Python 方法重写 - 派生类中的参数比基类更具体

Question

假设我想创建一个名为

Document

的抽象基类，并且希望其所有子类实现名为

from_paragraphs

的类方法，该方法从一系列

Paragraph

对象构造文档。但是，

LegalDocument

只能从

LegalParagraph

对象构造，而

AcademicDocument

只能从

AcademicParagraph

对象构造。

我的直觉是这样做：

from abc import ABC, abstractmethod
from typing import Sequence


class Document(ABC):
    @classmethod
    @abstractmethod
    def from_paragraphs(cls, paragraphs: Sequence["Paragraph"]):
        pass


class LegalDocument(Document):
    @classmethod
    def from_paragraphs(cls, paragraphs: Sequence["LegalParagraph"]):
        return  # some logic here...


class AcademicDocument(Document):
    @classmethod
    def from_paragraphs(cls, paragraphs: Sequence["AcademicParagraph"]):
        return  # some logic here...


class Paragraph:
    text: str


class LegalParagraph(Paragraph):
    pass


class AcademicParagraph(Paragraph):
    pass

然而，Pyright 对此有所抱怨，因为派生类上的

from_paragraphs

违反了里氏替换原则。如何表达我想要的？

Answer 1

这种模式称为工厂模式：根据输入，您会得到不同类型的对象。你那里的东西不起作用，因为：

# Because Document should not have knowledge of derived class:
doc = Document.from_paragraphs(...)

# Because type mismatch
doc = LegalDocument.from_paragraphs([AcademicParagraphs()]) 

这是我解决这个问题的方法：

class Document:
    def __init__(self, paragraphs):
        self.paragraphs = paragraphs

class LegalDocument(Document):
    pass

class AcademicDocument(Document):
    pass

class Paragraph:
    def __init__(self, text):
        self.text = text

class LegalParagraph(Paragraph):
    pass

class AcademicParagraph(Paragraph):
    pass

def create_document(*paragraphs):
    # assume that all paragraphs are of the same type
    if isinstance(paragraphs[0], LegalParagraph):
        klass = LegalDocument
    elif isinstance(paragraphs[0], AcademicParagraph):
        klass = AcademicDocument
    else:
        raise TypeError()

    return klass(paragraphs)


d1 = create_document(LegalParagraph("foo"), LegalParagraph("bar"))
assert isinstance(d1, LegalDocument)

d2 = create_document(AcademicParagraph("moo"))
assert isinstance(d2, AcademicDocument)

注释

• 我将有一套简单的课程，而不是搞乱ABC
• 没有类方法，

  __ini__()

  就足够了
• 我有一个工厂函数

  create_document

  ，它将创建一个文档，其中类型取决于输入
• 另一种方法是调整

  Document.__new__()

  方法，但这需要一些关于

  __new__()

  工作原理的知识，但并不是每个人都知道这一点。