pytest integration - how to properly import - python

I am using py.test to test my python code. The relevant structure of my project is
-myproject
file.py
file2.py
-test/
input.txt
input2.dat
test_part.py
-regress/
file2_old.py
__init__.py
-test/
test_file2_regression.py
test_part.py imports file and file2 and test_file2_regression.py imports file2 and regress.file2_old. If I run pytest in the console, I get the import errors, that the packages don't exist. Running python -m pytest on the other hand works perfectly fine, but only if I run it from the myproject/ directory.
What is the correct way to do this, to get it working from anywhere in my project? I already tried modifying the PYTHONPATH but I honestly have no idea how to properly do it.
Further information:
I don't have any setup files, and my __init__s are just empty files. If manipulating the PYTHONPATH is necessary it needs to be done relative to myproject as I use it on several machines. I am running python 2.7.
I already check out:
Importing correctly with pytest
Good Integration Practices
but it didn't really help me.

The solution which works with the pytest command from any directory in the project is to include before the imports in the test*.py files:
import os
from sys import path
PATH = os.path.abspath(os.path.dirname(__file__))
path.append(os.path.join(PATH, os.pardir, os.pardir))
where the proper number of os.pardir is used to navigate to the project directory, from there the __init__.py files allow to import the modules.
Both argv[0] and inspect.getsourcefile don't provide the necessary information. argv[0] contains the location of the used py.test module and the getsourcefile method simply returns None.
Edit: since Python 3.4 we can use instead of os.path the modern pathlib:
from pathlib import Path
from sys import path
PATH = Path(__file__).resolve()
path.append(PATH.parents[2])

Having the same issue and similar success in searching for "the best way of doing this", I concluded for myself to avoid the situation whenever possible (by running the actual scripts consequently from top level), but to answer your question, my current approach (for instance for unit testing from a parallel folder) is
from sys import argv, path
from os.path import dirname, join
path.append(join(dirname(argv[0]), ".."))
This makes the interpreter also search in the folder above where the script is started. Another approach (instead of using argv) is to use the introspect module to obtain the filename. These work better for me than using __file__, as the latter is not always defined.
Edit 29.10.:
Alternative to argv[0] is to use
from inspect import getsourcefile
from sys import path
from os.path import dirname, join
this_file = getsourcefile(lambda _: None)
path.append(join(dirname(this_file), ".."))
I hope this will work at least for the requested purpose, see also How do I get the path of the current executed file in Python?.
The simplest - if it works in your case - is of course:
from os.path import dirname, join
path.append(join(dirname(__file__), ".."))

Related

PEP-8: module at top of file

Desiring to improve my Python style, I ran a PEP-8 style checker on one of my script and it complained about something I don't know how to fix. The prologue of the script is something like:
#! /bin/env python3
import sys
import os
exe_name = os.path.basename(os.path.realpath(__file__))
bin_dir = os.path.dirname(os.path.realpath(__file__))
inst_dir = os.path.dirname(bin_dir)
sys.path.insert(0, inst_dir+'/path/to/packages')
import mypackage.mymodule
and the style checker complain on the import mymodule line, stating that it should be a top of file. But I obviously can't move it before setting the path where it will be found. Is there a good way to achieve this (mandating an environment variable or a shell wrapper are not what I find better than my current code) while respecting PEP-8 recommendations at the same time?
If you want to avoid path manipulation, you may be able to do so by using the under-known .pth feature.
sys.path should begin with the directory containing the main program either by name or by reference as ''. I assume that the file importing mymodule is not part of mypackage, so that the '' entry is not useful for importing mymodule.
sys.path should end with the site-packages directory for the executing binary. That is the normal place for added packages. If you do not want to move mypackage into site-packages, you can extend the latter 'vitually' by putting a mystuff.pth file in it. It should contain one line: the path to the directory containing mypackage. Call it myprojects. Then mypackage and any other package in myprojects can be imported as if they were in site-packages.
One advantage of .pth files is that you can put identical copies in multiple site-packages directories. For instance, I have multiple projects in F:/python. I have multiple versions of Python installed. So I have put python.pth containing that one line in the site-packages for each.
The best strategy would be to put the sys.path related code in separate file and import it in working code file.
So, I will split above code in two files. One named my_module.py and other named working_module.py
Now my_module will contain below lines
import sys
import os
exe_name = os.path.basename(os.path.realpath(__file__))
bin_dir = os.path.dirname(os.path.realpath(__file__))
inst_dir = os.path.dirname(bin_dir)
sys.path.insert(0, inst_dir+'/path/to/packages')
and working_module will contain
import my_module
import mypackage.mymodule
because we are importing my_module before mypackage, it will execute the path related code before and your package will be available in path.
You can use importlib module (python 3.5) or imp for python 2.7 to load mypackage.mymodule programatically. Both have the same purpose:
mechanisms used to implement the import statement
This question might help you more:
How to import a module given the full path?
https://docs.python.org/3/library/importlib.html#examples
https://docs.python.org/2/library/imp.html

