Pythonic解决方案从迭代器​​中删除N值

Question

是否存在n从迭代器中删除值的pythonic解决方案？您可以通过丢弃n值来执行此操作,如下所示:

def _drop(it, n):
    for _ in xrange(n):
        it.next()

但是,IMO并不像Python代码那样优雅.我在这里错过了更好的方法吗？

Answer 1

我相信你正在寻找"消费"食谱

http://docs.python.org/library/itertools.html#recipes

def consume(iterator, n):
    "Advance the iterator n-steps ahead. If n is none, consume entirely."
    # Use functions that consume iterators at C speed.
    if n is None:
        # feed the entire iterator into a zero-length deque
        collections.deque(iterator, maxlen=0)
    else:
        # advance to the empty slice starting at position n
        next(islice(iterator, n, n), None)

如果您在不需要特殊的行为n是None,你可以只使用

next(islice(iterator, n, n), None)

Answer 2

您可以创建一个从 element 开始的迭代切片n：

import itertools
def drop(it, n):
    return itertools.islice(it, n, None)