交错2个不等长度的列表[重复]

我希望能够交错两个长度可能不相等的列表。我所拥有的是：

  def interleave(xs,ys):
    a=xs
    b=ys
    c=a+b
    c[::2]=a
    c[1::2]=b
    return c

这对于长度等于或只是 +/-1 的列表非常有用。但如果假设 xs=[1,2,3] 和 ys= ["hi,"bye","no","yes","why"] 则会出现此消息：

c[::2]=a
ValueError: attempt to assign sequence of size 3 to extended slice of size 4

如何使用索引来解决这个问题？或者我必须使用for循环？ 编辑：我想要的是让额外的值出现在最后。

您可以使用itertools.izip_longest http://docs.python.org/2/library/itertools.html#itertools.izip_longest here:

>>> from itertools import izip_longest
>>> xs = [1,2,3]
>>> ys = ["hi","bye","no","yes","why"]
>>> s = object()
>>> [y for x in izip_longest(xs, ys, fillvalue=s) for y in x if y is not s]
[1, 'hi', 2, 'bye', 3, 'no', 'yes', 'why']

Using roundrobin食谱来自迭代工具 http://docs.python.org/2/library/itertools.html#recipes，这里不需要哨兵值：

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

Demo:

>>> list(roundrobin(xs, ys))
[1, 'hi', 2, 'bye', 3, 'no', 'yes', 'why']
>>> list(roundrobin(ys, xs))
['hi', 1, 'bye', 2, 'no', 3, 'yes', 'why']