Attempted relative import in non-package for optional package [duplicate]

Imagine this directory structure:
app/
__init__.py
sub1/
__init__.py
mod1.py
sub2/
__init__.py
mod2.py
I'm coding mod1, and I need to import something from mod2. How should I do it?
I tried from ..sub2 import mod2 but I'm getting an "Attempted relative import in non-package".
I googled around but found only "sys.path manipulation" hacks. Isn't there a clean way?
Edit: all my __init__.py's are currently empty
Edit2: I'm trying to do this because sub2 contains classes that are shared across sub packages (sub1, subX, etc.).
Edit3: The behaviour I'm looking for is the same as described in PEP 366 (thanks John B)
Everyone seems to want to tell you what you should be doing rather than just answering the question.
The problem is that you're running the module as '__main__' by passing the mod1.py as an argument to the interpreter.
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.
In Python 2.6, they're adding the ability to reference modules relative to the main module. PEP 366 describes the change.
Update: According to Nick Coghlan, the recommended alternative is to run the module inside the package using the -m switch.
Here is the solution which works for me:
I do the relative imports as from ..sub2 import mod2
and then, if I want to run mod1.py then I go to the parent directory of app and run the module using the python -m switch as python -m app.sub1.mod1.
The real reason why this problem occurs with relative imports, is that relative imports works by taking the __name__ property of the module. If the module is being directly run, then __name__ is set to __main__ and it doesn't contain any information about package structure. And, thats why python complains about the relative import in non-package error.
So, by using the -m switch you provide the package structure information to python, through which it can resolve the relative imports successfully.
I have encountered this problem many times while doing relative imports. And, after reading all the previous answers, I was still not able to figure out how to solve it, in a clean way, without needing to put boilerplate code in all files. (Though some of the comments were really helpful, thanks to #ncoghlan and #XiongChiamiov)
Hope this helps someone who is fighting with relative imports problem, because going through PEP is really not fun.
main.py
setup.py
app/ ->
__init__.py
package_a/ ->
__init__.py
module_a.py
package_b/ ->
__init__.py
module_b.py
You run python main.py.
main.py does: import app.package_a.module_a
module_a.py does import app.package_b.module_b
Alternatively 2 or 3 could use: from app.package_a import module_a
That will work as long as you have app in your PYTHONPATH. main.py could be anywhere then.
So you write a setup.py to copy (install) the whole app package and subpackages to the target system's python folders, and main.py to target system's script folders.
"Guido views running scripts within a package as an anti-pattern" (rejected
PEP-3122)
I have spent so much time trying to find a solution, reading related posts here on Stack Overflow and saying to myself "there must be a better way!". Looks like there is not.
This is solved 100%:
app/
main.py
settings/
local_setings.py
Import settings/local_setting.py in app/main.py:
main.py:
import sys
sys.path.insert(0, "../settings")
try:
from local_settings import *
except ImportError:
print('No Import')
explanation of nosklo's answer with examples
note: all __init__.py files are empty.
main.py
app/ ->
__init__.py
package_a/ ->
__init__.py
fun_a.py
package_b/ ->
__init__.py
fun_b.py
app/package_a/fun_a.py
def print_a():
print 'This is a function in dir package_a'
app/package_b/fun_b.py
from app.package_a.fun_a import print_a
def print_b():
print 'This is a function in dir package_b'
print 'going to call a function in dir package_a'
print '-'*30
print_a()
main.py
from app.package_b import fun_b
fun_b.print_b()
if you run $ python main.py it returns:
This is a function in dir package_b
going to call a function in dir package_a
------------------------------
This is a function in dir package_a
main.py does: from app.package_b import fun_b
fun_b.py does from app.package_a.fun_a import print_a
so file in folder package_b used file in folder package_a, which is what you want. Right??
def import_path(fullpath):
"""
Import a file with full path specification. Allows one to
import from anywhere, something __import__ does not do.
"""
path, filename = os.path.split(fullpath)
filename, ext = os.path.splitext(filename)
sys.path.append(path)
module = __import__(filename)
reload(module) # Might be out of date
del sys.path[-1]
return module
I'm using this snippet to import modules from paths, hope that helps
This is unfortunately a sys.path hack, but it works quite well.
I encountered this problem with another layer: I already had a module of the specified name, but it was the wrong module.
what I wanted to do was the following (the module I was working from was module3):
mymodule\
__init__.py
mymodule1\
__init__.py
mymodule1_1
mymodule2\
__init__.py
mymodule2_1
import mymodule.mymodule1.mymodule1_1
