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Mock.patch in python unittest could work for two paths

I have a dir which has such structure:
├── aaa.py
├── src
│   └── subsrc
│   ├── else.py
│   └── util.py (there is a "foo" function")
└── tests
├── __init__.py
└── unittests
├── __init__.py
└── test_aaa.py
so "aaa.py", "tests" dir and "src" dir are in project root. and in "test_aaa.py", I use mock to mock function in "util.py":
from src.subsrc.util import foo
import pytest
from unittest import mock
#mock.patch("src.subsrc.util.foo")
def test_foo(mock):
mock.return_value = 111
and then I run python3.7 -m pytest inside "unittests" dir, it worked. This makes sense to me since pytest will find the first dir without __init__.py and then add it to PATH(in this case project root dir will be added) so it could find "src.subsrc.util.foo".
But then I made a small change to "test_aaa.py", in its "mock.patch", I added "aaa" at the beginning:
from src.subsrc.util import foo
import pytest
from unittest import mock
#mock.patch("aaa.src.subsrc.util.foo")
def test_foo(mock):
mock.return_value = 111
it still worked, "aaa.py" is an executable, in "aaa.py":
#!python3.7
from src.subsrc.else import other
if __name__ = "__main__":
# ...
pass
I am very confused why #mock.patch("aaa.src.subsrc.util.foo") also worked, is Python so smart that it could ignore 'aaa' then go "src.subsrc.." to find what it needs? Thanks!
update:
I suspect if because "aaa.py"'s name is special so I changed it to different names, but it still worked. Like I change it to "bbb.py", then in mock.patch, "aaa.src..." does not work but "bbb.src..." still worked. So I am sure "mock.patch" find this executable first.
update:
I guess it could be related to how "mock.patch()" works?

Your example seems to be a bit too stripped-down, but I'll try to expand it in order to explain. When reading about mocks in Python, you will often encounter the phrase "mock it where it's used", which isn't really helpful if you are new to the topic (but here's an excellent article on this concept).
In your test_aaa.py you will probably want to test some functionality of your aaa.py module, which may call some function from src/subsrc/util.py. After importing your foo() function in the aaa.py module, that's the exact location where you should point #mock.patch to: #mock.patch("aaa.foo"). By doing this, your mock will have access to all invocations of foo() in the functions you are about to test, namely aaa.do_something(). I've expanded your example as follows:
# aaa.py
from src.subsrc.util import foo
def do_something():
return foo()
if __name__ == "__main__":
value = do_something()
print(f"value is {value}")
# src/subsrc/util.py
def foo():
return 222
# tests/unittests/test_aaa.py
from unittest import mock
from aaa import do_something
#mock.patch("aaa.foo")
def test_foo(foo_mocked):
foo_mocked.return_value = 111
value = do_something()
assert value == 111
When executing this like python aaa.py, I get the output as expected (value is 222) while the test passes with its assert value == 111.
In your example, #mock.patch("src.subsrc.util.foo") obviously worked, but probably didn't do what you intended. From your example code, I cannot see how #mock.patch("aaa.src.subsrc.util.foo") shouldn't have returned a ModuleNotFoundError.

Trouble instantiating class in a subdirectory

Have a module in a subdirectory and when I try to import it, I get a NameError: namefoois not defined. When I put the class code directly into the __main__.py file it works fine. __init__.py files are empty.
I've tried the following all with other errors:
MyProject/
├── __init__.py
├── __main__.py
├── foo/
│ ├── bar.py
│ ├── __init__.py
bar.py
class Bar:
def __init__(self):
print( 'am here' )
pass
__main__.py
from MyProject import foo
#from MyProject import bar # errors with cannot import bar from MyProject
#from foo import bar # errors with No module named foo
if __name__ == '__main__':
w = Bar()
Is there perhaps a better way to organise this?

The Bar class is in the file bar.py, so I think you'd need to do
from MyProject.foo import bar
w = bar.Bar()
or
from MyProject.foo.bar import Bar
w = Bar()

You didn't share your foo/__init__.py, but you could fix the situation by adding something like this to it:
from .bar import Bar
That adds Bar to the namespace of foo and causes Python to find Bar when you just import foo.
However, you would probably do well to look at a few standard examples for writing a package. For one, you probably shouldn't name your project MyProject, as that name signals it's a class (with TitleCase). Naming it 'project' further confuses the issue, as it appears you're writing a package, so my_package sounds about right.

