Python, faking module in Class definition - python

I suspect this is one of those questions that will result in "Why do you want to do that?" but here goes:
I need to find a way to fake the module of a class defined in __main__; ie:
Lets say we have a.py:
class A(object):
name = "a"
pass
And b.py:
import inspect
import a
class B(object):
__name__ = a.__name__
name = "b"
pass
print inspect.getmodule(B)
Calling b.py results in:
<module '__main__' from 'b.py'>
So, how do I get the module of class A taken to be b, without moving it into b ?

You can set the class's __module__ attribute, but there are caveats:
class A(object):
__module__ = 'b'
The main caveat is that this will cause getmodule to return None if the module b has not imported at the time you call getmodule(A). I think it's also possible for problems to arise if you import b in some nonstandard way, like importing it via an implicit relative import instead of as part of a package (if it's part of a package).
Conceptually, though, what you're doing is somewhat dubious. It's risky to set A's module to be some other module, because that module might not be known (i.e., imported) when the real module is, and in fact it's possible the "disguise" module might not even exist.

Related

Is there a way to monkey patch a class inside a module before the module body is executed in python?

I have file main.py which contains the following code:
class A:
def __init__(self, a):
self.a = a
def run(self):
return self.a+10
a = A(4)
print(a.run())
In file test.py, I tried to monkey patch class A in main.py as follows:
import main
class A:
def __init__(self, a):
self.a = a
def run(self):
return self.a+5
main.A = A
Unfortunately, when I run import test from a python interpreter, the module still prints out 14 as opposed to my expected output which is 9.
Is there a way to monkey patch a class inside a module before the module body is executed?
The problem here is that when you imported the main.py file, it executed the code a = A(4) using the real implementation of the class A. Then the rest of your test.py was executed and you replaced the A reference, but it was too late.
You can check that by adding in your test :
print(__name__) # __main__
print(main.__name__) # so70731368_main
print(A.__module__) # __main__
print(main.A.__module__) # __main__
print(main.a.__class__.__module__) # so70731368_main
Here, __main__ is a bit confusing but that's how Python call the first file you run (in your case test.py). The a instance is declared in the so70731368_main module, and it used the A class from the same module, you just changed A after the fact with the definition from the test file (__main__).
The fact that you need to patch two definitions (A and a) defined in the same file is very tricky. unittest.mock.patch is not powerful enough to patch inside an import (it patches after the import).
You can not, in a clean and simple way, prevent a to be instantiated as a main.A (real) class and get printed. What you can do is patch it after, for later uses, that is what you showed.
To answer directly your question : "patching" means replacing one reference by another, so the reference has to already be defined. In your example, it would require to patch between the class definition and the class instantiation (for the print to not use the real a), which is not supported.
There is no simple solution to this problem. If you have control over the code of the main.py file, then try to change it so that it does not instantiate a at import time.

python import nested class

I'm new to python. This is not my actual case, it's just my curiosity about importing nested class.
So I have a main.py and test.py.
test.py:
class one():
class two():
def twodef():
pass
so in my main.py, I can do 'import test' or 'from test import one'. But I got error when do: 'from test.one import two'.
error : ImportError: No module named one
Anyone can explain this?
You can only do from module import name for names that exist in the module's global scope - basically, names that are defined in top-level module code. Names that are defined in lower-level scopes - for example within classes as in your example - are not importable. You need to import the object that contains the name, and access it from there.
from test import one
my_two = one.two()
You could do the following:
from test import one
two_object = one.two()
two_object.twodef()
inside test.py
class One:
class Two:
#staticmethod
def twodef():
print("twodef")
My main logic:
from test import One
two_obj = One.Two
two_obj.twodef()
Remove the parenthesis at the end of class definition.
Coding standards:
Always declare your class names in upper camel case (class one => class One).
For more about scope
Preference:
And if you have a single class inside python file, always name that file in that class name but in lower camel case. (test.py => one.py). I have thought this as a standard earlier, but seems its practiced widely. I would use this because, if then project size grew and if you end up having a lot of classes and modules, it's logical and easy to access, rather than visiting the module to find the class name.
Ref1
Ref2

