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How to extend the list data structure in Python without violating Liskov substitution - supply an attribute instead of an instance?

I’m building a class that extends the list data structure in Python, called a Partitional. I’m adding a few methods that I find myself using frequently when dividing a list into partitions.
The class is initialized with a (nullable) list, which exists as an attribute on the class.
class Partitional(list):
"""Extends the list data type. Adds methods for dividing a list into partition sets
and returning data about those partition sets"""
def __init__(self, source_list: list=[]):
super().__init__()
self.source_list: list = source_list
self.n: int = len(source_list)
...
I want to be able to reliably replace list instances with Partitional instances without violating Liskov substitution. So for list’s methods, I wrote methods on the Partitional class that operate on self.source_list, e.g.
...
def remove(self, matched_item):
self.source_list.remove(matched_item)
self.__init__(self.source_list)
def pop(self, *args):
popped_item = self.source_list.pop(*args)
self.__init__(self.source_list)
return popped_item
def clear(self):
self.source_list.clear()
self.__init__(self.source_list)
...
(the __init__ call is there because the Partitional class builds some internal attributes based on self.source_list when it’s initialized, so these need to be rebuilt if source_list changes.)
And I also want Python’s built-in methods that take a list as an argument to work with a Partitional instance, so I set to work writing method overrides for those as well, e.g.
...
def __len__(self):
return len(self.source_list)
def __enumerate__(self):
return enumerate(self.source_list)
...
The relevant built-in methods are a finite set for any given Python version, but... is there not a simpler way to do this?
My question:
Is there a way to write a class such that, if an instance of that class is used as the argument for a function, the class provides an attribute to the function instead, by default?
That way I’d only need to override this default behaviour for a subset of built-in methods.
So for example, if a use case involving a list instance looks like this:
example_list: list = [1,2,3,4,5]
length = len(example_list)
we substitute a Partitional instance built from the same list:
example_list: list = [1,2,3,4,5]
example_partitional = Partitional(example_list)
length = len(example_partitional)
and what’s “actually” happening is this:
length = len(example_partitional.source_list)
i.e.
length = len([1,2,3,4,5])
Other notes:
In working on this, I’ve realized that there are two broad categories of Liskov substitution violation possible:
Inherent violation, where the structure of the child class will make it incompatible with any use case where the child class is used in place of the parent class, e.g. if you override some fundamental property or structure of the parent.
Context-dependent violation, where, for any given piece of software, so long as you never use the child class in a way that would violate Liskov substitution, you’re fine. E.g. You override a method on the parent class that would change how a built-in function acts when it takes an instance of the class as an argument, but you never use that built-in method with the class instance in your system. Or any system that depends on your system. Or... (you see how relying on this caveat is not foolproof)
What I’m looking to do is come up with a technique that will protect against both categories of violation, without having to worry about use cases and context.

How to use python metaclass for the following scenario?

I want to create a configuration class with cascading feature. What do I mean by this? let say we have a configuration class like this
class BaseConfig(metaclass=ConfigMeta, ...):
def getattr():
return 'default values provided by the metaclass'
class Config(BaseConfig):
class Embedding(BaseConfig, size=200):
class WordEmbedding(Embedding):
size = 300
when I use this in code I will access the configuration as follows,
def function(Config, blah, blah):
word_embedding_size = Config.Embedding.Word.size
char_embedding_size = Config.Embedding.Char.size
The last line access a property which does not exist in Embedding class 'Char'. That should invoke getattr() which should return 200 in this case. I am not familiar with metaclasses enough to make a good judgement, but I gues I need to define the __new__() of the metaclass.
does this approach makes sense or is there a better way to do it?
EDIT:
class Config(BaseConfig):
class Embedding(BaseConfig, size=200):
class WordEmbedding(Embedding):
size = 300
class Log(BaseConfig, level=logging.DEBUG):
class PREPROCESS(Log):
level = logging.INFO
#When I use
log = logging.getLogger(level=Config.Log.Model.level) #level should be INFO

