Python中除迭代器的最后N个元素外的所有元素 [英] All but the last N elements of iterator in Python

问题描述

在Python中获取除迭代器的最后N个元素之外的所有元素的最佳方法是什么?这是理论上的例子:

What is the best way to get all but the last N elements of an iterator in Python? Here is an example of it in theoretical action:

>>> list(all_but_the_last_n(range(10), 0))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> list(all_but_the_last_n(range(10), 2))
[0, 1, 2, 3, 4, 5, 6, 7]

推荐答案

仅出于乐趣，这是Ignacio解决方案的一种变体，不需要双端队列.

Just for the fun of it, here's a variation on Ignacio's solution that doesn't require a deque.

>>> def truncate(it, n):
...     cache = [next(it) for i in range(n)]
...     index = 0
...     for val in it:
...         val, cache[index] = cache[index], val
...         index = (index + 1) % n
...         yield val

当我写上面的文章时，我并不特别关心速度...但是也许这会更快一点:

I wasn't especially concerned with speed when I wrote the above... but perhaps this would be a tad faster:

def truncate(it, n):
    cache = [next(it) for i in range(n)]
    index = 0
    for val in it:
        yield cache[index]
        cache[index] = val
        index = (index + 1) % n

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