Using variables inside a function in another python file [duplicate]

This question already has answers here:
What does if __name__ == "__main__": do?
(45 answers)
Closed 8 months ago.
i cant seem to make this work.
I have 2 python files, lets say a.py and b.py
in a.py is this:
def foo():
global name
name = "foo"
import b
b.bar()
if __name__ == "__main__":
foo()
in b.py this:
import a
def bar():
name = a.name
print(name)
I have three different question in relation to this code:
Why do i get the error: AttributeError: 'module' object has no attribute 'name'
I know for certain that b.py cant access the variable defined in the function in b.py but how do i solve this?
does global in this case changes anything? if not, why?
i tried doing name = a.foo.name instead of name = a.name but this doesnt do the trick either and gives me: AttributeError: 'function' object has no attribute 'name',
is this even practicable in any case and what did i do wrong?
Thanks for taking the time and sorry if this seems obvious to some of you, i'am still getting into this.
Scripts aren't modules. Its something of a mind bender but when you run python3 a.py, a.py isn't a module (or more properly, its a module named __main__) and it will be reloaded when you import.
Suppose I have c.py
print('hello')
if __name__=="__main__":
import c
When I run it I get
hello
hello
To illustrate with your example, if I change b.py to
import __main__
def bar():
name = __main__.name
print(name)
It works. But only kindof because if somebody imports a you've still got two modules and two copies of the variable.
If you want shared data, they have to be in imported modules, not top level scripts. When I want shared data, I create a separate module that doesn't import anything (or only imports from the standard library) and it is either empty or has defaults for the variables I want to share.
1, 2) The specific purpose of if __name__ == "__main__": is to contain code that will not run when the file is imported. Since foo() is inside that block and nowhere else, the foo function does not run when import a happens; therefore the global name is not assigned to; therefore the a module does not have a name global; therefore b cannot refer to the nonexistent module global with a.name.
a.foo.name cannot fix the problem because, although a.foo exists (there is a foo function defined in the a module at the top level), it does not have a name attribute. The local variables in a function's code are not part of the function object; they exist only temporarily, while the code is running, and are not inherently part of any object. (Even if you call locals(), that creates a new dict, and assigns the local variables as values.)
It is possible to assign attributes to a function - for example, in a.py:
def foo():
pass
foo.name = "bar"
And such an attribute can be used from outside, including in another module:
# b.py
import a
print(a.foo.name)
But this is entirely orthogonal to what the function actually does, or to using it for any particular purpose.