This is a bit confuse. I am not sure if this would be the best notation to declare configurations with default parameters - it seems verbose. But yes, given the flexibility of metaclasses and magic methods in Python, it is possible for something like this to old all flexibility you need.
Just for the sake of it, I'd like to say that using nested classes as namespaces, like you are doing, is probably the only useful thing for them. (nested classes). It is common to see a lot of people that misunderstands Python OO at all trying to make use of nested classes.
So - for your problem, you need that in the final class, a __getattr__ method exists that can fetch default values for atributes. These attributes in turn are declared as keywords to nested classes - which also can have the same metaclass. Otherwise, the hierarchy of nested classes just work for you to fetch nested attributes, using the dot notation in Python.
Moreover, for each class in a nested set, one can pass in keyword parameters that are to be used as default, if the next level of nested classes is not defined. In the given example, trying to access Config.Embedding.Char.size with a non exisitng Char should return the default "size". Not that a __getattr__ in "Embedding" can return you a fake "Char" object - but that object is the one that have to yield a size attribute. So, our __getattr__ have yet to yield an object that has itself a propper __getattr__;
However, I will suggest a change to your requirements - instead of passing in the default values as keyword parameters, to have a reserved name - like _default inside which you can put your default attributes. That way, you can provide deeply nested default subtress, instead of just scalar values as well, and the implementation can possibly be simpler.
Actually - a lot simpler. By using keywords to the class as you propose, you'd actually need to have a metaclass set those default parameters in a data structure(it would be possible in either __new__ or __init__ though). But by just using the nested classes all the way, with a reserved name, a custom __getattr__ on the metac class will work. That will retrieve unexisting class attributes on the configuration classes themselves, and all one have to do, if a requested attribute does not exist, is try to retrieve the _default class I mentioned.
Thus, you can work with something like:
class ConfigMeta(type):
def __getattr__(cls, attr):
return cls._default
class Base(metaclass=ConfigMeta):
pass
class Config(Base):
class Embed(Base):
class _default(Base):
size = 200
class Word(Base):
size = 300
assert Config.Embed.Char.size == 200
assert Config.Embed.Word.size == 300
Btw - just last year I was working on a project to have configurations like this, with default values, but using a dictionary syntax - that is why I mentioned I am not sure the nested class would be a nice design. But since all the functionality can be provided by a metaclass with 3 LoC I guess this beats anything in the way.
Also, that is why I think being able to nest whole default subtrees can be useful for what you want - I've been there.

You can use a metaclass to set the attribute:
class ConfigMeta(type):
def __new__(mt, clsn, bases, attrs):
try:
_ = attrs['size']
except KeyError:
attrs['size'] = 300
return super().__new__(mt, clsn, bases, attrs)
Now if the class does not have the size attribute, it would be set to 300 (change this to meet your need).

When is the object() built-in useful?

I'm trying to figure out what I would use the object() built-in function for. It takes no arguments, and returns a "featureless object" of the type that is common to all Python classes, and has all the methods that are common to all Python classes.
To quote Jack Skellington, WHAT. IS. THIS?

Even if you do not need to program with it, object serves a purpose: it is the common class from which all other objects are derived. It is the last class listed by the mro (method resolution order) method. We need a name and object for this concept, and object serves this purpose.
Another use for object is to create sentinels.
sentinel = object()
This is often used in multithreaded programming -- passed through queues -- to signal a termination event. We might not want to send None or any other value since the queue handler may need to interpret those values as arguments to be processed. We need some unique value that no other part of the program may generate.
Creating a sentinel this way provides just such a unique object that is sure not to be a normal queue value, and thus can be tested for and used as a signal for some special event. There are other possibilities, such as creating a class, or class instance, or a function, but all those alternatives are bigger, more resource heavy, and not as pithy as object().

It is most useful if you are overriding the dot (especially __setattr__), it allows you to break recursion. For example:
class SomeClass(object):
def __setattr__(self, name, value):
if name not in ('attr1', 'attr2', 'attr3', 'attr4'):
object.__setattr__(self, name, value)
else:
do_something_else()

Python idiom for dict-able classes?

I want to do something like this:
class Dictable:
def dict(self):
raise NotImplementedError
class Foo(Dictable):
def dict(self):
return {'bar1': self.bar1, 'bar2': self.bar2}
Is there a more pythonic way to do this? For example, is it possible to overload the built-in conversion dict(...)? Note that I don't necessarily want to return all the member variables of Foo, I'd rather have each class decide what to return.
Thanks.