Note that I have already installed mymodule, but in my installation I do not have "mymodule1"
and I would get an ImportError because it was trying to import from my installed modules.
I tried to do a sys.path.append, and that didn't work. What did work was a sys.path.insert
if __name__ == '__main__':
sys.path.insert(0, '../..')
So kind of a hack, but got it all to work!
So keep in mind, if you want your decision to override other paths then you need to use sys.path.insert(0, pathname) to get it to work! This was a very frustrating sticking point for me, allot of people say to use the "append" function to sys.path, but that doesn't work if you already have a module defined (I find it very strange behavior)
Let me just put this here for my own reference. I know that it is not good Python code, but I needed a script for a project I was working on and I wanted to put the script in a scripts directory.
import os.path
import sys
sys.path.append(os.path.abspath(os.path.join(os.path.dirname(__file__), "..")))
As #EvgeniSergeev says in the comments to the OP, you can import code from a .py file at an arbitrary location with:
import imp
foo = imp.load_source('module.name', '/path/to/file.py')
foo.MyClass()
This is taken from this SO answer.
Take a look at http://docs.python.org/whatsnew/2.5.html#pep-328-absolute-and-relative-imports. You could do
from .mod1 import stuff
From Python doc,
In Python 2.5, you can switch import‘s behaviour to absolute imports using a from __future__ import absolute_import directive. This absolute- import behaviour will become the default in a future version (probably Python 2.7). Once absolute imports are the default, import string will always find the standard library’s version. It’s suggested that users should begin using absolute imports as much as possible, so it’s preferable to begin writing from pkg import string in your code
I found it's more easy to set "PYTHONPATH" enviroment variable to the top folder:
bash$ export PYTHONPATH=/PATH/TO/APP
then:
import sub1.func1
#...more import
of course, PYTHONPATH is "global", but it didn't raise trouble for me yet.
On top of what John B said, it seems like setting the __package__ variable should help, instead of changing __main__ which could screw up other things. But as far as I could test, it doesn't completely work as it should.
I have the same problem and neither PEP 328 or 366 solve the problem completely, as both, by the end of the day, need the head of the package to be included in sys.path, as far as I could understand.
I should also mention that I did not find how to format the string that should go into those variables. Is it "package_head.subfolder.module_name" or what?
You have to append the module’s path to PYTHONPATH:
export PYTHONPATH="${PYTHONPATH}:/path/to/your/module/"
A hacky way to do it is to append the current directory to the PATH at runtime as follows:
import pathlib
import sys
sys.path.append(pathlib.Path(__file__).parent.resolve())
import file_to_import # the actual intended import
In contrast to another solution for this question this uses pathlib instead of os.path.
This method queries and auto populates the path:
import os
import inspect
currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
parentdir = os.path.dirname(currentdir)
os.sys.path.insert(1, parentdir)
# print("currentdir = ", currentdir)
# print("parentdir=", parentdir)
What a debate!
Relative newcomer to python (but years of programming experience, and dislike of perl). Relative lay-person when it comes to the dark art of Apache setup, but I know what I (think I) need to get my little experimental projects working at home.
Here is my summary of what the situ seems to be.
If I use the -m 'module' approach, I need to:-
dot it all together;
run it from a parent folder;
lose the '.py';
create an empty (!) __init__.py file in every sub-folder.
How does that work in a cgi environment, where I have aliased my scripts directory, and want to run a script directly as /dirAlias/cgi_script.py??
Why is amending sys.path a hack? The python docs page states: "A program is free to modify this list for its own purposes." If it works, it works, right? The bean counters in Accounts don't care how it works.
I just want to go up one level and down into a 'modules' dir:-
.../py
/cgi
/build
/modules
so my 'modules' can be imported from either the cgi world or the server world.
I've tried the -m/modules approach but I think I prefer the following (and am not confused how to run it in cgi-space):-
Create XX_pathsetup.py in the /path/to/python/Lib dir (or any other dir in the default sys.path list). 'XX' is some identifier that declares an intent to setup my path according to the rules in the file.
In any script that wants to be able to import from the 'modules' dir in above directory config, simply import XX_pathsetup.py.
And here's my really simple XX_pathsetup.py:
import sys, os
pypath = sys.path[0].rsplit(os.sep,1)[0]
sys.path.insert( 0, pypath+os.sep+'modules' )
Not a 'hack', IMHO. 1 small file to put in the python 'Lib' dir, one import statement which declares intent to modify the path search order.

