I'm trying to use super in a subclass which is wrapped in another class using a class decorator:
def class_decorator(cls):
class WrapperClass(object):
def make_instance(self):
return cls()
return WrapperClass
class MyClass(object):
def say(self, x):
print(x)
#class_decorator
class MySubclass(MyClass):
def say(self, x):
super(MySubclass, self).say(x.upper())
However, the call to super fails:
>>> MySubclass().make_instance().say('hello')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 4, in say
TypeError: super(type, obj): obj must be an instance or subtype of type
The problem is that, when say is called, MySubclass doesn't refer to the original class anymore, but to the return value of the decorator.
One possible solution would be to store the value of MySubclass before decorating it:
class MySubclass(MyClass):
def say(self, x):
super(_MySubclass, self).say(x.upper())
_MySubclass = MySubclass
MySubclass = class_decorator(MySubclass)
This works, but isn't intuitive and would need to be repeated for each decorated subclass. I'm looking for a way that doesn't need additional boilerplate for each decorated subclass -- adding more code in one place (say, the decorator) would be OK.
Update: In Python 3 this isn't a problem, since you can use __class__ (or the super variant without arguments), so the following works:
#class_decorator
class MySubclass(MyClass):
def say(self, x):
super().say(x.upper())
Unfortunately, I'm stuck with Python 2.7 for this project.
The problem is that your decorator returns a different class than python (or anyone who uses your code) expects. super not working is just one of the many unfortunate consequences:
>>> isinstance(MySubclass().make_instance(), MySubclass)
False
>>> issubclass(MySubclass, MyClass)
False
>>> pickle.dumps(MySubclass().make_instance())
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
_pickle.PicklingError: Can't pickle <class '__main__.MySubclass'>: it's not the same object as __main__.MySubclass
This is why a class decorator should modify the class instead of returning a different one. The correct implementation would look like this:
def class_decorator(wrapped_cls):
#classmethod
def make_instance(cls):
return cls()
wrapped_cls.make_instance = make_instance
return wrapped_cls
Now super and everything else will work as expected:
>>> MySubclass().make_instance().say('hello')
HELLO
The problem occurs because at the time when MySubclass.say() is called, the global symbol MySubclass no longer refers to what's defined in your code as 'class MySubclass'. It is an instance of WrapperClass, which isn't in any way related to MySubclass.
If you are using Python3, you can get around this by NOT passing any arguments to 'super', like this:
super().say(x.upper())
I don't really know why you use the specific construct that you have, but it does look strange that a sub-class of MyClass that defines 'say()' - and has itself a 'say()' method in the source code would have to end up as something that does not have that method - which is the case in your code.
Note you could change the class WrapperClass line to make it read
class WrapperClass(cls):
this will make your wrapper a sub-class of the one you just decorated. This doesn't help with your super(SubClass, self) call - you still need to remove the args (which is OK only on Python3), but at least an instance created as x=MySubclass() would have a 'say' method, as one would expect at first glance.
EDIT: I've come up with a way around this, but it really looks odd and has the disadvantage of making the 'wrapped' class know that it is being wrapped (and it becomes reliant on that, making it unusable if you remove the decorator):
def class_decorator(cls):
class WrapperClass(object):
def make_instance(self):
i = cls()
i._wrapped = cls
return i
return WrapperClass
class MyClass(object):
def say(self, x):
print(x)
#class_decorator
class MySubclass(MyClass):
def say(self, x):
super(self._wrapped, self).say(x.upper())
# make_instance returns inst of the original class, non-decorated i = MySubclass().make_instance() i.say('hello')
In essence, _wrapped saves a class reference as it was at declaration time, consistent with using the regular super(this_class_name, self) builtin call.
Related
Can I actually use functions of another class as a parameter and attribute for a class instance/object?
I noticed if I do something like this there are lots of oddities (Note that I use Jupyter lab):
class ObjectClass:
#A class; I will insert a function into generalMethod
def __init__(self, generalMethod):
self.generalMethod = generalMethod
class GeneralMethods():
#Two different methods that I want to call
def method1(self):
add(2)
def method2(self):
print("Hey now, you're an all-star, get your game on, go play" )
return "Hey, I can return stuff, at least!",2
def add(input):
#A simple function that adds 1 to input
print(1 + input)
#Creating two objects with different methods as inputs from GeneralMethods
gm = GeneralMethods()
object1 = ObjectClass(gm.method1())
object2 = ObjectClass(gm.method2())
#Attempting to call anything from generalMethod; a getter method does the same
object1.generalMethod
object2.generalMethod
gm.method1() and gm.method2() does everything inside it, even when I simply declare it as a parameter of the object/instance!
