Python 迭代器自定义类，包装在OrderedDict上，对下一步感到困惑__

我意识到我仍然不太理解如何实现迭代器类。因此，一个类可以对其某些内容进行“for循环”

我已经看过这些答案（我仍然不明白）：

据我所知，iterable是一个实现

\uuuuu iter\uuuuuu

并返回迭代器的函数，迭代器是已实现

\uuuu next\uuuuu

的函数。 从这一点上，我不知何故理解了，如果我希望我的类是一个迭代器，并且是iterable。我必须定义

\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

以返回self，并定义

\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。到目前为止我错了吗

这是我班的脚手架：

class Wrapper:
    def __init__(self, low, high):
        self.store = OrderedDict()  # it is imported (ommited)

    def __getitem__(self, key):
        return self.store[key]

    def __iter__(self):
        return self
        # Option 3
        return self.store

    def __next__(self):
        # Option 1 
        for key in self.store:
            return key
        # Option 2
        for key in self.store.keys():
            return key

尝试了上述选项，但均无效：(

发生的事情是我准备了一些py.tests来测试迭代是否正确，只是一个简单的for循环，没有什么特别的。测试只是永远运行（或者比我的耐心<5分钟更长），但是mock类有5个项，所以不会花太长时间

我做错了什么？
您可以进行以下更改：

class Wrapper:
    def __init__(self, low, high):
        self.store = OrderedDict() 
        self.__iter = None  # maintain state of self as iterator

    def __iter__(self):
        if self.__iter is None:
            self.__iter = iter(self.store)
        return self

    def __next__(self):
        try:
            return next(self.__iter)
        except StopIteration:   # support repeated iteration
            self.__iter = None  
            raise

\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
方法应返回
self
，以正确实现协议。这可确保在循环期间调用
next
将迭

for x in wrapper:
    # do stuff with x
    next(wrapper)  # skip one

在大多数情况下，插入
\uuuu iter\uuu
就足够了：

class Wrapper:
    def __init__(self, low, high):
        self.store = OrderedDict() 

    def __iter__(self):
        return iter(self.store)

您可以进行以下更改：

class Wrapper:
    def __init__(self, low, high):
        self.store = OrderedDict() 
        self.__iter = None  # maintain state of self as iterator

    def __iter__(self):
        if self.__iter is None:
            self.__iter = iter(self.store)
        return self

    def __next__(self):
        try:
            return next(self.__iter)
        except StopIteration:   # support repeated iteration
            self.__iter = None  
            raise

\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
方法应返回
self
，以正确实现协议。这可确保在循环期间调用
next
将迭

for x in wrapper:
    # do stuff with x
    next(wrapper)  # skip one

在大多数情况下，插入
\uuuu iter\uuu
就足够了：

class Wrapper:
    def __init__(self, low, high):
        self.store = OrderedDict() 

    def __iter__(self):
        return iter(self.store)

你说，你的概念搞混了

我意识到我仍然不太理解如何实现一个迭代器类。因此，一个可以对其某些内容进行“for循环”的类

但事情不是这样的。如果您只想在类的实例上执行
for
循环，几乎可以肯定的是，您不应该让实例直接成为迭代器。您应该编写一个
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
方法，手动或通过
yield
ing返回迭代器，而不应该编写
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu下一步
：

# Returning an iterator manually:
class Wrapper:
    ...
    def __iter__(self):
        return iter(self.store)

# or with yield:
class Wrapper:
    ...
    def __iter__(self):
        for thing in self.store:
            yield thing

迭代器在技术上是可迭代的，但只能使用一次。要编写迭代器，您必须
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
返回
self
，并在每次调用时
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu.我将在半开放区间[0,10]上编写迭代器，因为为OrderedICT编写迭代器只会将所有内容委托给OrderedICT的迭代器，因为这太直接了，不具有指导意义：

class Iterator:
    def __init__(self):
        self.state = 0
    def __iter__(self):
        return self
    def __next__(self):
        if self.state == 10:
            raise StopIteration
        self.state += 1
        return self.state - 1

但是，即使您确实需要迭代器，手动编写迭代器也是不寻常的；使用
yield
：

def iterator():
    for i in range(10):
        yield i

# or managing the index manually, to show what it would be like without range:
def iterator():
    i = 0
    while i < 10:
        yield i
        i += 1

def迭代器（）：
对于范围（10）内的i：
产量一
#或者手动管理索引，以显示没有范围时的情况：
def迭代器（）：
i=0
当我<10时：
产量一
i+=1
你的概念混淆了。你说

我意识到我仍然不太理解如何实现一个迭代器类。因此，一个可以对其某些内容进行“for循环”的类

但事情不是这样的。如果您只想在类的实例上执行
for
循环，几乎可以肯定的是，您不应该让实例直接成为迭代器。您应该编写一个
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
方法，手动或通过
yield
ing返回迭代器，而不应该编写
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu下一步
：

# Returning an iterator manually:
class Wrapper:
    ...
    def __iter__(self):
        return iter(self.store)

# or with yield:
class Wrapper:
    ...
    def __iter__(self):
        for thing in self.store:
            yield thing

迭代器在技术上是可迭代的，但只能使用一次。要编写迭代器，您必须
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
返回
self
，并在每次调用时
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu.我将在半开放区间[0,10]上编写迭代器，因为为OrderedICT编写迭代器只会将所有内容委托给OrderedICT的迭代器，因为这太直接了，不具有指导意义：

class Iterator:
    def __init__(self):
        self.state = 0
    def __iter__(self):
        return self
    def __next__(self):
        if self.state == 10:
            raise StopIteration
        self.state += 1
        return self.state - 1

但是，即使您确实需要迭代器，手动编写迭代器也是不寻常的；使用
yield
：

def iterator():
    for i in range(10):
        yield i

# or managing the index manually, to show what it would be like without range:
def iterator():
    i = 0
    while i < 10:
        yield i
        i += 1

def迭代器（）：
对于范围（10）内的i：
产量一
#或者手动管理索引，以显示没有范围时的情况：
def迭代器（）：
i=0
当我<10时：
产量一
i+=1
谢谢，这工作得很好。在没有状态维护的情况下，它似乎也能工作。您能解释一下为什么需要它吗？您可以在包装器中为x执行
（隐式调用
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu），但是
下一个（包装器）
（隐式调用
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu如果不维护状态，则无法正常工作。
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
和
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu迭代器协议。因为这个类定义了迭代器，
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu>必须返回
self
。很可能，OPmethod@user2357112你到底是什么意思？是的我的类定义是错误的，但Schwobasegll解决方案似乎使它工作得非常完美。谢谢，这项工作