The Pythonic way depends on what you want to do. If your objects shouldn't be regarded as mappings in their own right, then a dict method is perfectly fine, but you shouldn't "overload" dict to handle dictables. Whether or not you need the base class depends on whether you want to do isinstance(x, Dictable); note that hasattr(x, "dict") would serve pretty much the same purpose.
If the classes are conceptually mappings of keys to values, then implementing the Mapping protocol seems appropriate. I.e., you'd implement
__getitem__
__iter__
__len__
and inherit from collections.Mapping to get the other methods. Then you get dict(Foo()) for free. Example:
class Foo(Mapping):
def __getitem__(self, key):
if key not in ("bar1", "bar2"):
raise KeyError("{} not found".format(repr(key))
return getattr(self, key)
def __iter__(self):
yield "bar1"
yield "bar2"
def __len__(self):
return 2

Firstly, look at collections.ABC, which describes the Python abstract base class protocol (equivalent to interfaces in static languages).
Then, decide if you want to write your own ABC or make use of an existing one; in this case, Mapping might be what you want.
Note that although the dict constructor (i.e. dict(my_object)) is not overrideable, if it encounters an iterable object that yields a sequence of key-value pairs, it will construct a dict from that; i.e. (Python 2; for Python 3 replace items with iteritems):
def __iter__(self):
return {'bar1': self.bar1, 'bar2': self.bar2}.iteritems()
However, if your classes are intended to behave like a dict you shouldn't do this as it's different from the expected behaviour of a Mapping instance, which is to iterate over keys, not key-value pairs. In particular it would cause for .. in to behave incorrectly.

Most of the answers here are about making your class behave like a dict, which isn't actually what you asked. If you want to express the idea, "I am a class that can be turned into a dict," I would simply define a bunch of classes and have them each implement .dict(). Python favors duck-typing (what an object can do) over what an object is. The ABC doesn't add much. Documentation serves the same purpose.

You can certainly overload dict() but you almost never want to! Too many aspects of the standard library depend upon dict being available and you will break most of its functionality. You cab probably do something like this though:
class Dictable:
def dict(self):
return self.__dict__

Python - Overwriting __getattribute__ for an instance?

This one seems a bit tricky to me. Sometime ago I already managed to overwrite an instance's method with something like:
def my_method(self, attr):
pass
instancemethod = type(self.method_to_overwrite)
self.method_to_overwrite = instancemethod(my_method, self, self.__class__)
which worked very well for me; but now I'm trying to overwrite an instance's __getattribute__() function, which doesn't work for me for the reason the method seems to be
<type 'method-wrapper'>
Is it possible to do anything about that? I couldn't find any decent Python documentation on method-wrapper.

You want to override the attribute lookup algorithm on an per instance basis? Without knowing why you are trying to do this, I would hazard a guess that there is a cleaner less convoluted way of doing what you need to do. If you really need to then as Aaron said, you'll need to install a redirecting __getattribute__ handler on the class because Python looks up special methods only on the class, ignoring anything defined on the instance.
You also have to be extra careful about not getting into infinite recursion:
class FunkyAttributeLookup(object):
def __getattribute__(self, key):
try:
# Lookup the per instance function via objects attribute lookup
# to avoid infinite recursion.
getter = object.__getattribute__(self, 'instance_getattribute')
return getter(key)
except AttributeError:
return object.__getattribute__(self, key)
f = FunkyAttributeLookup()
f.instance_getattribute = lambda attr: attr.upper()
print(f.foo) # FOO
Also, if you are overriding methods on your instance, you don't need to instanciate the method object yourself, you can either use the descriptor protocol on functions that generates the methods or just curry the self argument.
#descriptor protocol
self.method_to_overwrite = my_method.__get__(self, type(self))
# or curry
from functools import partial
self.method_to_overwrite = partial(my_method, self)

You can't overwrite special methods at instance level. For new-style classes, implicit invocations of special methods are only guaranteed to work correctly if defined on an object’s type, not in the object’s instance dictionary.

There are a couple of methods which you can't overwrite and __getattribute__() is one of them.

I believe method-wrapper is a wrapper around a method written in C.