Using relative imports in python with mutliple parent and children folders

I had a small problem with importing a script that was in a parent folder, but I managed to resolve it using:
import sys
sys.path.append("../")
My directory is like this:
Data
|->->code
|->->script1.py
|->->->->subfolder
|->->->script2.py
When I run script2 (which imports script1) from the subfolder directory, the script runs without problems. But if I try to run script2 from the code directory using:
:~ ./subfolder/script2.py
I get an error :
ImportError: No module named script1
I tried using relative imports but because my code is not structured in packages it doesn't work. Is there a way I can run script2 from both directories (the parent and the child) and still be able to import script1 everytime?
Thank you in advance,
Georgi Nikolov
EDIT: Ok, after I read through all the suggestions, I did a "simple" hack which is quite ugly in my opinion but works quite well:
import sys
parent_folder = sys.path[0].split("/subfolder")[0]
sys.path.append(parent_folder)
import script1
Now I can even call script2 from the root and it will manage to import script1
To import a module that is up a level, you can use this.
import os, sys
sys.path.append(os.path.join(os.path.dirname(__file__), ".."))
An Explanation
__file__ # The full path to your running file.
os.path.dirname # See below. (1)
os.path.join # See below. (2)
sys.path.append # See below. (3)
".." # Universal for "up a level".
os.path.dirname(path) -
Return the directory name of pathname path.
os.path.join(path, *paths) -
Join one or more path components intelligently.
sys.path -
A list of strings that specifies the search path for modules.
You can add a string with the append method.
You should use absolute paths:
import sys, os
HERE = os.path.abspath(os.path.dirname(__file__))
sys.path.append(os.path.join(HERE, ".."))
But better do not do this: Guido views running scripts within a package as an anti-pattern
You should put the standalone scripts in the root folder of the project. A script should not be used as a module and as the main script.
If you cannot move the script to the root, make another bootstrap script at the root which will import your service script running a dedicated function in it.

