从迭代器中删除 N 个值的 Pythonic 解决方案

Question

是否有从迭代器中删除 n 值的 pythonic 解决方案？您只需丢弃 n 值即可，如下所示:

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

但在我看来，这并不像 Python 代码应有的那样优雅。我在这里缺少更好的方法吗？

Answer 1

我相信您正在寻找“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)

如果您不需要 n 为 None 时的特殊行为，你可以使用

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