Python iter()函数如何工作?

问题描述

以下代码使我感到困惑:

The following code confuses me:

>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> zip(*([iter(a)]*2))
[(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)]
>>> iter(a)
<listiterator object at 0x7f3e9920cf50>
>>> iter(a).next()
0
>>> iter(a).next()
0
>>> iter(a).next()
0

next()始终返回0.那么，iter函数如何工作?

next() is always returning 0. So, how does the iter function work?

推荐答案

您每次都创建一个 new 迭代器.每个新的迭代器都从头开始，它们都是独立的.

You are creating a new iterator each time. Each new iterator starts at the beginning, they are all independent.

创建一次迭代器，然后对该一个实例进行迭代:

Create the iterator once, then iterate over that one instance:

>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> a_iter = iter(a)
>>> next(a_iter)
0
>>> next(a_iter)
1
>>> next(a_iter)
2

我使用了 next()函数，而不是调用了iterator.next()方法； Python 3将后者重命名为iterator.__next__()，但是next()函数将调用正确的拼写"，就像使用len()调用object.__len__一样.

I used the next() function rather than calling the iterator.next() method; Python 3 renames the latter to iterator.__next__() but the next() function will call the right 'spelling', just like len() is used to call object.__len__.

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