尝试通过交替值组合列表时卡在扩展切片上

Question

我有一段代码如下：

ascii_key = [None]*(len(ascii_username) + len(ascii_password))
ascii_key[0::2] = ascii_username
ascii_key[1::2] = ascii_password

我不断收到以下错误：

ValueError：尝试将大小为 8 的序列分配给大小为 6 的扩展切片

我认为这与它要我准确填写整个扩展列表这一事实有关，如果是这种情况，有没有办法在它发疯之前将两个列表组合起来？

理想的过程是列表 ["A", "B", "C"] 和 ["D", "E", "F"] 变成

["A", "D", "B", "E", "C", "F"]

我尝试按照此 post Pythonic way to combine two lists in an alternating fashion?

中的解决方案

-提前致歉，我是新手程序员。

Answer 1

对于该任务，您可以使用 roundrobin，如 itertools recipes

中所示

from itertools import cycle, islice
def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    num_active = len(iterables)
    nexts = cycle(iter(it).__next__ for it in iterables)
    while num_active:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            # Remove the iterator we just exhausted from the cycle.
            num_active -= 1
            nexts = cycle(islice(nexts, num_active))

然后：

t1 = ['A', 'B', 'C']
t2 = ['D', 'E', 'F']
interleaved = list(roundrobin(t1,t2))
print(interleaved)  # ['A', 'D', 'B', 'E', 'C', 'F']

请注意，由于 roundrobin(t1,t2) 是生成器，我将其输入 list 函数以获得 list。它将使用所有元素，例如 t1 = ['A', 'B', 'C']、t2 = ['D', 'E', 'F', 'G', 'H'] 结果是 ['A', 'D', 'B', 'E', 'C', 'F', 'G', 'H'].

Answer 2

你要做的是交织两个序列。您看到错误的原因是 ascii_username 和 ascii_password 的长度不同。当您尝试分配给列表的切片时，您必须提供切片中元素的确切数量。

在这个例子中很容易看出：以 2 的步长对 x 进行切片比 y.

具有更多的元素

x = [1, 2, 3, 4, 5, 6, 7, 8]
y = 'yes'
z = 'nope!'

print(x[::2])
# prints:
[1, 3, 5, 7]
print(list(y))
['y', 'e', 's']

尝试分配给 x[::2] 留下未重新分配的悬空 7。

要解决此问题，您可以使用 more_itertools 包中的 interleave_longest。

from more_itertools import interleave_longest

list(interleave_longest(x, y))
# returns:
['y', 'n', 'e', 'o', 's', 'p', 'e', '!']

或者如果你不想安装新包，函数的源代码非常小。

from itertools import chain, zip_longest

_marker = object()

def interleave_longest(*iterables):
    """Return a new iterable yielding from each iterable in turn,
    skipping any that are exhausted.

        >>> list(interleave_longest([1, 2, 3], [4, 5], [6, 7, 8]))
        [1, 4, 6, 2, 5, 7, 3, 8]

    This function produces the same output as :func:`roundrobin`, but may
    perform better for some inputs (in particular when the number of iterables
    is large).

    """
    i = chain.from_iterable(zip_longest(*iterables, fillvalue=_marker))
    return [x for x in i if x is not _marker]


interleave_longest(x, y)
# returns:
['y', 'n', 'e', 'o', 's', 'p', 'e', '!']

尝试通过交替值组合列表时卡在扩展切片上

Stuck on extended slices when attempting to combine lists by alternating values

python

list

sequence

slice