relative path not working even with __init__.py

I know that there are plenty of similar questions on stack overflow. But the common answer doesn't seem to be working for me.
I have a file structure like this
proj/
lib/
__init__.py
aa.py
bb.py
test/
__init__.py
aa_test.py
I figured that if I include the code in my test.py
import lib.aa
or
from lib import aa
I would be able to reference the modules in the lib/ directory. But that did not work.
So I tried to add to path, and it adds it correctly:
os.environ["PATH"] += ":%s" % os.path.abspath(os.path.join("..",""))
print os.environ["PATH"]
but even now when I try the import statements above... I keep getting the error
ImportError: No module named aa
or
ImportError: Importing from non-package <Something...>
Is there something obvious I am missing?
Is there a way to check if I have configured my __init__.py files correctly, or to see my package hierarchy?
You need to update your sys.path, which is where python looks for modules, as opposed to your system's path in the current environment, which is what os.environ["PATH"] is referring to.
Example:
import os, sys
sys.path.insert(0, os.path.abspath(".."))
import aa
After doing this, you can use your functions in aa like this: aa.myfunc()
There's some more information in the accepted answer for python: import a module from a directory
The lib directory needs to be in your python module search path, which isn't the same things as the search path used by your shell.
This will probably work for you:
import sys, os
sys.path.append(os.path.abspath(".."))
However, it is probably better to run your code from a context where the lib package is already on the path. Such as from the 'proj' directory.
Where is the code that you're trying to import lib.aa from? I'm guessing /proj/ is not your working directory and it would need to be as it's setup right now. Instead of PATH, you would want to add your directory to PYTHONPATH so it appears in the search path for an import. See http://docs.python.org/tutorial/modules.html#the-module-search-path
Also, please take a look at http://as.ynchrono.us/2007/12/filesystem-structure-of-python-project_21.html It strongly recommends you put an extra level of directory in place so instead of lib.aa, you would refer to it as my_proj.lib.aa.
I had similar problems and here is my advice.
Instead of changing sys.path, better run your test.py from being in proj (i.e. project root) directory. This way project dir will automatically be in sys.path and you will be able to import lib package.
And use absolute imports.
System PATH variable is not used by python import statement. It uses PYTHONPATH, but best way to add new directory to import search path is to modify sys.path.
If this does not help, add to the question your value of sys.path and value returned by os.getcwd().
In Sublime Text 3, abspath it didn't work for me.
I use this instead in the top "__ init __.py" file
Hope it works for you.
from os.path import dirname
from sys import path
path.insert( 0 , dirname( __file__ ) ) ;
from test import aa_test

How to do relative imports in Python?

