Copy assignment for class member functions - python

I have a class A with some member functions that all do the same thing.
class A:
def a():
... boilerplate code ...
b = c = d = a
For debugging reasons, I would like to know the name of each member function at runtime. But since they all point to the same memory address, they will have the same __name__ attribute and I cannot figure out a way to distinguish between A.a and A.b just by looking at the object.
a = A.a
b = A.b
a.__name__ == b.__name__ # this is true
# how do I tell the difference between a and b?
Is there a way to achieve this without manually creating the functions b, c and d with the same boilerplate code?

No. Objects and names in Python live in separate spaces. There's only one function object there, and the function object doesn't know through what name it was conjured.
If you were a masochist, I suppose it would be possible to get a traceback and look at the line of code that called you, but that's just not practical.
You could do something like:
def reala(self,me=None):
pass
def a(self):
return reala('a')
def b(self):
return reala('b')
...

Related

How to "wrap" object to automatically call superclass method instead of overriden ones?

Consider:
class A(object):
def f(self): print("A")
class B(A):
def f(self): print("B")
b = B()
I can call A.f on b by doing:
A.f(b)
Is there an easy way to "wrap" b such that wrap(b).f() calls A.f for any f?
Here is my solution which copies the methods from the most upper base class:
import types, copy
def get_all_method_names(clazz):
return [func for func in dir(clazz) if callable(getattr(clazz, func))]
def wrap(obj):
obj = copy.copy(obj)
obj_clazz = obj.__class__
base_clazz = obj_clazz.__bases__[-1] # the one which directly inherits from object
base_methods = get_all_method_names(base_clazz) # list of all method names in base_clazz
for base_method_name in base_methods:
base_method = getattr(base_clazz, base_method_name) # get the method object
if isinstance(base_method, types.FunctionType): # skip dunder methods like __class__, __init__
setattr(obj, base_method_name, base_method) # copy it into our object
return obj
# class declaration from question here
wrapped_b = wrap(b)
wrapped_b.f(wrapped_b) # prints A, unfortunately we have to pass the self parameter explicitly
b.f() # prints B, proof that the original object is untouched
This feels dirty to me, but it also seems to work. I'm not sure I'd rely on this for anything important.
import copy
def upcast(obj, clazz):
if not isinstance(obj, clazz): # make sure we're actually "upcasting"
raise TypeError()
wrapped = copy.copy(obj)
wrapped.__class__ = clazz
return wrapped
This results in
>>> a = A()
>>> a.f()
A
>>> b = B()
>>> b.f()
B
>>> upcast(b, A).f()
A
What I've really done here is essentially monkey-patch a clone of b and lied to it and told it it's actually an A, so when it comes time to resolve which version of f to call, it'll call the one from A.
Object Slicing is not supported in python the way it is done in C++ (The link you are pointing to takes a cpp example).
In Python Object Slicing is a rather different thing which means to slice up any object which supports sequence protocol (implements getitem() and len() methods).
Example :
A = [1,2,3,4,5,6,7,8]
print(A[1:3])
But in C++ Object Slicing is just cutting off the properties added by a base class instance when assigned to a parent class variable.

Order of classes matters when using variables from each other

I am new to Python and didn't find an answer to the following problem:
I have two classes and want them to use variables from each other. Is there a simple way to do this because if I do it like this class a does not know that class b exists.
class a:
y=1
print(b.x)
class b:
x=1
print(a.y)
And how do I use overwrite the variables, the following code does not work:
class a:
y=b.x
class b:
x=1
You are executing print as part of the class definition. It executes as soon as python sees that line of code, before it's read the part about class b.
Instead, use functions inside the classes to execute code after the classes have been defined:
class a:
y=1
def go():
print(b.x)
class b:
x=1
def go():
print(a.y)
a.go()
b.go()
As I said in a comment, your code isn't making effective use of classes. Here's what I think would be better approach that offers more flexibility in working around the circular reference issue.
First the class definitions (which follow the PEP 8 naming convention guidelines):
class A:
def __init__(self, value, linked_value=None):
self.y = value
if isinstance(linked_value, B):
self.linked_value = linked_value.x
def print_linked_value(self):
print(self.linked_value)
class B:
def __init__(self, value, linked_value=None):
self.x = value
if isinstance(linked_value, A):
self.linked_value = linked_value.y
def print_linked_value(self):
print(self.linked_value)
Definitions like that provide two ways to set up the circular references:
By creating them separately, then explicitly linking them:
# First create instances of each class.
a = A(1)
b = B(42)
# Then link them.
a.linked_value = b.x
b.linked_value = a.y
a.print_linked_value() # -> 42
b.print_linked_value() # -> 1
*OR* by creating the first one without a linked value and leaving only the second needing to be linked manually.
# First create instances of each class, but link the second to the first
# when it's created.
a = A(1)
b = B(42, a) # Create and link to first.
# Then link the first to the second to complete the circular references.
a.linked_value = b.x
# Same result.
a.print_linked_value() # -> 42
b.print_linked_value() # -> 1
Final note: Another, more advanced alternative that can also be applied in situations like this by using the built-in property() function as a decorator to create "descriptors". Here's an answer to a somewhat related question that illustrating its use.
class A:
y = 1
def foo(self):
print B.x
class B:
x = 1
def bar(self):
print A.y
>>> A().foo()
2
>>> B().bar()
1
Use 'print' in some function definition.

