如果python迭代器返回可迭代对象，我如何将这些对象链接到一个大迭代器？ [英] If a python iterator returns iterable objects, how can I chain those objects into one big iterator?

问题描述

我将在这里给出一个简化的例子。假设我在python中有一个迭代器，这个迭代器返回的每个对象本身都是可迭代的。我想获取这个迭代器返回的所有对象，并将它们链接到一个长迭代器中。是否有标准实用程序可以实现这一目标？

I'll give a simplified example here. Suppose I have an iterator in python, and each object that this iterator returns is itself iterable. I want to take all the objects returned by this iterator and chain them together into one long iterator. Is there a standard utility to make this possible?

这是一个人为的例子。

Here is a contrived example.

x = iter([ xrange(0,5), xrange(5,10)])

x 是一个返回迭代器的迭代器，我想链接所有x返回到一个大迭代器的迭代器。在这个例子中这样一个操作的结果应该是
等于xrange（0,10），它应该被懒惰地评估。

x is an iterator that returns iterators, and I want to chain all the iterators returned by x into one big iterator. The result of such an operation in this example should be equivalent to xrange(0,10), and it should be lazily evaluated.

推荐答案

您可以使用 itertools.chain.from_iterable

You can use itertools.chain.from_iterable

>>> import itertools
>>> x = iter([ xrange(0,5), xrange(5,10)])
>>> a = itertools.chain.from_iterable(x)
>>> list(a)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

如果您的版本没有（显然，它是2.6中的新功能），您可以手动执行：

If that's not available on your version (apparently, it's new in 2.6), you can just do it manually:

>>> x = iter([ xrange(0,5), xrange(5,10)])
>>> a = (i for subiter in x for i in subiter)
>>> list(a)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

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