Python迭代器协议__iter__在迭代器返回值？

Question

所以我最近在我的 python 类中了解了迭代器协议，我被告知 __iter__(self) 方法应该总是返回一个新的迭代器而不是 self。在 Fluent Python 书中，它谈到了在迭代器中返回 self，所以我不确定为什么我的导师告诉我不能使用 self 作为返回值。

这是我们在考试中的示例，我因使用 self 而不是 FileNamesIterator 而失分。

class FileNamesIterator:
    """
    Iterator that iterates over all the file names in the
    FileNamesIterable.
    """

    def __init__(self, filename_iterable: FileNamesIterable):
        self.filename_iterable = filename_iterable
        self.curr_index = 0

    def __next__(self) -> str:
        file_list = self.filename_iterable.file_list
        if self.curr_index == len(file_list):
            raise StopIteration()

        next_file_name = file_list[self.curr_index]
        self.curr_index += 1

        if ".png" in next_file_name:
            next_file_name = f"{next_file_name} - Portable Network Graphics " \
                             f"File"
        elif ".gif" in next_file_name:
            next_file_name = f"{next_file_name} - Graphics Interchange " \
                             f"Format File"

        return next_file_name

    def __iter__(self) -> FileNamesIterator:
        return self


class FileNamesIterable:
    def __init__(self, file_list: list):
        self.file_list = file_list

    def __iter__(self) -> FileNamesIterator:
        return FileNamesIterator(self)

Answer 1

要在容器类中实现迭代器协议，您需要向其添加__iter__ 方法。这允许使用for 循环迭代类实例，因为它在类实例上调用iter() 以获取迭代器对象（从instance.__iter__()，然后从返回的对象（迭代器）调用__next__ 方法，直到它耗尽（提高StopIteration 而不是返回另一个数据值）。这就是为什么迭代器类的__iter__ 方法应该返回一个带有__next__ 方法的对象。

如果实现该协议的类本身可以具有__next__ 方法，您可以从__iter__ 返回self：

class MyIterator:
    def __init__(self, data):
        self._data = data
        self._index = 0

    def __next__(self):
        if self._index == len(self._data):
            raise StopIteration
        self._index += 1
        return self._data[self._index-1]

    def __iter__(self):
        return self

如果将迭代器类与可迭代对象分开，则必须返回一个新的迭代器对象，而不是 self（也不是迭代器类名）：

class Iterator:
    def __init__(self, iterable):
        self._iterable = iterable

    # required for the iterator protocol
    def __next__(self):
        if not self._iterable:
            raise StopIteration
        value = self._iterable.pop(0)
        return value

    # required for the iterator protocol. It *has* to return self
    def __iter__(self):
        return self


class Container(list):
    """
    This container is a list. A list is already an iterable
    but we override the list __iter__ just to see how its code
    should be like
    """
    def __iter__(self):
        """
        Returns an iterator protocol compliant object. Note it is an
        instance, not the Iterator class. Otherwise it raises TypeError
        (no iterator type. Note also that it is a *new* iterator instance.
        """
        return Iterator(self)


print("Using a single class:")
it = MyIterator([1, 2, 3])
print(it)

for elem in it:
    print(elem)

print()

print("Separating the iterator class from the container class:")
container = Container((1, 2, 3))
it = iter(container)
print(it)

print(next(it))
print(next(it))
print(next(it))

print()

for elem in container:
    print(elem)

输出：

Using a single class:
<__main__.MyIterator object at 0x7f65a23b6fd0>
1
2
3

Separating the iterator class from the container class:
<__main__.Iterator object at 0x7f65a23b6ca0>
1
2
3