Python如何优雅的交错合并两个列表

Question

比如a = [1, 2, 3], b = [4, 5, 6]
合并为[1, 4, 2, 5, 3, 6]
你觉得怎么写比较优雅？

Answer 1

pythonfrom itertools import chain list(chain.from_iterable(zip(a, b))) # py2 list(chain(*zip(a, b)))

Answer 2

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

Answer 3

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))

Answer 4

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……

Answer 5

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

Answer 6

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)))

Answer 7

a.extend(b)

Answer 8

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)

Answer 9

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