将嵌套的字符串列表转换为特定顺序的嵌套字符列表 python

Question

我有一个类似这样的列表列表：

list_example = [['one', 'two', 'thr'], ['fou','fiv','six']]

目标是将字符串中的每个字符分解成一个新列表，其中每个列表都有字符在每个字符串中的出现顺序（解释起来很复杂，但像这样）：

new_list = [[['o', 't', 't'], ['n', 'w', 'h'], ['e', 'o', 'r']], [['f','f','s'],['o','i','i'],['u','v','x']]]

我编写了一个仅适用于一个块的代码：

list_one = ['one', 'two', 'thr']
broken_up_data = []
for i in range(0, 3):
     list_broken_up = [seq[i] for seq in list_one]
     broken_up_data.append(list_broken_up)
print(broken_up_data)
>[['o', 't', 't'], ['n', 'w', 'h'], ['e', 'o', 'r']]

但是当我尝试遍历所有列表时，我得到：

list_example = [['one', 'two', 'thr'], ['fou','fiv','six']]
broken_up_data = []
for x in list_example:
    for i in range(0, 3):
        list_broken_up = [seq[i] for seq in list_example]
        broken_up_data.append(list_broken_up)
print(broken_up_data)
>[['one', 'fou'], ['two', 'fiv'], ['thr', 'six'], ['one', 'fou'], ['two', 'fiv'], ['thr', 'six']]

我可以看到它在单独的字符串中按位置对字符串进行分组（即 'one' 和 'fou' 都排在第一位），但它不再分解字符串并且没有按照我的意愿正确循环。这可能是我设置方式的简单修复？

Answer 1

使用zip，您可以聚合元素。（获取成对的第一个元素，第二个元素，...）

>>> ['one', 'two', 'thr']
['one', 'two', 'thr']
>>> zip(*['one', 'two', 'thr'])
<zip object at 0x7f68226f4d80>
>>> list(zip(*['one', 'two', 'thr']))
[('o', 't', 't'), ('n', 'w', 'h'), ('e', 'o', 'r')]

# To get lists, not tuples
>>> list(map(list, zip(*['one', 'two', 'thr'])))
[['o', 't', 't'], ['n', 'w', 'h'], ['e', 'o', 'r']]
# or
>>> [list(x) for x in zip(*['one', 'two', 'thr'])]
[['o', 't', 't'], ['n', 'w', 'h'], ['e', 'o', 'r']]

将以上应用到给定列表（+列表理解）

>>> list_example = [['one', 'two', 'thr'], ['fou','fiv','six']]
>>> [list(zip(*x)) for x in list_example]
[[('o', 't', 't'), ('n', 'w', 'h'), ('e', 'o', 'r')], [('f', 'f', 's'), ('o', 'i', 'i'), ('u', 'v', 'x')]]
>>> [list(map(list, zip(*x))) for x in list_example]
[[['o', 't', 't'], ['n', 'w', 'h'], ['e', 'o', 'r']], [['f', 'f', 's'], ['o', 'i', 'i'], ['u', 'v', 'x']]]

将嵌套的字符串列表转换为特定顺序的嵌套字符列表 python

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