Define #property on function - python

In JavaScript, we can do the following to any object or function
const myFn = () => {};
Object.defineProperties(myFn, {
property: {
get: () => console.log('property accessed')
}
});
This will allow for a #property like syntax by defining a getter function for the property property.
myFn.property
// property accessed
Is there anything similar for functions in Python?
I know we can't use property since it's not a new-style class, and assigning a lambda with setattr will not work since it'll be a function.
Basically what I want to achieve is that whenever my_fn.property is to return a new instance of another class on each call.
What I currently have with setattr is this
setattr(my_fn, 'property', OtherClass())
My hopes are to design an API that looks like this my_fn.property.some_other_function().
I would prefer using a function as my_fn and not an instance of a class, even though I realize that it might be easier to implement.
Below is the gist of what I'm trying to achieve
def my_fn():
pass
my_fn = property('property', lambda: OtherClass())
my_fn.property
// will be a new instance of OtherClass on each call

It's not possible to do exactly what you want. The descriptor protocol that powers the property built-in is only invoked when:
The descriptor is defined on a class
The descriptor's name is accessed on an instance of said class
Problem is, the class behind functions defined in Python (aptly named function, exposed directly as types.FunctionType or indirectly by calling type() on any function defined at the Python layer) is a single shared, immutable class, so you can't add descriptors to it (and even if you could, they'd become attributes of every Python level function, not just one particular function).
The closest you can get to what you're attempting would be to define a callable class (defining __call__) that defines the descriptor you're interested in as well. Make a single instance of that class (you can throw away the class itself at this point) and it will behave as you expect. Make __call__ a staticmethod, and you'll avoid changing the signature to boot.
For example, the behavior you want could be achieved with:
class my_fn:
# Note: Using the name "property" for a property has issues if you define
# other properties later in the class; this is just for illustration
#property
def property(self):
return OtherClass()
#staticmethod
def __call__(...whatever args apply; no need for self...):
... function behavior goes here ...
my_fn = my_fn() # Replace class with instance of class that behaves like a function
Now you can call the "function" (really a functor, to use C++ parlance):
my_fn(...)
or access the property, getting a brand new OtherClass each time:
>>> type(my_fn.property) is type(my_fn.property)
True
>>> my_fn.property is my_fn.property
False
No, this isn't what you asked for (you seem set on having a plain function do this for you), but you're asking for a very JavaScript specific thing which doesn't exist in Python.

What you want is not currently possible, because the property would have to be set on the function type to be invoked correctly. And you are not allowed to monkeypatch the function type:
>>> type(my_fn).property = 'anything else'
TypeError: can't set attributes of built-in/extension type 'function'
The solution: use a callable class instead.
Note: What you want may become possible in Python 3.8 if PEP 575 is accepted.

