如何交错两个不同长度的列表？

Question

我是Python的新手,我仍然很难将语言本身用于我的程序.这是我到目前为止所拥有的:

# Purpose: 'twolists' = takes 2 lists, & returns a new list containing
# alternating elements of lists. 
# Return = final_list
# Parameter = list1, list2

def twolists(list1, list2): # don't forget to return final_list
    alt_list = []
    a1 = len(list1)
    a2 = len(list2)

    for i in range(# ? ):
        # append one thing from list1 to alt_list - How?
        # append one thing from list2 to alt_list - How?

现在该程序应该产生如下输出:

outcome = twolists([ ], ['w', 'x', 'y', 'z'])
print(outcome)
['w', 'x', 'y', 'z']

outcome = twolists([0, 1], ['w', 'x'])
print(outcome)
[0, 'w', 1, 'x']

outcome = twolists([0, 1], ['w', 'x', 'y', 'z'])
print(outcome)
[0, 'w', 1, 'x', 'y', 'z']

outcome = twolists([0, 1, 2, 3], ['w', 'x'])
print(outcome)
[0, 'w', 1, 'x', 2, 3]

Answer 1

def twolists(list1, list2):
    newlist = []
    a1 = len(list1)
    a2 = len(list2)

    for i in range(max(a1, a2)):
        if i < a1:
            newlist.append(list1[i])
        if i < a2:
            newlist.append(list2[i])

    return newlist

Answer 2

这使用zip_longest来自itertools(它是标准库的一部分)的列表推导来将两个列表中的项交织成a tuple,默认情况下None用作fillvalue.

这还使用chain也itertools趋于平缓列表.

最后,它会过滤None列表中的项目:

from itertools import chain, zip_longest
def twolists(l1, l2):
    return [x for x in chain(*zip_longest(l1, l2)) if x is not None]

或者根据@EliKorvigo的推荐,itertools.chain.from_iterable用于懒惰迭代:

def twolists(l1, l2):
    return [x for x in chain.from_iterable(zip_longest(l1, l2)) if x is not None]

测试

In [56]: twolists([0, 1], ['w', 'x'])
Out[56]: [0, 'w', 1, 'x']

In [57]: twolists([0, 1], ['w', 'x', 'y', 'z'])
Out[57]: [0, 'w', 1, 'x', 'y', 'z']

In [74]: twolists([0, 1, 2, 3], ['w', 'x'])
Out[74]: [0, 'w', 1, 'x', 2, 3]