参见英文答案 how to interleaving lists 2个 我希望能够交错两个可能长度不等的列表.我有的是: def interleave(xs,ys): a=xs b=ys c=a+b c[::2]=a c[1::2]=b return c 这适用于长度相等或只是/ -1的列表.但是如
我希望能够交错两个可能长度不等的列表.我有的是:
def interleave(xs,ys): a=xs b=ys c=a+b c[::2]=a c[1::2]=b return c
这适用于长度相等或只是/ -1的列表.但是如果让我们说xs = [1,2,3]和ys = [“hi,”bye“,”no“,”yes“,”why“]这条消息出现:
c[::2]=a ValueError: attempt to assign sequence of size 3 to extended slice of size 4
如何使用索引修复此问题?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.
itertools.izip_longest
:
>>> 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']
使用itertools的roundrobin配方,此处不需要哨兵值:
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))
演示:
>>> list(roundrobin(xs, ys)) [1, 'hi', 2, 'bye', 3, 'no', 'yes', 'why'] >>> list(roundrobin(ys, xs)) ['hi', 1, 'bye', 2, 'no', 3, 'yes', 'why']