But anyObject.generalMethod doesn't do anything besides return whatever is in the return when I call it, and if a function is in there, it will return None.
So can I actually call a function from the attribute and it perform like gm.method1() does when calling it from the attribute (anyObjectIChoose.generalMethod).
You can pass a function as parameter:
def foo():
print('hello')
def bar(_f):
_f()
bar(_f=foo)
'hello'
Note that when you add () to function's name, you invoke it. To pass as param you need just the name, not invoke it.
I'm trying to use a subclass that enhances instead of overriding the base class. I'm using the super method to call the base class. I find that I need to use the name mangling feature in __init__ (but only in init?) to make the code work. So for the heck of it I made the this print example. Since I didn't use name mangling I expected it to call subclass twice when I did the init, instead it calls the base class
It seems that __init__ sometimes sees the base class and sometimes sees the subclass. I'm sure it's just an incomplete understanding on my part, but was do I need name mangling for the real code, when in the print example it calls the base and subclass just fine?
the code
class base:
def __init__(self):
self.print()
def print(self):
print("base")
class subclass(base):
def __init__(self):
super(subclass, self).__init__()
self.print()
def print(self):
super(subclass, self).print()
print("subclass")
x = base()
x.print()
print("--")
y = subclass()
y.print()
the output - why doesn't y = subclass() print subclass instead of base since I didn't use name mangling?
> ./y.py
base
base
--
base
subclass
base
subclass
base
subclass
broken code when I don't use name mangling, works when I use self.__set and __set = set (the commented code). It gets the following error when I don't use __set:
File "./x.py", line 5, in __init__
self.set(arg)
TypeError: set() missing 1 required positional argument: 'arg2'
the code:
class base:
def __init__(self, arg):
self.set(arg)
# self.__set(arg)
# __set = set
def set(self, arg):
self.arg = arg
def print(self):
print("base",self.arg)
class subclass(base):
def __init__(self, arg1, arg2):
super(subclass, self).__init__(arg1)
self.set(arg1, arg2)
def set(self, arg1, arg2):
super(subclass, self).set(arg1)
self.arg2 = arg2
def print(self):
super(subclass, self).print()
print("subclass", self.arg2, self.arg)
x = base(1)
x.print()
x.set(11)
x.print()
y = subclass(2,3)
y.print()
y.set(4,5)
y.print()
======= update =======
I rewrote the code to look like this:
class base:
def __init__(self):
print("base init")
self.print()
def print(self):
print("base print")
class subclass(base):
def __init__(self):
print("sc init")
super(subclass, self).__init__()
print("sc after super")
self.print()
def print(self):
print("subclass print start")
super(subclass, self).print()
print("subclass print")
y = subclass()
print("--")
y.print()
when I run I get this output:
sc init
base init
subclass print start <<<< why is the subclass print called here
base print
subclass print
sc after super
subclass print start
base print
subclass print
--
subclass print start
base print
subclass print
why does the self.print in the base init call the subclass print when I'm initing the subclass? I was expecting that to call the base print. it does call the base print when I call it outside of the init.
Your subclass print explicitly calls the superclass one. So every time subclass.print is called, both "base" and "subclass" will be printed. This happens three times, because you call the print method three times: in subclass.__init__, in base.__init__ (which is called by subclass.__init__), and in subclass.print (which calls the superclass version).
In your "set" example, subclass.__init__ calls base.__init__, which tries to call self.set with just one argument. But since you are instantiating subclass, self.set is subclass.set, which takes two arguments.
It's unclear what you're trying to achieve with these examples. Your subclass doesn't really need to call base.__init__, because all that would do is call base.set, and you're already calling that from subclass.set. So even if you succeeded with all your calls, it would result in some methods getting called multiple times, just like with the print example.
My impression is that you're getting a bit carried away and trying to have every method call its superclass version. That's not always a good idea. If you write a subclass, and it calls a superclass method, you need to make sure that the subclass still provides an interface that's compatible with what the superclass expects. If it doesn't, you may need to not call the superclass method and instead have the subclass incorporate its functionality "inline" (although this may be more risky if other classes out in the world have made assumptions about how the base class works). The upshot is that you always need to think about what methods call which others; you can't just call every superclass method everywhere and expect that to work.
I want to call a method from the parent class in a child class.