Calling class variable with self

How would you I came up with this interesting (at least to me) example.
import numpy as np
class Something(object):
a = np.random.randint(low=0, high=10)
def do(self):
self.a += 1
print(self.a)
if __name__ == '__main__':
something = Something()
print(something.__str__())
something.do()
something2 = Something()
print(something2.__str__())
something2.do()
something3 = Something()
print(something3.__str__())
something3.do()
The above prints the following in the console:
$ python test.py
<__main__.Something object at 0x7f03a80e0518>
1
<__main__.Something object at 0x7f03a80cfcc0>
1
<__main__.Something object at 0x7f03a80cfcf8>
1
I'm a bit confused because I (wrongly) assumed the value of a would have increased.
I'm able to obtain the behaviour I would expect if I use the #classmethod decorator.
import numpy as np
class Something(object):
a = np.random.randint(low=0, high=10)
#classmethod
def do(cls):
cls.a += 1
print(cls.a)
if __name__ == '__main__':
something = Something()
print(something.__str__())
something.do()
something2 = Something()
print(something2.__str__())
something2.do()
something3 = Something()
print(something3.__str__())
something3.do()
This correctly prints the following in the console.
python test.py
<__main__.Something object at 0x7faac77becc0>
3
<__main__.Something object at 0x7faac77becf8>
4
<__main__.Something object at 0x7faac77c3978>
5
Now, I'm wondering in the first example, when I'm calling self.a, what I'm accessing? It's not a class variable since I don't seem to be able to change its value. It's not an instance variable either, since this seems to be shared among different objects of the same class. How would you call it?
Is this a class variable that I'm using in the wrong way? I know the cls name if a convention, so maybe I'm truly accessing a class variable, but I'm not able to change its value because I haven't decorate the method with the #classmethod decorator.
Is this a sort of illegitimate use of the language? I mean something it's best practice to not do in order to avoid introduce a bug on a later stage?
What is happening is that self.a refers to two things at different times.
When no instance variable exists for a name, Python will lookup the value on the class. So the value retrieved for self.a will be the class variable.
But when setting an attribute via self, Python will always set an instance variable. So now self.a is a new instance variable whose value is equal to the class variable + 1. This attribute shadows the class attribute, which you can no longer access via self but only via the class.
(One minor point, which has nothing to do with the question: you should never access double-underscore methods directly. Instead of calling something2.__str__(), call str(something2) etc.)
Answer by Daniel Roseman clearly explains the problem. Here are some additional points and hope it helps.
You can use type(self).a instead of self.a. Also look at the discussions
Python: self vs type(self) and the proper use of class variables and
Python: self.__class__ vs. type(self)
import numpy as np
class Something(object):
a = np.random.randint(low=0, high=10)
def do(self):
type(self).a += 1
print(type(self).a)
if __name__ == '__main__':
something = Something()
print(str(something ))
something.do()
something2 = Something()
print(str(something2))
something2.do()
something3 = Something()
print(str(something3))
something3.do()

"Undeclared variable" declaration in python

The following code is an example:
class A(object):
def f(self):
pass
A.f.b = 42
How is this variable being allocated? If I declare A.f.a, A.f.b, and A.f.c variables am I creating 3 different objects of A? Can someone explain what's going on in memory (as this does not appear to be something easily coded in C)?
The following only works in Python 3:
class A(object):
def f(self):
pass
A.f.a = 41
A.f.b = 42
A.f.c = 43
A.f is an object of type function, and you have always been able to add new attributes to a function object. No instances of A have been created; the three attributes are referenced from the function f itself.
If you had two instances a1 = A() and a2 = A(), however, neither a1.f.b and a2.f.b are defined, because a1.f is not a reference to A.f; it is a reference to an object of type method. This results from how Python's descriptor protocol is used to implement instance methods.
A.b = 42 adds a class variable to A, and thus makes it visible instantly for each instance of A (but only 1 entry in memory)
You can add attributes to classes and instances anytime you like in Python. The cleanest way would be to do it a declare time or this could be misleading.
class A:
b = 12
But for quick "extensions" of classes or instances you could choose to dynamically add them.
ex:
class A(object):
pass
a = A()
print('b' in dir(a)) # False
A.b = 42
print('b' in dir(a)) # True even if instanciated before creation of `b`

