如果a是较长的列表，b则较短
from itertools import groupby
len_ab = len(a) + len(b)
groups = groupby(((a[len(a)*i//len_ab], b[len(b)*i//len_ab]) for i in range(len_ab)),
key=lambda x:x[0])
[j[i] for k,g in groups for i,j in enumerate(g)]
例如
>>> a = range(8)
>>> b = list("abc")
>>> len_ab = len(a) + len(b)
>>> groups = groupby(((a[len(a)*i//len_ab], b[len(b)*i//len_ab]) for i in range(len_ab)), key=lambda x:x[0])
>>> [j[i] for k,g in groups for i,j in enumerate(g)]
[0, 'a', 1, 2, 'b', 3, 4, 5, 'c', 6, 7]
您可以使用此技巧来确保a比b
b, a = sorted((a, b), key=len)
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