I am attempting to modify a value in a class __dict__ directly using something like X.__dict__['x'] += 1. It is impossible to do the modification like that because a class __dict__ is actually a mappingproxy object that does not allow direct modification of values. The reason for attempting direct modification or equivalent is that I am trying to hide the class attribute behind a property defined on the metaclass with the same name. Here is an example:
class Meta(type):
def __new__(cls, name, bases, attrs, **kwargs):
attrs['x'] = 0
return super().__new__(cls, name, bases, attrs)
#property
def x(cls):
return cls.__dict__['x']
class Class(metaclass=Meta):
def __init__(self):
self.id = __class__.x
__class__.__dict__['x'] += 1
This is example shows a scheme for creating an auto-incremented ID for each instance of Class. The line __class__.__dict__['x'] += 1 can not be replaced by setattr(__class__, 'x', __class__.x + 1) because x is a property with no setter in Meta. It would just change a TypeError from mappingproxy into an AttributeError from property.
I have tried messing with __prepare__, but that has no effect. The implementation in type already returns a mutable dict for the namespace. The immutable mappingproxy seems to get set in type.__new__, which I don't know how to avoid.
I have also attempted to rebind the entire __dict__ reference to a mutable version, but that failed as well: https://ideone.com/w3HqNf, implying that perhaps the mappingproxy is not created in type.__new__.
How can I modify a class dict value directly, even when shadowed by a metaclass property? While it may be effectively impossible, setattr is able to do it somehow, so I would expect that there is a solution.
My main requirement is to have a class attribute that appears to be read only and does not use additional names anywhere. I am not absolutely hung up on the idea of using a metaclass property with an eponymous class dict entry, but that is usually how I hide read only values in regular instances.
EDIT
I finally figured out where the class __dict__ becomes immutable. It is described in the last paragraph of the "Creating the Class Object" section of the Data Model reference:
When a new class is created by type.__new__, the object provided as the namespace parameter is copied to a new ordered mapping and the original object is discarded. The new copy is wrapped in a read-only proxy, which becomes the __dict__ attribute of the class object.
Probably the best way: just pick another name. Call the property x and the dict key '_x', so you can access it the normal way.
Alternative way: add another layer of indirection:
class Meta(type):
def __new__(cls, name, bases, attrs, **kwargs):
attrs['x'] = [0]
return super().__new__(cls, name, bases, attrs)
#property
def x(cls):
return cls.__dict__['x'][0]
class Class(metaclass=Meta):
def __init__(self):
self.id = __class__.x
__class__.__dict__['x'][0] += 1
That way you don't have to modify the actual entry in the class dict.
Super-hacky way that might outright segfault your Python: access the underlying dict through the gc module.
import gc
class Meta(type):
def __new__(cls, name, bases, attrs, **kwargs):
attrs['x'] = 0
return super().__new__(cls, name, bases, attrs)
#property
def x(cls):
return cls.__dict__['x']
class Class(metaclass=Meta):
def __init__(self):
self.id = __class__.x
gc.get_referents(__class__.__dict__)[0]['x'] += 1
This bypasses critical work type.__setattr__ does to maintain internal invariants, particularly in things like CPython's type attribute cache. It is a terrible idea, and I'm only mentioning it so I can put this warning here, because if someone else comes up with it, they might not know that messing with the underlying dict is legitimately dangerous.