I use XX.__init__() in my child class and call the press function from the parent class. But it fails when I run the following code:
Func.py
class PC:
def __init__(self):
PCKeyDis = {}
self.PCKeyDis = PCKeyDis
def Press(self,key):
KeyDis = self.PCKeyDis
if len(key)==1 and key.islower():
key = key.upper()
win32api.keybd_event(KeyDis[key],0,0,0)
time.sleep(0.1)
win32api.keybd_event(KeyDis[key],0,win32con.KEYEVENTF_KEYUP,0)
class PCFunc(PC):
def __init__(self):
pass
def Sentence(self,string):
PC.__init__()
strlist = list(string)
for i in xrange(len(strlist)):
if strlist[i] == ' ':
strlist[i] = 'Space'
PC.Press(strlist[i]) #use this function
action.py
import Func
import win32gui
PC = Func.PC()
PCFunc = Func.PCFunc ()
win32gui.SetForegroundWindow(win32gui.FindWindow(winclass,winnm))
PCFunc.Sentence(path)
I get:
unbound method Sentence() must be called with PCFunc instance as first argument (got str instance instead)
If you want to call the constructor of the base class, then you do it on instantiation in the __init__() method, not in the Sentence() method:
def __init__(self):
super(self.__class__, self).__init__()
Since Sentence() is an instance method, you need to call it via an instance of the class (like the error tells you):
pc_func = PCFunc()
pc_func.Sentence(var)
Here you are calling the method with an undefined variable:
PCFunc.Sentence(path)
Instead you need to give a string as parameter, so either write Sentence('path'), or define the variable first:
path = 'my path'
pc_func.Sentence(path)
Do not use the same name as the class name for an instance of the class:
PCFunc = Func.PCFunc ()
Otherwise the variable name storing the instance overwrites the class name.
Apart from that, it is unclear what your code is actually supposed to do. Have a look at the Python code conventions for a first step to making your code more readible. Then do some research about classes and inheritance.
The code you posted does not produce the error you posted. Here is an example that will produce that error:
class Dog:
def do_stuff(self, string):
print string
d = Dog()
d.do_stuff('hello')
Dog.do_stuff(d, 'goodbye')
Dog.do_stuff('goodbye')
--output:--
hello
goodbye
Traceback (most recent call last):
File "1.py", line 9, in <module>
Dog.do_stuff('goodbye')
TypeError: unbound method do_stuff() must be called with Dog instance as first argument (got str instance instead)
An __init__() function can also produce that error:
class Dog:
def __init__(self):
pass
def do_stuff(self, string):
print(string)
Dog.__init__()
--output:--
Traceback (most recent call last):
File "1.py", line 7, in <module>
Dog.__init__()
TypeError: unbound method __init__() must be called with Dog instance as first argument (got nothing instead)
In the line:
d.do_stuff('hello')
the fragment d.do_stuff causes python to create and return a bound method object--which is then immediately executed by the function execution operator () in the fragment ('hello’). The bound method is bound to the instance d, hence the reason it is called a bound method. A bound method automatically passes the instance it contains to the method when the method is executed.
On the other hand, when you write:
Dog.do_stuff(....)
the fragment Dog.do_stuff causes python to create and return an unbound method. An unbound method does not contain an instance, so when an unbound method is executed by the function execution operator (), you must manually pass an instance. (In python3, things changed and you can pass anything as the first argument--an instance of the class isn't required.)
I am trying to debug a multi-threaded program that uses a third-party package.
At some point, one of the attributes of an object (that is not created directly by me) is changed and I can't figure out what changed it. I could not find anything in my code that changes it.
Since this is a third-party package, I prefer not to change its code directly, but rather patch it from the outside as necessary.
My plan was to somehow tap into or wrap the code that sets the attribute and set a breakpoint or print the stack trace from there.
I tried monkey-patching the __setattr__ method of the instance, but it was not triggered.
I also tried to patch the class itself:
def patch_class(target):
def method(self, name, value):
print(name, value)
print("called from", target)
setattr(self, name, value) # break or print trace here
target.__setattr__ = types.MethodType(method, target)
patch_class(WebSocket)
but then all of the attributes are set on the class itself, as the method is bound to it.
Wrapping the class with a proxy does not really help either, since I am not instantiating it myself, but rather get the instance at some point after its creation.
If it matters, the said class is ws4py's WebSocket that is created by another third-party package, but I consider this an exercise in general debugging techniques.