Related

Explicit call to __call__ works and uses __init__

I'm learning overloading in Python 3.X and to better understand the topic, I wrote the following code that works in 3.X but not in 2.X. I expected the below code to fail since I've not defined __call__ for class Test. But to my surprise, it works and prints "constructor called". Demo.
class Test:
def __init__(self):
print("constructor called")
#Test.__getitem__() #error as expected
Test.__call__() #this works in 3.X(but not in 2.X) and prints "constructor called"! WHY THIS DOESN'T GIVE ERROR in 3.x?
So my question is that how/why exactly does this code work in 3.x but not in 2.x. I mean I want to know the mechanics behind what is going on.
More importantly, why __init__ is being used here when I am using __call__?
In 3.x:
About attribute lookup, type and object
Every time an attribute is looked up on an object, Python follows a process like this:
Is it directly a part of the actual data in the object? If so, use that and stop.
Is it directly a part of the object's class? If so, hold onto that for step 4.
Otherwise, check the object's class for __getattr__ and __getattribute__ overrides, look through base classes in the MRO, etc. (This is a massive simplification, of course.)
If something was found in step 2 or 3, check if it has a __get__. If it does, look that up (yes, that means starting over at step 1 for the attribute named __get__ on that object), call it, and use its return value. Otherwise, use what was returned directly.
Functions have a __get__ automatically; it is used to implement method binding. Classes are objects; that's why it's possible to look up attributes in them. That is: the purpose of the class Test: block is to define a data type; the code creates an object named Test which represents the data type that was defined.
But since the Test class is an object, it must be an instance of some class. That class is called type, and has a built-in implementation.
>>> type(Test)
<class 'type'>
Notice that type(Test) is not a function call. Rather, the name type is pre-defined to refer to a class, which every other class created in user code is (by default) an instance of.
In other words, type is the default metaclass: the class of classes.
>>> type
<class 'type'>
One may ask, what class does type belong to? The answer is surprisingly simple - itself:
>>> type(type) is type
True
Since the above examples call type, we conclude that type is callable. To be callable, it must have a __call__ attribute, and it does:
>>> type.__call__
<slot wrapper '__call__' of 'type' objects>
When type is called with a single argument, it looks up the argument's class (roughly equivalent to accessing the __class__ attribute of the argument). When called with three arguments, it creates a new instance of type, i.e., a new class.
How does type work?
Because this is digging right at the core of the language (allocating memory for the object), it's not quite possible to implement this in pure Python, at least for the reference C implementation (and I have no idea what sort of magic is going on in PyPy here). But we can approximately model the type class like so:
def _validate_type(obj, required_type, context):
if not isinstance(obj, required_type):
good_name = required_type.__name__
bad_name = type(obj).__name__
raise TypeError(f'{context} must be {good_name}, not {bad_name}')
class type:
def __new__(cls, name_or_obj, *args):
# __new__ implicitly gets passed an instance of the class, but
# `type` is its own class, so it will be `type` itself.
if len(args) == 0: # 1-argument form: check the type of an existing class.
return obj.__class__
# otherwise, 3-argument form: create a new class.
try:
bases, attrs = args
except ValueError:
raise TypeError('type() takes 1 or 3 arguments')
_validate_type(name, str, 'type.__new__() argument 1')
_validate_type(bases, tuple, 'type.__new__() argument 2')
_validate_type(attrs, dict, 'type.__new__() argument 3')
# This line would not work if we were actually implementing
# a replacement for `type`, as it would route to `object.__new__(type)`,
# which is explicitly disallowed. But let's pretend it does...
result = super().__new__()
# Now, fill in attributes from the parameters.
result.__name__ = name_or_obj
# Assigning to `__bases__` triggers a lot of other internal checks!
result.__bases__ = bases
for name, value in attrs.items():
setattr(result, name, value)
return result
del __new__.__get__ # `__new__`s of builtins don't implement this.
def __call__(self, *args):
return self.__new__(self, *args)
# this, however, does have a `__get__`.
What happens (conceptually) when we call the class (Test())?
Test() uses function-call syntax, but it's not a function. To figure out what should happen, we translate the call into Test.__class__.__call__(Test). (We use __class__ directly here, because translating the function call using type - asking type to categorize itself - would end up in endless recursion.)
Test.__class__ is type, so this becomes type.__call__(Test).
type contains a __call__ directly (type is its own class, remember?), so it's used directly - we don't go through the __get__ descriptor. We call the function, with Test as self, and no other arguments. (We have a function now, so we don't need to translate the function call syntax again. We could - given a function func, func.__class__.__call__.__get__(func) gives us an instance of an unnamed builtin "method wrapper" type, which does the same thing as func when called. Repeating the loop on the method wrapper creates a separate method wrapper that still does the same thing.)
This attempts the call Test.__new__(Test) (since self was bound to Test). Test.__new__ isn't explicitly defined in Test, but since Test is a class, we don't look in Test's class (type), but instead in Test's base (object).
object.__new__(Test) exists, and does magical built-in stuff to allocate memory for a new instance of the Test class, make it possible to assign attributes to that instance (even though Test is a subtype of object, which disallows that), and set its __class__ to Test.
Similarly, when we call type, the same logical chain turns type(Test) into type.__class__.__call__(type, Test) into type.__call__(type, Test), which forwards to type.__new__(type, Test). This time, there is a __new__ attribute directly in type, so this doesn't fall back to looking in object. Instead, with name_or_obj being set to Test, we simply return Test.__class__, i.e., type. And with separate name, bases, attrs arguments, type.__new__ instead creates an instance of type.
Finally: what happens when we call Test.__call__() explicitly?
If there's a __call__ defined in the class, it gets used, since it's found directly. This will fail, however, because there aren't enough arguments: the descriptor protocol isn't used since the attribute was found directly, so self isn't bound, and so that argument is missing.
If there isn't a __call__ method defined, then we look in Test's class, i.e., type. There's a __call__ there, so the rest proceeds like steps 3-5 in the previous section.
In Python 3.x, every class is implicitely a child of the builtin class object. And at least in the CPython implementation, the object class has a __call__ method which is defined in its metaclass type.
That means that Test.__call__() is exactly the same as Test() and will return a new Test object, calling your custom __init__ method.
In Python 2.x classes are by default old-style classes and are not child of object. Because of that __call__ is not defined. You can get the same behaviour in Python 2.x by using new style classes, meaning by making an explicit inheritance on object:
# Python 2 new style class
class Test(object):
...