Imagine this directory structure:
app/
__init__.py
sub1/
__init__.py
mod1.py
sub2/
__init__.py
mod2.py
I'm coding mod1, and I need to import something from mod2. How should I do it?
I tried from ..sub2 import mod2 but I'm getting an "Attempted relative import in non-package".
I googled around but found only "sys.path manipulation" hacks. Isn't there a clean way?
Edit: all my __init__.py's are currently empty
Edit2: I'm trying to do this because sub2 contains classes that are shared across sub packages (sub1, subX, etc.).
Edit3: The behaviour I'm looking for is the same as described in PEP 366 (thanks John B)
Everyone seems to want to tell you what you should be doing rather than just answering the question.
The problem is that you're running the module as '__main__' by passing the mod1.py as an argument to the interpreter.
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.
In Python 2.6, they're adding the ability to reference modules relative to the main module. PEP 366 describes the change.
Update: According to Nick Coghlan, the recommended alternative is to run the module inside the package using the -m switch.
Here is the solution which works for me:
I do the relative imports as from ..sub2 import mod2
and then, if I want to run mod1.py then I go to the parent directory of app and run the module using the python -m switch as python -m app.sub1.mod1.
The real reason why this problem occurs with relative imports, is that relative imports works by taking the __name__ property of the module. If the module is being directly run, then __name__ is set to __main__ and it doesn't contain any information about package structure. And, thats why python complains about the relative import in non-package error.
So, by using the -m switch you provide the package structure information to python, through which it can resolve the relative imports successfully.
I have encountered this problem many times while doing relative imports. And, after reading all the previous answers, I was still not able to figure out how to solve it, in a clean way, without needing to put boilerplate code in all files. (Though some of the comments were really helpful, thanks to #ncoghlan and #XiongChiamiov)
Hope this helps someone who is fighting with relative imports problem, because going through PEP is really not fun.
main.py
setup.py
app/ ->
__init__.py
package_a/ ->
__init__.py
module_a.py
package_b/ ->
__init__.py
module_b.py
You run python main.py.
main.py does: import app.package_a.module_a
module_a.py does import app.package_b.module_b
Alternatively 2 or 3 could use: from app.package_a import module_a
That will work as long as you have app in your PYTHONPATH. main.py could be anywhere then.
So you write a setup.py to copy (install) the whole app package and subpackages to the target system's python folders, and main.py to target system's script folders.
"Guido views running scripts within a package as an anti-pattern" (rejected
PEP-3122)
I have spent so much time trying to find a solution, reading related posts here on Stack Overflow and saying to myself "there must be a better way!". Looks like there is not.
This is solved 100%:
app/
main.py
settings/
local_setings.py
Import settings/local_setting.py in app/main.py:
main.py:
import sys
sys.path.insert(0, "../settings")
try:
from local_settings import *
except ImportError:
print('No Import')
explanation of nosklo's answer with examples
note: all __init__.py files are empty.
main.py
app/ ->
__init__.py
package_a/ ->
__init__.py
fun_a.py
package_b/ ->
__init__.py
fun_b.py
app/package_a/fun_a.py
def print_a():
print 'This is a function in dir package_a'
app/package_b/fun_b.py
from app.package_a.fun_a import print_a
def print_b():
print 'This is a function in dir package_b'
print 'going to call a function in dir package_a'
print '-'*30
print_a()
main.py
from app.package_b import fun_b
fun_b.print_b()
if you run $ python main.py it returns:
This is a function in dir package_b
going to call a function in dir package_a
------------------------------
This is a function in dir package_a
main.py does: from app.package_b import fun_b
fun_b.py does from app.package_a.fun_a import print_a
so file in folder package_b used file in folder package_a, which is what you want. Right??
def import_path(fullpath):
"""
Import a file with full path specification. Allows one to
import from anywhere, something __import__ does not do.
"""
path, filename = os.path.split(fullpath)
filename, ext = os.path.splitext(filename)
sys.path.append(path)
module = __import__(filename)
reload(module) # Might be out of date
del sys.path[-1]
return module
I'm using this snippet to import modules from paths, hope that helps
This is unfortunately a sys.path hack, but it works quite well.
I encountered this problem with another layer: I already had a module of the specified name, but it was the wrong module.
what I wanted to do was the following (the module I was working from was module3):
mymodule\
__init__.py
mymodule1\
__init__.py
mymodule1_1
mymodule2\
__init__.py
mymodule2_1
import mymodule.mymodule1.mymodule1_1
Note that I have already installed mymodule, but in my installation I do not have "mymodule1"
