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Leer Challenge: Refactoring an ABC to a Protocol | Protocols vs ABCs
Python Abstract Base Classes
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Challenge: Refactoring an ABC to a Protocol

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You are given an existing ABC-based interface and two classes that currently inherit from it. Your task is to create an equivalent Protocol, verify that both classes satisfy it without inheritance, and compare isinstance() behavior between the two approaches.

You are given the following existing code:

from abc import ABC, abstractmethod

class Transformer(ABC):
    @abstractmethod
    def transform(self, data):
        pass

    @abstractmethod
    def get_name(self):
        pass

class UpperCaseTransformer(Transformer):
    def transform(self, data):
        return [item.upper() for item in data]

    def get_name(self):
        return "UpperCase"

class TrimTransformer(Transformer):
    def transform(self, data):
        return [item.strip() for item in data]

    def get_name(self):
        return "Trim"
Taak

Veeg om te beginnen met coderen

  1. Import Protocol and runtime_checkable from typing.
  2. Define a @runtime_checkable Protocol called TransformerProtocol with two methods: transform(self, data) and get_name(self), both with ... as their body.
  3. Define a new class ReverseTransformer (do not inherit from Transformer or TransformerProtocol) with:
    • transform(self, data) that returns a list of reversed strings: [item[::-1] for item in data];
    • get_name(self) that returns "Reverse".
  4. Create instances: upper = UpperCaseTransformer(), trim = TrimTransformer(), reverse = ReverseTransformer().
  5. Store the following in variables:
    • upper_abc = isinstance(upper, Transformer);
    • reverse_abc = isinstance(reverse, Transformer);
    • upper_proto = isinstance(upper, TransformerProtocol);
    • reverse_proto = isinstance(reverse, TransformerProtocol).
  6. Print all four variables.

Oplossing

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