Intercept magic method calls in python class

I am trying to make a class that wraps a value that will be used across multiple other objects. For computational reasons, the aim is for this wrapped value to only be calculated once and the reference to the value passed around to its users. I don't believe this is possible in vanilla python due to its object container model. Instead, my approach is a wrapper class that is passed around, defined as follows:
class DynamicProperty():
def __init__(self, value = None):
# Value of the property
self.value: Any = value
def __repr__(self):
# Use value's repr instead
return repr(self.value)
def __getattr__(self, attr):
# Doesn't exist in wrapper, get it from the value
# instead
return getattr(self.value, attr)
The following works as expected:
wrappedString = DynamicProperty("foo")
wrappedString.upper() # 'FOO'
wrappedFloat = DynamicProperty(1.5)
wrappedFloat.__add__(2) # 3.5
However, implicitly calling __add__ through normal syntax fails:
wrappedFloat + 2 # TypeError: unsupported operand type(s) for
# +: 'DynamicProperty' and 'float'
Is there a way to intercept these implicit method calls without explicitly defining magic methods for DynamicProperty to call the method on its value attribute?
Talking about "passing by reference" will only confuse you. Keep that terminology to languages where you can have a choice on that, and where it makes a difference. In Python you always pass objects around - and this passing is the equivalent of "passing by reference" - for all objects - from None to int to a live asyncio network connection pool instance.
With that out of the way: the algorithm the language follows to retrieve attributes from an object is complicated, have details - implementing __getattr__ is just the tip of the iceberg. Reading the document called "Data Model" in its entirety will give you a better grasp of all the mechanisms involved in retrieving attributes.
That said, here is how it works for "magic" or "dunder" methods - (special functions with two underscores before and two after the name): when you use an operator that requires the existence of the method that implements it (like __add__ for +), the language checks the class of your object for the __add__ method - not the instance. And __getattr__ on the class can dynamically create attributes for instances of that class only.
But that is not the only problem: you could create a metaclass (inheriting from type) and put a __getattr__ method on this metaclass. For all querying you would do from Python, it would look like your object had the __add__ (or any other dunder method) in its class. However, for dunder methods, Python do not go through the normal attribute lookup mechanism - it "looks" directly at the class, if the dunder method is "physically" there. There are slots in the memory structure that holds the classes for each of the possible dunder methods - and they either refer to the corresponding method, or are "null" (this is "viewable" when coding in C on the Python side, the default dir will show these methods when they exist, or omit them if not). If they are not there, Python will just "say" the object does not implement that operation and period.
The way to work around that with a proxy object like you want is to create a proxy class that either features the dunder methods from the class you want to wrap, or features all possible methods, and upon being called, check if the underlying object actually implements the called method.
That is why "serious" code will rarely, if ever, offer true "transparent" proxy objects. There are exceptions, but from "Weakrefs", to "super()", to concurrent.futures, just to mention a few in the core language and stdlib, no one attempts a "fully working transparent proxy" - instead, the api is more like you call a ".value()" or ".result()" method on the wrapper to get to the original object itself.
However, it can be done, as I described above. I even have a small (long unmaintained) package on pypi that does that, wrapping a proxy for a future.
The code is at https://bitbucket.org/jsbueno/lelo/src/master/lelo/_lelo.py
The + operator in your case does not work, because DynamicProperty does not inherit from float. See:
>>> class Foo(float):
pass
>>> Foo(1.5) + 2
3.5
So, you'll need to do some kind of dynamic inheritance:
def get_dynamic_property(instance):
base = type(instance)
class DynamicProperty(base):
pass
return DynamicProperty(instance)
wrapped_string = get_dynamic_property("foo")
print(wrapped_string.upper())
wrapped_float = get_dynamic_property(1.5)
print(wrapped_float + 2)
Output:
FOO
3.5