It is very easy to end up with dangling references doing this, and I have segfaulted Python quite a few times experimenting with this. Here's one simple case that crashed on Ideone:
import gc
class Foo(object):
x = []
Foo().x
gc.get_referents(Foo.__dict__)[0]['x'] = []
print(Foo().x)
Output:
*** Error in `python3': double free or corruption (fasttop): 0x000055d69f59b110 ***
======= Backtrace: =========
/lib/x86_64-linux-gnu/libc.so.6(+0x70bcb)[0x2b32d5977bcb]
/lib/x86_64-linux-gnu/libc.so.6(+0x76f96)[0x2b32d597df96]
/lib/x86_64-linux-gnu/libc.so.6(+0x7778e)[0x2b32d597e78e]
python3(+0x2011f5)[0x55d69f02d1f5]
python3(+0x6be7a)[0x55d69ee97e7a]
python3(PyCFunction_Call+0xd1)[0x55d69efec761]
python3(PyObject_Call+0x47)[0x55d69f035647]
... [it continues like that for a while]
And here's a case with wrong results and no noisy error message to alert you to the fact that something has gone wrong:
import gc
class Foo(object):
x = 'foo'
print(Foo().x)
gc.get_referents(Foo.__dict__)[0]['x'] = 'bar'
print(Foo().x)
Output:
foo
foo
I make absolutely no guarantees as to any safe way to use this, and even if things happen to work out on one Python version, they may not work on future versions. It can be fun to fiddle with, but it's not something to actually use. Seriously, don't do it. Do you want to explain to your boss that your website went down or your published data analysis will need to be retracted because you took this bad idea and used it?
This probably counts as an "additional name" you don't want, but I've implemented this using a dictionary in the metaclass where the keys are the classes. The __next__ method on the metaclass makes the class itself iterable, such that you can just do next() to get the next ID. The dunder method also keeps the method from being available through the instances. The dictionary storing the next id has a name starting with a double underscore, so it's not easily discoverable from any of the classes that use it. The incrementing ID functionality is thus entirely contained in the metaclass.
I tucked the assignment of the id into a __new__ method on a base class, so you don't have to worry about it in __init__. This also allows you to del Meta so all the machinery is a little harder to get to.
class Meta(type):
__ids = {}
#property
def id(cls):
return __class__.__ids.setdefault(cls, 0)
def __next__(cls):
id = __class__.__ids.setdefault(cls, 0)
__class__.__ids[cls] += 1
return id
class Base(metaclass=Meta):
def __new__(cls, *args, **kwargs):
self = object.__new__(cls)
self.id = next(cls)
return self
del Meta
class Class(Base):
pass
class Brass(Base):
pass
c0 = Class()
c1 = Class()
b0 = Brass()
b1 = Brass()
assert (b0.id, b1.id, c0.id, c1.id) == (0, 1, 0, 1)
assert (Class.id, Brass.id) == (2, 2)
assert not hasattr(Class, "__ids")
assert not hasattr(Brass, "__ids")
Note that I've used the same name for the attribute on both the class and the object. That way Class.id is the number of instances you've created, while c1.id is the ID of that specific instance.
My main requirement is to have a class attribute that appears to be read only and does not use additional names anywhere. I am not absolutely hung up on the idea of using a metaclass property with an eponymous class dict entry, but that is usually how I hide read only values in regular instances.
What you are asking for is a contradiction: If your example worked, then __class__.__dict__['x'] would be an "additional name" for the attribute. So clearly we need a more specific definition of "additional name." But to come up with that definition, we need to know what you are trying to accomplish (NB: The following goals are not mutually exclusive, so you may want to do all of these things):
You want to make the value completely untouchable, except within the Class.__init__() method (and the same method of any subclasses): This is unPythonic and quite impossible. If __init__() can modify the value, then so can anyone else. You might be able to accomplish something like this if the modifying code lives in Class.__new__(), which the metaclass dynamically creates in Meta.__new__(), but that's extremely ugly and hard to understand.
You want the code that manipulates the value to be "nicely encapsulated": Write a method in the metaclass that increments the private value (or does whatever other modification you need), and provide a read-only metaclass property that accesses it under the public name.
You are concerned about a subclass accidentally clashing names with the private name: Prefix the private name with a double underscore to invoke automatic name mangling. While this is usually seen as a bit unPythonic, it is appropriate for cases where name collisions may be less obvious to subclass authors, such as the internal names of a metaclass colliding with the internal names of a regular class instantiated from it.