If you dont know from what direcrory file will be runed then use . to show where package is
from .foo.bar import Bar
or
from .foo import bar
w = bar.Bar()
or
import .foo
w = foo.bar.Bar()
. before package name means that package lockated in same directory with current file

How to execute python file from other directory?

I've got this structure:
│
├ main.py
├ dir
| ├─ data.txt
| └─ other.py
Contents from other.py:
print(open('data.txt', 'utf-8').read())
I run main.py. It must start dir/other.py.
But other.py for works needs data.txt. Is there a way to start other.py from main.py, not editing other.py?
Note
User must be able to start other.py manualy without any errors

For this purpose you can use the import keyword. All you have to do is create an __init__.py script under the dir directory which will define the directory as a library. Then you can just use import others in the main script.
It is recommended to modify the others.py script with the below snippet
if __name__ == '__main__':
// do stuff
otherwise it will execute the library each time you import it
update
It is far more simple. You just have to change directory with the os.chdir("./dir") call. After that you can run a simple import and the script will be executed.
./dir/other.py:
print("Module starts")
print(open('data', 'r').read())
print("Module ends")
./main.py
print("Main start")
import os
os.chdir("./dir")
from others import other
print("Main end" )

You can import other in main file like from dir.other import *

Importlib.import_module will not import the module even though the param is the abs path

I have my .py module which is in C:\Python_Projects\MyModules\ with the name button_generator.py.
My code goes something like this:
module_path='C:\\Python_Projects\\MyModules'
module_name='button_generator.py'
sys.path.append(module_path)
try:
limp=importlib.import_module(module_name.split('.')[0])
except:
print 'module import error'
I have tried other versions aswell:
importlib.import_module(module_name) without the split
importlib.import_module('C:\Python_Projects\MyModules\button_generator.py')
importlib.import_module('C:\Python_Projects\MyModules\button_generator')
The folder C:\Python_Projects\MyModules is in my sys.path as I checked during debug.
Why wouldn't the module import?

I suggest you to reorder your project directories and avoid calling other modules which are not in your current directory project. You'll avoid those kind of errors.
For example, let's organize our project directories and folders to look something like this:
MyProjectFolder/
├── main.py
└── modules
├── __init__.py
└── MyLib.py
NB: Don't forget to add an empty file called __init__.py
MyLib.py :
#!/usr/bin/python3
class MyLib:
def __init__(self):
self.say_hello = "Hello i'm in modules/MyLib"
def print_say_hello(self):
print(self.say_hello)
main.py:
#!/usr/bin/python3
# from folder.file import class
from modules.MyLib import MyLib
class MainClass:
def __init__(self):
my_lib = MyLib() # load MyLib class
my_lib.print_say_hello() # access to MyLib methods
### Test
if __name__ == '__main__':
app = MainClass()
In terminal when i run:
$ python3 main.py
output:
Hello i'm in modules/MyLib
So here we have successfully imported the class in modules/MyLib.py into our main.py file.

I found the error:
After treating the ImportError exception by printing it's args, I noticed that button_generator.py had an Import that was not resolving. Basically, button_generator.py could not be imported because it had a wrong import.

python import modules inside folder

Want to learn the proper way for importing modules with folder
/garden
__init__.py
utilities.py
someTools.py
/plants
__init__.py
carrot.py
corn.py
Inside /plants/__init__.py I have
__all__ = ['carrot', 'corn']
from . import *
inside carrot.py
def info():
print "I'm in carrot.py"
When I do
import garden
garden.carrot.info()
# got correct result
I'm in carrot.py
My question is how do I correct namespace for utilities.py for example inside carrot.py and corn.py. I want to use function in utilities.py
Inside carrot.py when I
import utilities as util
# try use it and got global name error
util.someFunc()
# NameError: global name 'util' is not defined #
Can I configure __init__.py in plants folder so that it import all the modules inside garden? Like utilities and sometools ? so I don't have to import utilities in both carrot.py and corn.py and be able to use utilities ?

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
If you have app in your PYTHONPATH, then from anywhere you can >>> from app.package_...
so file in folder package_b used file in folder package_a, which is what you want. Right??