What is the use of class attribute which is a method/function

In Python when we define class all its members including variables and methods also becomes attributes of that class. In following example MyClass1.a and MyClass1.mydef1 are attributes of class MyClass1.
class MyClass1:
a = 10
def mydef1(self):
return 0
ins1 = MyClass1() # create instance
print(MyClass1.a) # access class attribute which is class variable
print(MyClass1.mydef1) # No idea what to do with it so just printing
print(ins1.mydef1) # No idea what to do with it so just printing
Output
10
<function MyClass1.mydef1 at 0x0000000002122EA0>
<bound method MyClass1.mydef1 of <__main__.MyClass1 object at 0x000000000212D0F0>>
Here attribute a is a variable and it can be used like any other variable.
But mydef1 is a method, if it is not invoked and just used like MyClass1.mydef1 or ins1.mydef1, it returns object for that method(correct me if I am wrong).
So my question is, what can we do with the Class/instance methods without invoking it? Are there any use cases for it or is it just good to know thing?
An attribute of a class that happens to be a function becomes a method for instances or that class:
inst.foo(params, ...)
is internally translated into:
cls.foo(inst, params, ...)
That means that what is actually invoked is the attribute from the class of the instance, and the instance itself is prepended to the argument list. It is just Python syntax to invoke methods on objects.
In your example the correct uses would be:
print(MyClass1.mydef1(ins1)) # prints 0
print(ins1.mydef1()) # also prints 0
Well instance methods can be called with the appropriate parameters of course:
print(ins1.mydef1()) # no parameters, so empty parenthesis, this call should print "0" in your example instead of the method description
If you use it without the parenthesis, you are playing with reference to the function, I don't think you can have any use of it, except checking the list of methods available in a class or something like that.

Why doesn't Python allow referencing a class inside its definition?