Related
TL;DR -
I have a class that uses a metaclass.
I would like to access the parameters of the object's constructor from the metaclass, just before the initialization process, but I couldn't find a way to access those parameters.
How can I access the constructor's parameters from the metaclass function __new__?
In order to practice the use of metaclasses in python, I would like to create a class that would be used as the supercomputer "Deep Thought" from the book "The Hitchhiker's Guide to the Galaxy".
The purpose of my class would be to store the various queries the supercomputer gets from users.
At the bottom line, it would just get some arguments and store them.
If one of the given arguments is number 42 or the string "The answer to life, the universe, and everything", I don't want to create a new object but rather return a pointer to an existing object.
The idea behind this is that those objects would be the exact same so when using the is operator to compare those two, the result would be true.
In order to be able to use the is operator and get True as an answer, I would need to make sure those variables point to the same object. So, in order to return a pointer to an existing object, I need to intervene in the middle of the initialization process of the object. I cannot check the given arguments at the constructor itself and modify the object's inner-variables accordingly because it would be too late: If I check the given parameters only as part of the __init__ function, those two objects would be allocated on different portions of the memory (they might be equal but won't return True when using the is operator).
I thought of doing something like that:
class SuperComputer(type):
answer = 42
def __new__(meta, name, bases, attributes):
# Check if args contains the number "42"
# or has the string "The answer to life, the universe, and everything"
# If so, just return a pointer to an existing object:
return SuperComputer.answer
# Else, just create the object as it is:
return super(SuperComputer, meta).__new__(meta, name, bases, attributes)
class Query(object):
__metaclass__ = SuperComputer
def __init__(self, *args, **kwargs):
self.args = args
for key, value in kwargs.items():
setattr(self, key, value)
def main():
number = Query(42)
string = Query("The answer to life, the universe, and everything")
other = Query("Sunny", "Sunday", 123)
num2 = Query(45)
print number is string # Should print True
print other is string # Should print False
print number is num2 # Should print False
if __name__ == '__main__':
main()
But I'm stuck on getting the parameters from the constructor.
I saw that the __new__ method gets only four arguments:
The metaclass instance itself, the name of the class, its bases, and its attributes.
How can I send the parameters from the constructor to the metaclass?
What can I do in order to achieve my goal?
You don't need a metaclass for that.
The fact is __init__ is not the "constructor" of an object in Python, rather, it is commonly called an "initializator" . The __new__ is closer to the role of a "constructor" in other languages, and it is not available only for the metaclass - all classes have a __new__ method. If it is not explicitly implemented, the object.__new__ is called directly.
And actually, it is object.__new__ which creates a new object in Python. From pure Python code, there is no other possible way to create an object: it will always go through there. That means that if you implement the __new__ method on your own class, you have the option of not creating a new instance, and instead return another pre-existing instance of the same class (or any other object).
You only have to keep in mind that: if __new__ returns an instance of the same class, then the default behavior is that __init__ is called on the same instance. Otherwise, __init__ is not called.
It is also worth noting that in recent years some recipe for creating "singletons" in Python using metaclasses became popular - it is actually an overkill approach,a s overriding __new__ is also preferable for creating singletons.
In your case, you just need to have a dictionary with the parameters you want to track as your keys, and check if you create a new instance or "recycle" one whenever __new__ runs. The dictionary may be a class attribute, or a global variable at module level - that is your pick:
class Recycler:
_instances = {}
def __new__(cls, parameter1, ...):
if parameter1 in cls._instances:
return cls._instances[parameter1]
self = super().__new__(cls) # don't pass remaining parameters to object.__new__
_instances[parameter1] = self
return self
If you'd have any code in __init__ besides that, move it to __new__ as well.
You can have a baseclass with this behavior and have a class hierarchy without needing to re-implement __new__ for every class.
