python通过返回嵌入式iterable使类可迭代 [英] python make class iterable by returning embedded iterable

问题描述

我在python中有一个类，该类具有可迭代的实例变量.我想通过迭代嵌入式可迭代对象来迭代类的实例.

I have a class in python, which has an iterable as instance variable. I want to iterate the instances of the class by iterating over the embedded iterable.

我实现了以下步骤:

def __iter__(self):
    return self._iterable.__iter__()

我真的不太喜欢在迭代器上调用__iter__()方法，因为它是一种特殊的方法.这是您将如何在python中解决此问题的方法，还是有更优雅的解决方案?

I don't really feel that comfortable calling the __iter__() method on the iterable, as it is a special method. Is this how you would solve this problem in python or is there a more elegant solution?

推荐答案

委派__iter__的最佳"方法是:

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

或者，可能值得了解:

def __iter__(self):
    for item in self._iterable:
        yield item

在退还商品之前，您可以先摆弄些什么(例如，如果需要yield item * 2).

Which will let you fiddle with each item before returning it (ex, if you wanted yield item * 2).

正如@Lattyware在评论中提到的那样，PEP380(计划包含在Python 3.3中)将允许:

And as @Lattyware mentions in the comments, PEP380 (slated for inclusion in Python 3.3) will allow:

def __iter__(self):
    yield from self._iterable

请注意，执行以下操作可能很诱人:

Note that it may be tempting to do something like:

def __init__(self, iterable):
    self.__iter__ = iterable.__iter__

但是无效:iter(foo)直接绕过foo.__iter__调用type(foo)上的__iter__方法.考虑例如:

But this won't work: iter(foo) calls the __iter__ method on type(foo) directly, bypassing foo.__iter__. Consider, for example:

class SurprisingIter(object):
    def __init__(self):
        self.__iter__ = lambda self: iter("abc")

    def __iter__(self):
        return iter([1, 2, 3])

您希望list(SurprisingIter())返回["a", "b", "c"]，但实际上返回[1, 2, 3].

You would expect that list(SurprisingIter()) would return ["a", "b", "c"], but it actually returns [1, 2, 3].

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