Python (3 and 2) doesn't allow you to reference a class inside its body (except in methods):
class A:
static_attribute = A()
This raises a NameError in the second line because 'A' is not defined, while this
class A:
def method(self):
return A('argument')
works fine.
In other languages, for example Java, the former is no problem and it is advantageous in many situations, like implementing singletons.
Why isn't this possible in Python? What are the reasons for this decision?
EDIT:
I edited my other question so it asks only for ways to "circumvent" this restriction, while this questions asks for its motivation / technical details.
Python is a dynamically typed language, and executes statements as you import the module. There is no compiled definition of a class object, the object is created by executing the class statement.
Python essentially executes the class body like a function, taking the resulting local namespace to form the body. Thus the following code:
class Foo(object):
bar = baz
translates roughly to:
def _Foo_body():
bar = baz
return locals()
Foo = type('Foo', (object,), _Foo_body())
As a result, the name for the class is not assigned to until the class statement has completed executing. You can't use the name inside the class statement until that statement has completed, in the same way that you can't use a function until the def statement has completed defining it.
This does mean you can dynamically create classes on the fly:
def class_with_base(base_class):
class Foo(base_class):
pass
return Foo
You can store those classes in a list:
classes = [class_with_base(base) for base in list_of_bases]
Now you have a list of classes with no global names referring to them anywhere. Without a global name, I can't rely on such a name existing in a method either; return Foo won't work as there is no Foo global for that to refer to.
Next, Python supports a concept called a metaclass, which produces classes just like a class produces instances. The type() function above is the default metaclass, but you are free to supply your own for a class. A metaclass is free to produce whatever it likes really, even things that are bit classes! As such Python cannot, up front, know what kind of object a class statement will produce and can't make assumptions about what it'll end up binding the name used to. See What is a metaclass in Python?
All this is not something you can do in a statically typed language like Java.
A class statement is executed just like any other statement. Your first example is (roughly) equivalent to
a = A()
A = type('A', (), {'static_attribute': a})
The first line obviously raises a NameError, because A isn't yet bound to anything.
In your second example, A isn't referenced until method is actually called, by which time A does refer to the class.
Essentially, a class does not exist until its entire definition is compiled in its entirety. This is similar to end blocks that are explicitly written in other languages, and Python utilizes implicit end blocks which are determined by indentation.
The other answers are great at explaining why you can't reference the class by name within the class, but you can use class methods to access the class.
The #classmethod decorator annotes a method that will be passed the class type, instead of the usual class instance (self). This is similar to Java's static method (there's also a #staticmethod decorator, which is a little different).
For a singleton, you can access a class instance to store an object instance (Attributes defined at the class level are the fields defined as static in a Java class):
class A(object):
instance = None
#classmethod
def get_singleton(cls):
if cls.instance is None:
print "Creating new instance"
cls.instance = cls()
return cls.instance
>>> a1 = A.get_singleton()
Creating new instance
>>> a2 = A.get_singleton()
>>> print a1 is a2
True
You can also use class methods to make java-style "static" methods:
class Name(object):
def __init__(self, name):
self.name = name
#classmethod
def make_as_victoria(cls):
return cls("Victoria")
#classmethod
def make_as_stephen(cls):
return cls("Stephen")
>>> victoria = Name.make_as_victoria()
>>> stephen = Name.make_as_stephen()
>>> print victoria.name
Victoria
>>> print stephen.name
Stephen
The answer is "just because".
It has nothing to do with the type system of Python, or it being dynamic. It has to do with the order in which a newly introduced type is initialized.
Some months ago I developed an object system for the language TXR, in which this works:
1> (defstruct foo nil (:static bar (new foo)))
#
2> (new foo)
#S(foo)
3> *2.bar
#S(foo)
Here, bar is a static slot ("class variable") in foo. It is initialized by an expression which constructs a foo.
Why that works can be understood from the function-based API for the instantiation of a new type, where the static class initialization is performed by a function which is passed in. The defstruct macro compiles a call to make-struct-type in which the (new foo) expression ends up in the body of the anonymous function that is passed for the static-initfun argument. This function is called after the type is registered under the foo symbol already.
We could easily patch the C implementation of make_struct_type so that this breaks. The last few lines of that function are:
sethash(struct_type_hash, name, stype);
if (super) {
mpush(stype, mkloc(su->dvtypes, super));
memcpy(st->stslot, su->stslot, sizeof (val) * su->nstslots);
}
call_stinitfun_chain(st, stype);
return stype;
}
The call_stinifun_chain does the initialization which ends up evaluating (new foo) and storing it in the bar static slot, and the sethash call is what registers the type under its name.
If we simply reverse the order in which these functions are called, the language and type system will still be the same, and almost everything will work as before. Yet, the (:static bar (new foo)) slot specifier will fail.
I put the calls in that order because I wanted the language-controlled aspects of the type to be as complete as possible before exposing it to the user-definable initializations.
I can't think of any reason for foo not to be known at the time when that struct type is being initialized, let alone a good reason. It is legitimate for static construction to create an instance. For example, we could use it to create a "singleton".
This looks like a bug in Python.