Python: update value of an instance and use it in a class from imported module

I have been studying Python for three months and now I have a question that I could not solve by using google, but luckily I am able to simplify it here:
I have a var_class module:
#var_class.py
class A(object):
def __init__(self, x):
self.x = x+2
zz = A(10)
class B():
b = 0
def __init__(self):
pass
def update(self):
B.b = zz.x
and in main program I have:
#main.py
from var_class import *
b_class = B()
b_class.b # I get 0 as expected
zz = A(100)
b_class.update()
b_class.b # Instead of 102, I get 12 unexpectedly
You see my goal is to change "zz" frequently for every data input and then update a class variable 'b', and the reason I wrote zz = A(10) in var_class.py is that otherwise when I import it, module 'var_class' is missing 'zz' in 'class B', it would give error "global name zz is not defined".
However, as I write it like this now, looks like the value '10' is stuck to the class, and I am not able to change it in the main program. Don't know how to overcome this.
Thanks in advance for any help.
Answer:
alKid wrote the whole answer first, have to thanks to alexvassel and Mr. C too, would like to know if there is way to thanks them, others helped me with the knowledge, also appreciate a lot.
Easy understanding:
You can't do that, you're inside var_class module, so zz is A(10).
How about passing zz as a parameter? Like this!
class A(object):
def __init__(self, x):
self.x = x+2
zz = A(10)
class B():
b = 0
def __init__(self):
pass
def update(self, zz):
B.b = zz.x
Hope this helps!
When you do B.b = zz.x (update method) you are inside the module var_class, so zz is A(10)
python runtime finds variables by namespace. Namespace is something like scope. When B.b = zz.x executes, the runtime first searches the local namespace(the function namespace -- update), it sees no zz. Then, it goes to the module space(var_class), well we get the variable and stop searching.
The python namespace search order:
1. local namespace, the function scope
2. global namespace, the module scope
3. built-in namespace
Better not use global variables around.
your code may like this:
class B():
b = 0
def __init__(self):
pass
def update(self, zz):
B.b = zz.x
That is a matter of scope! you are using zz = A(100) in your main.py. But when you call b_class.update(), b_class has to get "some2 variable called "zz" - and the one which is available is the one you defined in your class.py - and this one still has the value A(10)!!!
To work around this, you have different options.
The dirty hack (but closest to your solution) would be to define a global variable and use ist (google "python globals" for more info on this approach).
I would suggest passing A (or A.x) to your "update" method b_class.update(A.x). This avoids globals and is more readable. Of course, you'll have to adjust B.update for the new parameter.
when you import a variable from a module, you get a copy of the variable, not the original. you need to write to the original directly.
from a.b import c
from a.b.c import var
a.b.c.var = 1
var = 2
a.b.c.var
1
var
2
Edit: So, more correctly, in python, there are objects in memory, and then there are names for the objects. When you import a module, you create two separate names, but they both point to the same object - i.e. they have a reference to the same object. It's very similar to the below code, which doesn't require any imports at all:
>>> a = 4
>>> b = a
>>> b
4
>>> a
4
>>> b = 'something different'
>>> b
'something different'
>>> a
4
Why did changing b not also change a? The sequence is like this: First, we create an object (the 4) and point the name 'a' at it. Then, we create a name 'b', and we point it at the same object. So, now looking up those names returns the same object. Now, we then go back and point the name 'b' at a different object, a string. The object 'a' is pointing to still exists, and has not changed.
New users more often trip up on this the other way, with objects like lists:
>>> a = [1,2,3,4,5]
>>> b = a
>>> b
[1,2,3,4,5]
>>> a
[1,2,3,4,5]
>>> b.append(6)
>>> b
[1,2,3,4,5,6]
>>> a
[1,2,3,4,5,6]
>>> b = b[1:3]
>>> b
[2,3]
>>> a
[1,2,3,4,5,6]
What's going on here? Well, first we create a list object, and then point the name 'a' at it. Then we create the name 'b' and point it to the same object. So, 'a' and 'b' both point to the same list. Then, we use the reference b to get the object and modify it. Note that we haven't changed what 'b' points to in this case - we grabbed the reference, and then modified the object it points to directly. So, in this case, both 'a' and 'b' will see the change - they both point to the modified object. Then, we extract a slice, and assign it to 'b'. Now, this actually creates a new object, and points 'b' at it - 'b' is no longer pointing to the original object. So now 'a' and 'b' point to different objects, and now updates to one are no longer reflected in the other.
The import case is just a special case of this.
maybe you can do this
class B():
b = 0
def __init__(self):
pass
def update(self,value):
self.b =self.b+value

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