and I would get an ImportError because it was trying to import from my installed modules.
I tried to do a sys.path.append, and that didn't work. What did work was a sys.path.insert
if __name__ == '__main__':
sys.path.insert(0, '../..')
So kind of a hack, but got it all to work!
So keep in mind, if you want your decision to override other paths then you need to use sys.path.insert(0, pathname) to get it to work! This was a very frustrating sticking point for me, allot of people say to use the "append" function to sys.path, but that doesn't work if you already have a module defined (I find it very strange behavior)
Let me just put this here for my own reference. I know that it is not good Python code, but I needed a script for a project I was working on and I wanted to put the script in a scripts directory.
import os.path
import sys
sys.path.append(os.path.abspath(os.path.join(os.path.dirname(__file__), "..")))
As #EvgeniSergeev says in the comments to the OP, you can import code from a .py file at an arbitrary location with:
import imp
foo = imp.load_source('module.name', '/path/to/file.py')
foo.MyClass()
This is taken from this SO answer.
Take a look at http://docs.python.org/whatsnew/2.5.html#pep-328-absolute-and-relative-imports. You could do
from .mod1 import stuff
From Python doc,
In Python 2.5, you can switch import‘s behaviour to absolute imports using a from __future__ import absolute_import directive. This absolute- import behaviour will become the default in a future version (probably Python 2.7). Once absolute imports are the default, import string will always find the standard library’s version. It’s suggested that users should begin using absolute imports as much as possible, so it’s preferable to begin writing from pkg import string in your code
I found it's more easy to set "PYTHONPATH" enviroment variable to the top folder:
bash$ export PYTHONPATH=/PATH/TO/APP
then:
import sub1.func1
#...more import
of course, PYTHONPATH is "global", but it didn't raise trouble for me yet.
On top of what John B said, it seems like setting the __package__ variable should help, instead of changing __main__ which could screw up other things. But as far as I could test, it doesn't completely work as it should.
I have the same problem and neither PEP 328 or 366 solve the problem completely, as both, by the end of the day, need the head of the package to be included in sys.path, as far as I could understand.
I should also mention that I did not find how to format the string that should go into those variables. Is it "package_head.subfolder.module_name" or what?
You have to append the module’s path to PYTHONPATH:
export PYTHONPATH="${PYTHONPATH}:/path/to/your/module/"
A hacky way to do it is to append the current directory to the PATH at runtime as follows:
import pathlib
import sys
sys.path.append(pathlib.Path(__file__).parent.resolve())
import file_to_import # the actual intended import
In contrast to another solution for this question this uses pathlib instead of os.path.
This method queries and auto populates the path:
import os
import inspect
currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
parentdir = os.path.dirname(currentdir)
os.sys.path.insert(1, parentdir)
# print("currentdir = ", currentdir)
# print("parentdir=", parentdir)
What a debate!
Relative newcomer to python (but years of programming experience, and dislike of perl). Relative lay-person when it comes to the dark art of Apache setup, but I know what I (think I) need to get my little experimental projects working at home.
Here is my summary of what the situ seems to be.
If I use the -m 'module' approach, I need to:-
dot it all together;
run it from a parent folder;
lose the '.py';
create an empty (!) __init__.py file in every sub-folder.
How does that work in a cgi environment, where I have aliased my scripts directory, and want to run a script directly as /dirAlias/cgi_script.py??
Why is amending sys.path a hack? The python docs page states: "A program is free to modify this list for its own purposes." If it works, it works, right? The bean counters in Accounts don't care how it works.
I just want to go up one level and down into a 'modules' dir:-
.../py
/cgi
/build
/modules
so my 'modules' can be imported from either the cgi world or the server world.
I've tried the -m/modules approach but I think I prefer the following (and am not confused how to run it in cgi-space):-
Create XX_pathsetup.py in the /path/to/python/Lib dir (or any other dir in the default sys.path list). 'XX' is some identifier that declares an intent to setup my path according to the rules in the file.
In any script that wants to be able to import from the 'modules' dir in above directory config, simply import XX_pathsetup.py.
And here's my really simple XX_pathsetup.py:
import sys, os
pypath = sys.path[0].rsplit(os.sep,1)[0]
sys.path.insert( 0, pypath+os.sep+'modules' )
Not a 'hack', IMHO. 1 small file to put in the python 'Lib' dir, one import statement which declares intent to modify the path search order.

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