以一种交替的方式将两个列表结合起来?

这里有一种方法可以通过切片来实现:

>>> list1 = ['f', 'o', 'o']
>>> list2 = ['hello', 'world']
>>> result = [None]*(len(list1)+len(list2))
>>> result[::2] = list1
>>> result[1::2] = list2
>>> result
['f', 'hello', 'o', 'world', 'o']

这里面有一个食谱 itertools documentation :

from itertools import cycle, islice

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))

编辑:

对于大于3的python版本:

from itertools import cycle, islice

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))

这应该是你想要的:

>>> iters = [iter(list1), iter(list2)]
>>> print list(it.next() for it in itertools.cycle(iters))
['f', 'hello', 'o', 'world', 'o']

import itertools
print [x for x in itertools.chain.from_iterable(itertools.izip_longest(list1,list2)) if x]

我认为这是最有毒瘾的方法。

>>> sum(zip(l1, l2+[0]), ())[:-1]
('f', 'hello', 'o', 'world', 'o')

使用itertools并假设列表不包含任何内容:

>>> filter(None, sum(itertools.izip_longest(l1, l2), ()))
('f', 'hello', 'o', 'world', 'o')

这里是 Duncan's solution

list1 = ['f', 'o', 'o', 'b', 'a', 'r']
list2 = ['hello', 'world']
num = min(len(list1), len(list2))
result = [None]*(num*2)
result[::2] = list1[:num]
result[1::2] = list2[:num]
result.extend(list1[num:])
result.extend(list2[num:])
result

这将输出:

['f', 'hello', 'o', 'world', 'o', 'b', 'a', 'r'] 

[x for y in zip(list1, list2) for x in y]

由于第一个列表还有一个元素,您可以将其添加到后期:

[x for y in zip(list1, list2) for x in y] + [list1[-1]]

list3 = [ item for pair in zip(list1, list2 + [0]) for item in pair][:-1]

def combine(list1, list2):
    lst = []
    len1 = len(list1)
    len2 = len(list2)

    for index in range( max(len1, len2) ):
        if index+1 <= len1:
            lst += [list1[index]]

        if index+1 <= len2:
            lst += [list2[index]]

    return lst

这是基于卡洛斯·瓦林特的贡献

A=["a","b","c","d"]
B=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]

def cyclemix(xs, ys, n=1):
    for p in range(0,int((len(ys)+len(xs))/n)):
        for g in range(0,min(len(ys),n)):
            yield ys[0]
            ys.append(ys.pop(0))
        for g in range(0,min(len(xs),n)):
            yield xs[0]
            xs.append(xs.pop(0))

print [x for x in cyclemix(A, B, 3)]

这将按每组3个值交错列出A和B:

['a', 'b', 'c', 1, 2, 3, 'd', 'a', 'b', 4, 5, 6, 'c', 'd', 'a', 7, 8, 9, 'b', 'c', 'd', 10, 11, 12, 'a', 'b', 'c', 13, 14, 15]

可能有点晚了再买一条蟒蛇一行。当两个列表的大小相等或不相等时,这种方法就起作用了。值得一提的是,它将修改a和b。如果这是一个问题,您需要使用其他解决方案。

a = ['f', 'o', 'o']
b = ['hello', 'world']
sum([[a.pop(0), b.pop(0)] for i in range(min(len(a), len(b)))],[])+a+b
['f', 'hello', 'o', 'world', 'o']

a = "hlowrd"
b = "el ol"

def func(xs, ys):
    ys = iter(ys)
    for x in xs:
        yield x
        yield ys.next()

print [x for x in func(a, b)]

这是一个使用列表理解的单行程序,没有其他库:

list3 = [sub[i] for i in range(len(list2)) for sub in [list1, list2]] + [list1[-1]]

如果您允许通过副作用更改初始列表1,还有另一种方法:

[list1.insert((i+1)*2-1, list2[i]) for i in range(len(list2))]

from itertools import chain
list(chain(*zip('abc', 'def')))  # Note: this only works for lists of equal length
['a', 'd', 'b', 'e', 'c', 'f']

在最短时间内停车:

def interlace(*iters, next = next) -> collections.Iterable:
    """
    interlace(i1, i2, ..., in) -> (
        i1-0, i2-0, ..., in-0,
        i1-1, i2-1, ..., in-1,
        .
        .
        .
        i1-n, i2-n, ..., in-n,
    )
    """
    return map(next, cycle([iter(x) for x in iters]))

当然,解析next/\uu next\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。

list3 = [element for element in list(itertools.chain.from_iterable([val for val in itertools.izip_longest(list1, list2)])) if element != None]

灵感来源于 answers to another question

import itertools

list(itertools.chain.from_iterable(itertools.izip_longest(list1, list2, fillvalue=object)))[:-1]

[i for l in itertools.izip_longest(list1, list2, fillvalue=object) for i in l if i is not object]

[item for sublist in map(None, list1, list2) for item in sublist][:-1]

努比呢?它也适用于字符串:

import numpy as np

np.array([[a,b] for a,b in zip([1,2,3],[2,3,4,5,6])]).ravel()

结果:

array([1, 2, 2, 3, 3, 4])

from itertools import zip_longest
from functools import reduce

reduce(lambda lst, zipped: [*lst, *zipped] if zipped[1] != None else [*lst, zipped[0]], zip_longest(list1, list2),[])

我要做的很简单:

chain.from_iterable( izip( list1, list2 ) )

它将提供一个迭代器,而不需要创建任何额外的存储需求。

我年纪太大了,无法理解列表,所以:

import operator
list3 = reduce(operator.add, zip(list1, list2))