As for a metaclass, none of its methods are called when actually creating a new instance of the classes created with that metaclass. It would only be of use to automatically insert this behavior, by decorating or creating a fresh __new__ method, on classes created with that metaclass. Since this behavior is easier to track, maintain, and overall to combine with other classes just using ordinary inheritance, no need for a metaclass at all.
Say I have a third-party library where a metaclass requires me to implement something. But I want to have an intermediate "abstract" subclass that doesn't. How can I do this?
Consider this to be a very minimal example of what third-party library has:
class ServingMeta(type):
def __new__(cls, name, bases, classdict):
if any(isinstance(b, ServingMeta) for b in bases):
if "name" not in classdict:
# Actual code fails for a different reason,
# but the logic is the same.
raise TypeError(f"Class '{name}' has no 'name' attribute")
return super().__new__(cls, name, bases, classdict)
class Serving(object, metaclass=ServingMeta):
def shout_name(self):
return self.name.upper()
I cannot modify the code above. It's an external dependency (and I don't want to fork it).
The code is meant to be used this way:
class Spam(Serving):
name = "SPAM"
spam = Spam()
print(spam.shout_name())
However, I happen to have a lot of spam, and I want to introduce a base class with the common helper methods. Something like this:
class Spam(Serving):
def thrice(self):
return " ".join([self.shout_name()] * 3)
class LovelySpam(Spam):
name = "lovely spam"
class WonderfulSpam(Spam):
name = "wonderful spam"
Obviously, this doesn't work and fails with the well-expected TypeError: Class 'SpamBase' has no 'name' attribute declared. Would third-party library had a SpamBase class without a metaclass, I could've subclassed that - but no such luck this time (I've mentioned this inconvenience to the library authors).
I can make it a mixin:
class SpamMixin(object):
def thrice(self):
return " ".join([self.shout_name()] * 3)
class LovelySpam(SpamMixin, Serving):
name = "lovely spam"
class WonderfulSpam(SpamMixin, Serving):
name = "wonderful spam"
However, this makes me and my IDE cringe a little, as it quickly becomes cumbersome to repeat SpamMixin everywhere and also because object has no shout_name attribute (and I don't want to silence analysis tools). In short, I just don't like this approach.
What else can I do?
Is there a way to get a metaclass-less version of Serving? I think of something like this:
ServingBase = remove_metaclass(Serving)
class Spam(ServingBase, metaclass=ServingMeta):
...
But don't know how to actually implement remove_metaclass and whenever it's even reasonably possible (of course, it must be doable, with some introspection, but it could require more arcane magic than I can cast).
Any other suggestions are also welcomed. Basically, I want to have my code DRY (one base class to rule them all), and have my linter/code analysis icons all green.
The mixin approach is the correct way to go. If your IDE "cringe" that is a deffect on that tool - just disable a little of the "features" that are obviously incorrect tunning when coding for a dynamic language like Python.
And this is not even about creating things dynamically, it is merely multiple-inheritance, which is supported by the language since forever. And one of the main uses of multiple-inheritance is exactly being able to create mixins just as this one you need.
Another inheritance-based workaround is to make your hierarchy one level deeper, and just introduce the metaclass after you come up with your mixin methods:
class Mixin(object):
def mimixin(self): ...
class SpamBase(Mixin, metaclass=ServingMeta):
name = "stub"
Or just addd the mixin in an intermediate subclass:
class Base(metaclass=Serving Meta):
name = "stub"
class MixedBase(Mixin, Base):
name = "stub"
class MyLovingSpam(MixedBase):
name = "MyLovingSpam"
If you don't want to be repeating the mixin=-base name in every class, that is the way to go.
"Removing" a metaclass just for the sake of having a late mixin is way over the top. Really. Broken. The way to do it wol e re-create the class dynamically, just as #vaultah mentions in the other answer, but doing that in an intermediate class is a thing you should not do. Doing that to please the IDE is something you should not do twice: messing with metaclasses is hard enough already. Removing things on inheritance/class creation that the language puts there naturally is something nasty (cf. this answer: How to make a class attribute exclusive to the super class ) . On the other hand, mixins and multiple inheritance are just natural.
Are you still there? I told you not to do so:
Now, onto your question - instead of "supressing the metaclass" in an intermediate class, it would be more feasible to inherit the metaclass you have there and change its behavior - so that it does not check for the constraints in specially marked classes - create an attribute for your use, like _skip_checking
class MyMeta(ServingMeta):
def __new__(metacls, name, bases, namespace):
if namespace.get("_skip_checking", False):
# hardcode call to "type" metaclass:
del namespace["_skip_checking"]
cls = type.__new__(metacls, name, bases, namespace)
else:
cls = super().__new__(metacls, name, bases, namespace)
return cls
# repeat for __init__ if needed.
class Base(metaclass=MyMeta):
_skip_checking = True
# define mixin methods
class LoveSpam(Base):
name = "LoveSpam"
There's really no direct way to remove the metaclass from a Python class, because the metaclass created that class. What you can try is re-create the class using a different metaclass, which doesn't have unwanted behaviour. For example, you could use type (the default metaclass).
In [6]: class Serving(metaclass=ServingMeta):
...: def shout_name(self):
...: return self.name.upper()
...:
In [7]: ServingBase = type('ServingBase', Serving.__bases__, dict(vars(Serving)))
Basically this takes the __bases__ tuple and the namespace of the Serving class, and uses them to create a new class ServingBase. N.B. this means that ServingBase will receive all bases and methods/attributes from Serving, some of which may have been added by ServingMeta.
What's the correct idiom for this please?
I want to define an object containing properties which can (optionally) be initialized from a dict (the dict comes from JSON; it may be incomplete). Later on I may modify the properties via setters.
There are actually 13+ properties, and I want to be able to use default getters and setters, but that doesn't seem to work for this case:
But I don't want to have to write explicit descriptors for all of prop1... propn
Also, I'd like to move the default assignments out of __init__() and into the accessors... but then I'd need expicit descriptors.
What's the most elegant solution? (other than move all the setter calls out of __init__() and into a method/classmethod _make()?)
[DELETED COMMENT The code for badprop using default descriptor was due to comment by a previous SO user, who gave the impression it gives you a default setter. But it doesn't - the setter is undefined and it necessarily throws AttributeError.]
class DubiousPropertyExample(object):
def __init__(self,dct=None):
self.prop1 = 'some default'
self.prop2 = 'other default'
#self.badprop = 'This throws AttributeError: can\'t set attribute'
if dct is None: dct = dict() # or use defaultdict
for prop,val in dct.items():
self.__setattr__(prop,val)
# How do I do default property descriptors? this is wrong
##property
#def badprop(self): pass
# Explicit descriptors for all properties - yukk
#property
def prop1(self): return self._prop1
#prop1.setter
def prop1(self,value): self._prop1 = value
#property
def prop2(self): return self._prop2
#prop2.setter
def prop2(self,value): self._prop2 = value
dub = DubiousPropertyExample({'prop2':'crashandburn'})
print dub.__dict__
# {'_prop2': 'crashandburn', '_prop1': 'some default'}
If you run this with line 5 self.badprop = ... uncommented, it fails:
self.badprop = 'This throws AttributeError: can\'t set attribute'
AttributeError: can't set attribute
[As ever, I read the SO posts on descriptors, implicit descriptors, calling them from init]
I think you're slightly misunderstanding how properties work. There is no "default setter". It throws an AttributeError on setting badprop not because it doesn't yet know that badprop is a property rather than a normal attribute (if that were the case it would just set the attribute with no error, because that's now normal attributes behave), but because you haven't provided a setter for badprop, only a getter.
Have a look at this:
>>> class Foo(object):
#property
def foo(self):
return self._foo
def __init__(self):
self._foo = 1
>>> f = Foo()
>>> f.foo = 2
Traceback (most recent call last):
File "<pyshell#12>", line 1, in <module>
f.foo = 2
AttributeError: can't set attribute
You can't set such an attribute even from outside of __init__, after the instance is constructed. If you just use #property, then what you have is a read-only property (effectively a method call that looks like an attribute read).
If all you're doing in your getters and setters is redirecting read/write access to an attribute of the same name but with an underscore prepended, then by far the simplest thing to do is get rid of the properties altogether and just use normal attributes. Python isn't Java (and even in Java I'm not convinced of the virtue of private fields with the obvious public getter/setter anyway). An attribute that is directly accessible to the outside world is a perfectly reasonable part of your "public" interface. If you later discover that you need to run some code whenever an attribute is read/written you can make it a property then without changing your interface (this is actually what descriptors were originally intended for, not so that we could start writing Java style getters/setters for every single attribute).
If you're actually doing something in the properties other than changing the name of the attribute, and you do want your attributes to be readonly, then your best bet is probably to treat the initialisation in __init__ as directly setting the underlying data attributes with the underscore prepended. Then your class can be straightforwardly initialised without AttributeErrors, and thereafter the properties will do their thing as the attributes are read.
If you're actually doing something in the properties other than changing the name of the attribute, and you want your attributes to be readable and writable, then you'll need to actually specify what happens when you get/set them. If each attribute has independent custom behaviour, then there is no more efficient way to do this than explicitly providing a getter and a setter for each attribute.
If you're running exactly the same (or very similar) code in every single getter/setter (and it's not just adding an underscore to the real attribute name), and that's why you object to writing them all out (rightly so!), then you may be better served by implementing some of __getattr__, __getattribute__, and __setattr__. These allow you to redirect attribute reading/writing to the same code each time (with the name of the attribute as a parameter), rather than to two functions for each attribute (getting/setting).
It seems like the easiest way to go about this is to just implement __getattr__ and __setattr__ such that they will access any key in your parsed JSON dict, which you should set as an instance member. Alternatively, you could call update() on self.__dict__ with your parsed JSON, but that's not really the best way to go about things, as it means your input dict could potentially trample members of your instance.
As to your setters and getters, you should only be creating them if they actually do something special other than directly set or retrieve the value in question. Python isn't Java (or C++ or anything else), you shouldn't try to mimic the private/set/get paradigm that is common in those languages.
I simply put the dict in the local scope and get/set there my properties.
class test(object):
def __init__(self,**kwargs):
self.kwargs = kwargs
#self.value = 20 asign from init is possible
#property
def value(self):
if self.kwargs.get('value') == None:
self.kwargs.update(value=0)#default
return self.kwargs.get('value')
#value.setter
def value(self,v):
print(v) #do something with v
self.kwargs.update(value=v)
x = test()
print(x.value)
x.value = 10
x.value = 5
Output
0
10
5
I have a weird and unusual use case for metaclasses where I'd like to change the __metaclass__ of a base class after it's been defined so that its subclasses will automatically use the new __metaclass__. But that oddly doesn't work:
class MetaBase(type):
def __new__(cls, name, bases, attrs):
attrs["y"] = attrs["x"] + 1
return type.__new__(cls, name, bases, attrs)
class Foo(object):
__metaclass__ = MetaBase
x = 5
print (Foo.x, Foo.y) # prints (5, 6) as expected
class MetaSub(MetaBase):
def __new__(cls, name, bases, attrs):
attrs["x"] = 11
return MetaBase.__new__(cls, name, bases, attrs)
Foo.__metaclass__ = MetaSub
class Bar(Foo):
pass
print(Bar.x, Bar.y) # prints (5, 6) instead of (11, 12)
What I'm doing may very well be unwise/unsupported/undefined, but I can't for the life of me figure out how the old metaclass is being invoked, and I'd like to least understand how that's possible.
