Question

我有两个清单。我想像这样加入他们：

[1, 2, 3], [1, 2, 3] => [1, 1, 2, 2, 3, 3]
[1], [2, 3] => [1, 2]
[], [1, 2, 3] => []
[4, 5, 6, 7], [1] => [4, 1]

在没有itertools?的情况下，有没有简单的方法可以做到这一点似乎我想出了如何做到这一点：

new_list = first_list+second_list
new_list[::2] = first_list
new_list[1::2] = second_list
return new_list

但它只适用于相同长度的列表

Answer 1

使用zip将您的列表列表（您可以拥有任意数量的子列表）作为参数使用*解包并展平项目（zip停靠在最短的清单）：

s = [1, 2, 3], [1, 2, 3]

result = [y for x in zip(*s) for y in x]

相当于（对于那些不习惯参数解包的人）：

first_list = [1, 2, 3]
second_list = [1, 2, 3]
result = [y for x in zip(first_list,second_list) for y in x]

结果：

[1, 1, 2, 2, 3, 3]

使用s = [1], [2, 3]，您可以按预期获得[1,2]

Answer 2

您可以将zip与列表理解结合使用：

>>> [x for row in zip([1, 2, 3], [1, 2, 3]) for x in row]
[1, 1, 2, 2, 3, 3]
>>> [x for row in zip([1], [2, 3]) for x in row]
[1, 2]
>>> [x for row in zip([], [1, 2, 3]) for x in row]
[]
>>> [x for row in zip([4, 5, 6, 7], [1]) for x in row]
[4, 1]

如果您想从最长的列表中获取每个元素，可以使用itertools.zip_longest：

>>> from itertools import zip_longest
>>> [x for row in zip_longest([4, 5, 6, 7], [1]) for x in row]
[4, 1, 5, None, 6, None, 7, None]

它将None元素添加到最短列表中。

Answer 3

此功能适用于上述及以外的这些案例！

def joo(l1,l2):
    l = []
    for i in range(min(len(l1), len(l2))):
        l.append(l1[i])
        l.append(l2[i])
    return l
l = joo(l1, l2)

感谢Adirio的建议！

Answer 4

您可以使用zip：

def zip_lists(list1, list2):
    return [a for b in zip(list1, list2) for a in b]

print(zip_lists([1, 2, 3], [1, 2, 3]))  # => [1, 1, 2, 2, 3, 3]
print(zip_lists([1], [2, 3]))  # => [1, 2]
print(zip_lists([], [1, 2, 3]))  # => []
print(zip_lists([4, 5, 6, 7], [1]))  # => [4, 1]

将两个列表加在一起会遇到麻烦

4 个答案: