python3中zip的替代方案？

Question

python3中zip的替代方案？你知道吗

from itertools import zip_longest 
list_1 = [["ele1"],["ele_2"],["ele_3"]]
list_2 = [["ele4"],["ele_5"]]

result = [[x for x in t if x is not None] for t in zip_longest(list_1,list_2)]
print(result)

我的输出是

[[['ele1'], ['ele4']], [['ele_2'], ['ele_5']], [['ele_3']]]

预期产量：

[['ele1'], ['ele4']], [['ele_2'], ['ele_5']], [['ele_3']]

Answer 1

如果要避免使用压缩两个列表，我的方法是在try/except子句中追加两个列表中的值，并在迭代时追加定义为迭代器的list_2（或list_1中最短的）中的值，以这种方式避免必须zip两个列表：

# iterate over the longest list. Define other as iterator
l2 = iter(list_2)
out = [[] for _ in range(len(list_1))]
for ix, i in enumerate(list_1): 
    try:
        out[ix].append(i)
        out[ix].append(next(l2))
    except StopIteration:
        break

提供：

print(out)
[[['ele1'], ['ele4']], [['ele_2'], ['ele_5']], [['ele_3']]]

Answer 2

您还可以在另一个列表中收集两个（或更多）列表，并使用嵌套列表理解来模拟zip_longest的行为。你知道吗

>>> lists = [list_1, list_2] # an also be more than two lists
>>> [[lst[i] for lst in lists if i < len(lst)]
...  for i in range(max(map(len, lists)))]
...
[[['ele1'], ['ele4']], [['ele_2'], ['ele_5']], [['ele_3']]]

（如果在上面的表达式中用max替换min，则得到zip。）

如果要打印不带最外层[...]的结果：

>>> print(', '.join(map(str, _)))                                          
[['ele1'], ['ele4']], [['ele_2'], ['ele_5']], [['ele_3']]