EDIT: Based on a suggestion made by jsbueno, I replaced the line Foo.__metaclass__ = MetaSub with the following line, which did exactly what I wanted:
Foo = type.__new__(MetaSub, "Foo", Foo.__bases__, dict(Foo.__dict__))
The problem is the __metaclass__ attribute is not used when inherited, contrary to what you might expect. The 'old' metaclass isn't called either for Bar. The docs say the following about how the metaclass is found:
The appropriate metaclass is determined by the following precedence
rules:
If dict['__metaclass__'] exists, it is used.
Otherwise, if there is at least one base class, its metaclass is used (this looks for a __class__ attribute first and if not found,
uses its type).
Otherwise, if a global variable named __metaclass__ exists, it is used.
Otherwise, the old-style, classic metaclass (types.ClassType) is used.
So what is actually used as metaclass in your Bar class is found in the parent's __class__ attribute and not in the parent's __metaclass__ attribute.
More information can be found on this StackOverflow answer.
The metaclass information for a class is used at the moment it is created (either parsed as a class block, or dynamically, with a explicit call to the metaclass). It can't be changed because the metaclass usually does make changes at class creation time - the created class type is the metaclasse. Its __metaclass__ attribute is irrelevant once it is created.
However, it is possible to create a copy of a given class, and have the copy bear a different metclass than the original class.
On your example, if instead of doing:
Foo.__metaclass__ = MetaSub
you do:
Foo = Metasub("Foo", Foo.__bases__, dict(Foo.__dict__))
You will achieve what you intended. The new Foo is for all effects equal its predecessor, but with a different metaclass.
However, previously existing instances of Foo won't be considered an instance of the new Foo - if you need that, you better create the Foo copy with a different name instead.
Subclasses use the __metaclass__ of their parent.
The solution to your use-case is to program the parent's __metaclass__ so that it will have different behaviors for the parent than for its subclasses. Perhaps have it inspect the class dictionary for a class variable and implement different behaviors depending on its value
(this is the technique type uses to control whether or not instances are given a dictionary depending on the whether or not __slots__ is defined).
I know that I can dynamically add an instance method to an object by doing something like:
import types
def my_method(self):
# logic of method
# ...
# instance is some instance of some class
instance.my_method = types.MethodType(my_method, instance)
Later on I can call instance.my_method() and self will be bound correctly and everything works.
Now, my question: how to do the exact same thing to obtain the behavior that decorating the new method with #property would give?
I would guess something like:
instance.my_method = types.MethodType(my_method, instance)
instance.my_method = property(instance.my_method)
But, doing that instance.my_method returns a property object.
The property descriptor objects needs to live in the class, not in the instance, to have the effect you desire. If you don't want to alter the existing class in order to avoid altering the behavior of other instances, you'll need to make a "per-instance class", e.g.:
def addprop(inst, name, method):
cls = type(inst)
if not hasattr(cls, '__perinstance'):
cls = type(cls.__name__, (cls,), {})
cls.__perinstance = True
inst.__class__ = cls
setattr(cls, name, property(method))
I'm marking these special "per-instance" classes with an attribute to avoid needlessly making multiple ones if you're doing several addprop calls on the same instance.
Note that, like for other uses of property, you need the class in play to be new-style (typically obtained by inheriting directly or indirectly from object), not the ancient legacy style (dropped in Python 3) that's assigned by default to a class without bases.
Since this question isn't asking about only adding to a spesific instance,
the following method can be used to add a property to the class, this will expose the properties to all instances of the class YMMV.
cls = type(my_instance)
cls.my_prop = property(lambda self: "hello world")
print(my_instance.my_prop)
# >>> hello world
Note: Adding another answer because I think #Alex Martelli, while correct, is achieving the desired result by creating a new class that holds the property, this answer is intended to be more direct/straightforward without abstracting whats going on into its own method.