contents directly. It is the same reason you can import any other package that you have installed... they're all included in your path.
Edit: 2020-05-08: Is seems the website I quoted is no longer controlled by the person who wrote the advice, so I'm removing the link to the site. Thanks for letting me know baxx.
If someone's still struggling a bit after the great answers already provided, I found advice on a website that no longer is available.
Essential quote from the site I mentioned:
"The same can be specified programmatically in this way:
import sys
sys.path.append('..')
Of course the code above must be written before the other import
statement.
It's pretty obvious that it has to be this way, thinking on it after the fact. I was trying to use the sys.path.append('..') in my tests, but ran into the issue posted by OP. By adding the import and sys.path defintion before my other imports, I was able to solve the problem.
Just remove .. in test.py
For me pytest works fine with that
Example:
from A import foo
if you have an __init__.py in an upper folder, you can initialize the import as
import file/path as alias in that init file. Then you can use it on lower scripts as:
import alias
In my case, I had to change to this:
Solution 1(more better which depend on current py file path. Easy to deploy)
Use pathlib.Path.parents make code cleaner
import sys
import os
import pathlib
target_path = pathlib.Path(os.path.abspath(__file__)).parents[3]
sys.path.append(target_path)
from utils import MultiFileAllowed
Solution 2
import sys
import os
sys.path.append(os.getcwd())
from utils import MultiFileAllowed
In my humble opinion, I understand this question in this way:
[CASE 1] When you start an absolute-import like
python -m test_A.test
or
import test_A.test
or
from test_A import test
you're actually setting the import-anchor to be test_A, in other word, top-level package is test_A . So, when we have test.py do from ..A import xxx, you are escaping from the anchor, and Python does not allow this.
[CASE 2] When you do
python -m package.test_A.test
or
from package.test_A import test
your anchor becomes package, so package/test_A/test.py doing from ..A import xxx does not escape the anchor(still inside package folder), and Python happily accepts this.
In short:
Absolute-import changes current anchor (=redefines what is the top-level package);
Relative-import does not change the anchor but confines to it.
Furthermore, we can use full-qualified module name(FQMN) to inspect this problem.
Check FQMN in each case:
[CASE2] test.__name__ = package.test_A.test
[CASE1] test.__name__ = test_A.test
So, for CASE2, an from .. import xxx will result in a new module with FQMN=package.xxx, which is acceptable.
While for CASE1, the .. from within from .. import xxx will jump out of the starting node(anchor) of test_A, and this is NOT allowed by Python.
[2022-07-19] I think this "relative-import" limitation is quite an ugly design, totally against (one of) Python's motto "Simple is better than complex".
Not sure in python 2.x but in python 3.6, assuming you are trying to run the whole suite, you just have to use -t
-t, --top-level-directory directory
Top level directory of project (defaults to start directory)
So, on a structure like
project_root
|
|----- my_module
| \
| \_____ my_class.py
|
\ tests
\___ test_my_func.py
One could for example use:
python3 unittest discover -s /full_path/project_root/tests -t /full_path/project_root/
And still import the my_module.my_class without major dramas.
Having
package/
__init__.py
A/
__init__.py
foo.py
test_A/
__init__.py
test.py
in A/__init__.py import foo:
from .foo import foo
when importing A/ from test_A/
import sys, os
sys.path.append(os.path.abspath('../A'))
# then import foo
import foo
I know this is a dumb question but i'm stumped. My directory structure used to look like this:
-src
|
-module.py
-program.py
when this what my directory structure, I referenced module from program and all was well.
I've since changed my directory structure to this:
-src
|
-__init.py
-module.py
|
-programDir
|
-__init.py
-program.py
now, of course, I can't reach the module from program. How can I reference src as a package. I tried to create an
__init__.py
file in the src directory, but no luck.
Moar deets:
import statements i've tried in program.py:
import module
and
from src import module
the first one worked when the other module and program were in the same directory.
error i'm getting:
ImportError: No module named module
and just for the record: No, my module and program are not called module OR program
update: I've tried this in my program.py file:
from ...src import module
and
from ..src import module
both are giving me:
ValueError: Attempted relative import in non-package
For starters, I recommend reading the entry Modifying Python's Search Path in the docs.
It might be frowned upon by some, but if you wish to modify the PYTHONPATH from within your program, according to the documentation's standard modules entry you can use the sys.path.append method:
import sys
sys.path.append('..')
import module
Couldn't you use PEP 328 to solve this?
If you run program.py directly, with python program.py or with #!, then module.py's directory should be in the PYTHONPATH for import module to work. This can be achieved using a helper shell script that's kept in programDir, for instance, and looks something like:
#!/bin/bash
script_dir=`dirname $0`
# Add the script's parent directory to the PYTHONPATH
export PYTHONPATH=$PYTHONPATH:$script_dir/..
python $script_dir/program.py
Another, probably better, way would be to have program.py export a "main()" function, and create a helper python script at src/program that looks like:
#!/usr/bin/env python
from programDir.program import main
main()
In this case, you can use relative imports in src/programDir/program.py, so this should work:
from .. import module
The first one worked because Python's sys.path's first entry is '' which means it will look for module names in the current working directory from which you've executed the Python interpreter.
The issue you seem to have is that the directory located at src is not set on your PYTHONPATH. So, you can do is set the PYTHONPATH environment variable explicitly.
Here's an example using bash:
export PYTHONPATH=PATH_TO_SRC:${PYTHONPATH}
then run your program as normal
Another approach is that you can explicitly set sys.path by appending to it upon execution of your program.
So, in your program.py, you would have:
if __name__ == '__main__':
import os
import sys
sys.path.append(os.path.dirname(os.path.dirname(__file__)))
your_main_function()
Lastly, for serious python development, you should consider virtualenv and virtualenvwrapper as it will take care of most of these things for you.
You need to add __init__.py to /programDir to interpret the directory as a package. Once a package, you can import the package's contents.
So, in your case, if /src is on the PYTHONPATH, from module.py you can import program.py with from programDir import program.
If you use program as part of a package, in another python module, such as
import src.programDir.program as p
p.some_method()
you can use relative import in program.py, assuming you are creating a package with src (__init__.py in both src and programDir)
from .. import module
If not, for example you are calling program.py from the command line, you must add the directory containing src to your search path either by modifying sys.path or the PYTHONPATH env var, before importing.