Is there a more "pythonic" way of tapping into the mutation of an existing instance? (hack-ish ways will be appreciated as well)
I ended up creating a __setattr__ for the class.
def setter_fun(self, name, value):
print('setting', name, value)
self.__dict__[name] = value
if name is 'problematic_prop' and value is 'problematicValue':
traceback.print_stack()
# and set the class setter magic method
instance.__class__.__setattr__ = setter_fun
It is also possible to use setattr instead of using the __dict__ magic property:
setattr(self, name, value)
Now, when something sets the instance's problematic_prop to problematicValue, the stack trace will be printed:
>>> class A(object):
def __init__(self):
self.foo = 1
def set_problematic(self):
self.problematic_prop = 'problematicValue'
>>> a = A()
>>> a.__class__.__setattr__ = setter_fun
>>> a.foo = 2
setting foo 2
>>> print(a.foo)
2
>>> a.set_problematic()
setting problematic_prop problematicValue
Traceback (most recent call last):
File "<input>", line 1, in <module>
File "<input>", line 6, in set_problematic
File "<input>", line 5, in setter_fun
NameError: name 'traceback' is not defined
My failed attempts included either trying to attach the __setattr__ to the instance instead of the class, or trying to attach a bound method:
class MyClass(object):
def setter_fun(self, name, value):
print('setting', name, value)
self.__dict__[name] = value
if name is 'problematic_prop' and value is 'problematicValue':
traceback.print_stack()
def set_my_function(self):
# won't work, the function is bound to the current instance (self)
some.instace.__class__.__setattr__ = self.setter_fun
For one of the project I am currently working I was thinking of creating a class that could not be instantiate by a client and only be supplied an instance of through a particular interface i.e. the client would not be able create further instance out of it by some hackery such as:
>>> try:
... raise WindowsError
... except:
... foo = sys.exc_info()
...
>>> foo
(<type 'exceptions.WindowsError'>, WindowsError(), <traceback object at 0x0000000005503A48>)
>>> type(foo[2])()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: cannot create 'traceback' instances
once he has one.
I was successfully able to create a class that couldn't be instantiated. i.e.
>>> class Foo():
... def __init__(self):
... raise TypeError("cannot create 'Foo' instances")
...
>>> bar = Foo()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in __init__
TypeError: cannot create 'Foo' instances
>>> bar
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'bar' is not defined
But how could I use this every same definition to create an instance of the class?
Of course I could do something like this:
>>> class Foo():
... def __init__(self, instantiate = False):
... if not instantiate:
... raise TypeError("cannot create 'Foo' instances")
but I don't find it elegant enough nor does it completely prevent the client from further instantiating it. And no I aint going down the road of building a C++ module for it.
Any suggestions on how to achieve such a thing? import abc?
A brief rational to answer Martijn's question and for completeness:
Actual you could consider the instance of the particular, and related, classes, in question, as nodes in a tree and that both the parent and the children to remain connected, dependent on and cognizant of each other and have a single unique root throughout any instance of python(insured by the use package). Any state changes in a particular node would cause others to update themselves and the database to which they are connect, accordingly. Apart from that I was being curious to know how such a thing could be put in place (the traceback class was teasing me).
What you're doing is a bad idea, you shouldn't do it.
I'm sure there's an other, better solution.
If you do decide to go with your way anyways (you shouldn't), here's how you can create an object without using __init__():
Objects in python are created with the __new__() method. The method __init__() only edits the object which was created by __new__(). For example, __init__() usually initializes some attributes for the object.
When declaring something like x = Foo() what happens is this:
x = object.__new__(Foo) gets called first and creates the object.
Foo.__init__(x) gets called second, it simply initializes some attributes etc. to the already existing object.
This means that you are not required to call Foo() (and as a result, call __init__() too). Instead, you can just call __new__() directly:
class Foo(object):
def __init__(self):
raise TypeError("Cannot create 'Foo' instances.")
>>> x = object.__new__(Foo)
>>> x
<__main__.Foo object at 0x02B074F0>
Our x is now an instance of Foo, without any attributes that is, and it can use any methods defined in Foo class.
If you want, you can create your own replacement function of __init__ for initializing attributes:
def init_foo(foo, name):
foo.name = name
>>> init_foo(x, "Mike")
>>> x.name
'Mike'
This could of course be Foo's instance method too:
class Foo(object):
def __init__(self):
raise TypeError("Cannot create 'Foo' instances.")
def init(self, name):
self.name = name
>>> x = object.__new__(Foo)
>>> x.init("Mike")
>>> x.name
'Mike'
Going even step further, you can even use a classmethod for creating your object with only one call:
class Foo(object):
def __init__(self):
raise TypeError("Cannot create 'Foo' instances.")
#classmethod
def new(cls, name):
obj = object.__new__(cls)
obj.name = name
return obj
>>> x = Foo.new("Mike")
>>> x.name
'Mike'