Question

我通过循环遍历源列表来构建列表。如果我遇到特定值，我想跳过它以及列表中的下一个n项。

有没有更好的方法来编写下面的process_things函数？

def test_skip(thing):
    return thing == 'b'

def num_to_skip(thing):
    return 3

def process_things(things):
    """
    build list from things where we skip every test_skip() and the next num_to_skip() items
    """
    result = []
    skip_count = 0
    for thing in things:
        if test_skip(thing):
            skip_count = num_to_skip(thing)
        if skip_count > 0:
            skip_count -= 1
            continue
        result.append(thing)
    return result

source = list('abzzazyabyyab')
intended_result = list('aazyaa')

assert process_things(source) == intended_result

Answer 1

您可以使用迭代器和itertools中的consume食谱。致电consume(iterator, n)推进iterator个n个项目。如果在for循环遍历迭代器时调用它，则下一次迭代将从第一个consume没有消耗的项开始。

import collections
from itertools import islice

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)

def process_things(things):
    """
    build list from things where we skip every test_skip() and the next num_to_skip() items
    """
    result = []
    things = iter(things)
    for thing in things:
        if test_skip(thing):
            consume(things, num_to_skip(thing) - 1)
        else:
            result.append(thing)
    return result

请注意我写的process_things的方式，当它遇到test_skip要跳过的项目时，它会计算要跳过的num_to_skip(thing)项目中的项目。这与您的代码所做的相符，但它与您的描述完全匹配，后者表示＆＃34;跳过它以及列表中的下n个项目＆＃34;。

跳过for循环中的列表项

1 个答案: