セクション 3. 章 4
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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"
タスク
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- Import
Protocolandruntime_checkablefromtyping. - Define a
@runtime_checkableProtocol calledTransformerProtocolwith two methods:transform(self, data)andget_name(self), both with...as their body. - Define a new class
ReverseTransformer(do not inherit fromTransformerorTransformerProtocol) with:transform(self, data)that returns a list of reversed strings:[item[::-1] for item in data];get_name(self)that returns"Reverse".
- Create instances:
upper = UpperCaseTransformer(),trim = TrimTransformer(),reverse = ReverseTransformer(). - 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).
- Print all four variables.
解答
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セクション 3. 章 4
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