conditionally import module to shadow local implementation

I am writing a Python script where some of the core functionalities can be done by another existing library. Unfortunately, while that library has more features, it is also slower, so I'd like if the user could at runtime select whether they want to use that library or my own fast and simple implementation. Unfortunately I'm stuck at a point where I don't understand some of the workings of Python's module system.
Suppose that my main program was main.py, that the (optional) external module is in module_a.py and that my own fast and simple implementation of module_a together with the actual program code that uses either my own implementation or the one of module_a is in the file module_x.py:
main.py:
import module_x
module_x.test(True)
module_x.test(False)
module_a.py:
class myclass():
def __init__(self):
print("i'm myclass in module_a")
module_x.py:
class myclass():
def __init__(self):
print("i'm myclass in module_x")
def test(enable_a):
if enable_a:
try:
from module_a import myclass
except ImportError:
global myclass
enable_a = False
else:
global myclass
i = myclass()
When I now execute main.py I get:
$ python3 main.py
i'm myclass in module_a
i'm myclass in module_a
But why is this? If False is passed to test() then the import of the module_a implementation should never happen. Instead it should only see myclass from the local file. Why doesn't it? How do I make test() use the local definition of myclass conditionally?
My solution is supposed to run in Python3 but I see the same effect when I use Python2.7.
An import statement is permanent within the thread of execution unless it is explicitly undone. Furthermore, once the from ... import statement is executed in this case, it replaces the variable myclass in the global scope (at which point the class it was previously referencing defined in the same file is no longer referenced and can in theory be garbage collected)
So what is happening here is whenever you run test(True) the first time, your myclass in module_x is effectively deleted and replaced with the myclass from module_a. All subsequent calls to test(False) then call global myclass which is effectively a no-op since the global myclass now refers to the one imported from the other class (and besides the global call is unneeded when not changing the global variable from a local scope as explained here).
To work around this, I would strongly suggest encapsulating the desired module-switching behavior in a class that is independent of either module you would like to switch. You can then charge that class with holding a reference to both modules and providing the rest of you client code with the correct one. E.g.
module_a_wrapper.py
import module_x
import module_a
class ModuleAWrapper(object):
_target_module = module_x # the default
#classmethod
def get_module(cls):
return cls._target_module
def set_module(enable_a):
if enable_a:
ModuleAWrapper._target_module = module_a
else:
ModuleAWrapper._target_module = module_x
def get_module():
return ModuleAWrapper.get_module()
main.py:
from module_a_wrapper import set_module, get_module
set_module(True)
get_module().myclass()
set_module(False)
get_module().myclass()
Running:
python main.py
# Outputs:
i'm myclass in module_a
i'm myclass in module_x
You can read more about the guts of the python import system here
The answer by lemonhead properly explains why this effect happens and gives a valid solution.
The general rule seems to be: wherever and however you import a module, it will always replace any variables of the same name from the global scope.
Funnily, when I use the import foo as bar construct, then there must neither be a global variable named foo nor one named bar!
So while lemonhead's solution worked it adds lots of complexity and will lead to my code being much longer because every time I want to get something from either module I have to prefix that call with the getter function.
This solution allows me to solve the problem with a minimal amount of changed code:
module_x.py:
class myclass_fast():
def __init__(self):
print("i'm myclass in module_x")
def test(enable_a):
if enable_a:
try:
from module_a import myclass
except ImportError:
enable_a = False
myclass = myclass_fast
else:
myclass = myclass_fast
i = myclass()
So the only thing I changed was to rename the class I had in global scope from myclass to myclass_fast. This way it will not be overwritten anymore by the import of myclass from module_a. Then, on demand, I change the local variable myclass to either be the imported module or myclass_fast.

Masquerading real module of a class

Suppose you have the following layout for a python package
./a
./a/__init__.py
./a/_b.py
inside __init__.py you have
from _b import *
and inside _b.py you have
class B(object): pass
If you import from interactive prompt
>>> import a
>>> a.B
<class 'a._b.B'>
>>>
How can I completely hide the existence of _b ?
The problem I am trying to solve is the following: I want a facade package importing "hidden" modules and classes. The classes available from the facade (in my case a) are kept stable and guaranteed for the future. I want, however, freedom to relocate classes "under the hood", hence the hidden modules. This is all nice, but if some client code pickles an object provided by the facade, this pickled data will refer to the hidden module nesting, not to the facade nesting. In other words, if I reposition the B class in a module _c.py, client codes will not be able to unpickle because the pickled classes are referring to a._b.B, which has been moved. If they referred to a.B, I could relocate the B class as much as I want under the hood, without ruining pickled data.
try:
B.__module__= 'a'
Incidentally you probably want an absolute import:
from a._b import *
as relative imports without the new explicit dot syntax are going away (see PEP 328).
ETA re comment:
I would have to set the module explicitly for every class
Yes, I don't think there's a way around that but you could at least automate it, at the end of __init__:
for value in globals().values():
if inspect.isclass(value) and value.__module__.startswith('a.'):
value.__module__= 'a'
(For new-style classes only you could get away with isinstance(value, type) instead of inspect. If the module doesn't have to run as __main__ you could use __name__ instead of hard-coding 'a'.)
You could set the __module__ variable for Class B
class B(object): pass
B.__module__ = 'a'
For classes, functions, and methods, this attribute contains the name of the module in which the object was defined.
Or define it once in your __init__.py:
from a._b import B # change this line, when required, e.g. from a._c import B
B.__module__ = 'a'

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