a.extend(b)

python# Python 3
def interpolate(*seqs):
    for items in zip(*seqs):
        yield from items

>>> list(interpoltae('string', 'asdfgh'))
['s', 'a', 't', 's', 'r', 'd', 'i', 'f', 'n', 'g', 'g', 'h']

有个问题，如果这几个序列不都等长该怎么办。上面的解法是按长度最小的来：

python>>> list(interpolate('string', 'asdf'))
['s', 'a', 't', 's', 'r', 'd', 'i', 'f']

但是如果不按最小的来该怎么办？补字符？补什么字符？

原来stackoverflow已经讨论过，写法都很吊哦。我个人很喜欢这个：
调用的cycle/islice函数都来自itertools

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))

pythondef xmerge(a, b):
  tmp = (list(a), list(b));
  return [tmp[i%2].pop(0) if tmp[i%2] else tmp[1-i%2].pop(0) for i in xrange(0, len(a) + len(b))]


print xmerge([1,2,3], [5,6,7,8,9])
print xmerge([1,2,3,4,5], [7,8,9])

这样？

//写到一半搜了下python的三元表达式想起以前也是这种神兽飞过的心情用python的……
//不过还有三元写成if a then b else c的语言，现在还是能理解了……

修正: 之前的代码有问题, 更新一次..

不知道算不算优雅, 但应该省内存:

def xmerge(a, b):
    alen, blen = len(a), len(b)
    mlen = min(alen, blen)
    for i in xrange(mlen):
        yield a[i]
        yield b[i]

    if alen > blen:
        for i in xrange(mlen, alen):
            yield a[i]
    else:
        for i in xrange(mlen, blen):
            yield b[i]

a = [1, 2, 3]
b = [5, 6, 7, 8, 9, 10]

c = [i for i in xmerge(a, b)]
print c

c = [i for i in xmerge(b, a)]
print c

a = [1, 2, 3]
b = [4, 5, 6]

一般的方法

def slove(a, b):
c = []
i = 0
j = 0
while i<len(a) and j<len(b):
c.append(a[i])
c.append(b[j])
i += 1
j += 1
while i < len(a):
c.append(a[i])
i += 1
while j < len(b):
c.append(b[j])
j += 1

print(c)

if name == 'main':
slove(a, b)

a = [1,2,3]
b = [4,5,6]
s = []
for i in zip(a,b):s.extend(list(i))
print s

result:

[1, 4, 2, 5, 3, 6]

python3.3
result = [list(zip(a, b))[i][j] for i in range(len(a)) for j in range(len(list(zip(a, b))[0]))]

虽然是一行，但是有点牵强，小括号太多

a=[1,2,3]
b=[4,5,6]
a=set(a)
b=set(b)
c=list(a|b)

如果只是列表的merge,是不是可以转成set()，再做交集操作
set作list merge时比list快。
但 @lohocla4dam 帮忙指出了不足

pythonfrom itertools import chain
list(chain.from_iterable(zip(a, b)))

# py2
list(chain(*zip(a, b)))

优雅的做数据处理，scipy系列库还是需要的。
有现成matplotlib中的flatten函数可以用。

from matplotlib.cbook import flatten
a = [1, 2, 3]
b = [4, 5, 6]
list(flatten(zip(a,b)))