Python: how does inspect.ismethod work?

I'm trying to get the name of all methods in my class.
When testing how the inspect module works, i extraced one of my methods by obj = MyClass.__dict__['mymethodname'].
But now inspect.ismethod(obj) returns False while inspect.isfunction(obj) returns True, and i don't understand why. Is there some strange way of marking methods as methods that i am not aware of? I thought it was just that it is defined in the class and takes self as its first argument.
You are seeing some effects of the behind-the-scenes machinery of Python.
When you write f = MyClass.__dict__['mymethodname'], you get the raw implementation of "mymethodname", which is a plain function. To call it, you need to pass in an additional parameter, class instance.
When you write f = MyClass.mymethodname (note the absence of parentheses after mymethodname), you get an unbound method of class MyClass, which is an instance of MethodType that wraps the raw function you obtained above. To call it, you need to pass in an additional parameter, class instance.
When you write f = MyClass().mymethodname (note that i've created an object of class MyClass before taking its method), you get a bound method of an instance of class MyClass. You do not need to pass an additional class instance to it, since it's already stored inside it.
To get wrapped method (bound or unbound) by its name given as a string, use getattr, as noted by gnibbler. For example:
unbound_mth = getattr(MyClass, "mymethodname")
or
bound_mth = getattr(an_instance_of_MyClass, "mymethodname")
Use the source
def ismethod(object):
"""Return true if the object is an instance method.
Instance method objects provide these attributes:
__doc__ documentation string
__name__ name with which this method was defined
__func__ function object containing implementation of method
__self__ instance to which this method is bound"""
return isinstance(object, types.MethodType)
The first argument being self is just by convention. By accessing the method by name from the class's dict, you are bypassing the binding, so it appears to be a function rather than a method
If you want to access the method by name use
getattr(MyClass, 'mymethodname')
Well, do you mean that obj.mymethod is a method (with implicitly passed self) while Klass.__dict__['mymethod'] is a function?
Basically Klass.__dict__['mymethod'] is the "raw" function, which can be turned to a method by something called descriptors. This means that every function on a class can be both a normal function and a method, depending on how you access them. This is how the class system works in Python and quite normal.
If you want methods, you can't go though __dict__ (which you never should anyways). To get all methods you should do inspect.getmembers(Klass_or_Instance, inspect.ismethod)
You can read the details here, the explanation about this is under "User-defined methods".
From a comment made on #THC4k's answer, it looks like the OP wants to discriminate between built-in methods and methods defined in pure Python code. User defined methods are of types.MethodType, but built-in methods are not.
You can get the various types like so:
import inspect
import types
is_user_defined_method = inspect.ismethod
def is_builtin_method(arg):
return isinstance(arg, (type(str.find), type('foo'.find)))
def is_user_or_builtin_method(arg):
MethodType = types.MethodType
return isinstance(arg, (type(str.find), type('foo'.find), MethodType))
class MyDict(dict):
def puddle(self): pass
for obj in (MyDict, MyDict()):
for test_func in (is_user_defined_method, is_builtin_method,
is_user_or_builtin_method):
print [attr
for attr in dir(obj)
if test_func(getattr(obj, attr)) and attr.startswith('p')]
which prints:
['puddle']
['pop', 'popitem']
['pop', 'popitem', 'puddle']
['puddle']
['pop', 'popitem']
['pop', 'popitem', 'puddle']
You could use dir to get the name of available methods/attributes/etc, then iterate through them to see which ones are methods. Like this:
[ mthd for mthd in dir(FooClass) if inspect.ismethod(myFooInstance.__getattribute__(mthd)) ]
I'm expecting there to be a cleaner solution, but this could be something you could use if nobody else comes up with one. I'd like if I didn't have to use an instance of the class to use getattribute.

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