从List [type]继承时的Pylint错误>> [

Question

我编写了以下代码：

from typing import List

class Foo():
    def __init__(self, _a : int, _b : bool):
        self.a = _a
        self.b = _b

class Bar(List[Foo]):
    def print(self):
        for entry in self: # Non-iterable value self is used in an iterating contextpylint(not-an-iterable)
            print(entry.a)
            print(entry.b)

foo0 = Foo(0, True)
foo1 = Foo(1, False)
foo2 = Foo(2, True)
foo3 = Foo(3, False)
bar = Bar()
bar.append(foo0) # Instance of 'Bar' has no 'append' memberpylint(no-member)
bar.append(foo1) # Instance of 'Bar' has no 'append' memberpylint(no-member)
bar.append(foo2) # Instance of 'Bar' has no 'append' memberpylint(no-member)
bar.append(foo3) # Instance of 'Bar' has no 'append' memberpylint(no-member)

bar.print()

它运行得很好，似乎可以完成预期的工作，但Pylint似乎并不喜欢它（注释中的错误消息。）>
有没有办法使此停止？

我编写了以下代码：输入import List类Foo（）：def __init __（self，_a：int，_b：bool）：self.a = _a self.b = _b class Bar（List [Foo] ）：def打印（...

Answer 1

这将解决所有Pylint错误

from typing import List
class Foo():
    def __init__(self, _a : int, _b : bool):
        self.a = _a
        self.b = _b

class Bar(List[Foo]):
    def print(self):
        for entry in list(self):
            print(entry.a)
            print(entry.b)

foo0 = Foo(0, True)
foo1 = Foo(1, False)
foo2 = Foo(2, True)
foo3 = Foo(3, False)

bar = Bar([foo0,foo1,foo2,foo3])
bar.print()

Answer 2

正如其他人在评论中指出的那样，List[Foo]不是list类型，而是某些类型检查器（例如mypy）将用来分析代码的

类型提示