如何转置列表列表？

Question

令ll为列表的列表，tt为元组的元组

输入：ll =[["a1","a2"],["b1","b2"],["c1","c2"]]

期望的输出： tt =(("a1","b1","c1"),("a2","b2","c2"))

我已经设法解决了一个二元素列表的问题，这意味着内部列表每个只包含两个元素。

def list_of_list_to_tuple_of_tuple(ll):
    first_elements = [i[0] for i in ll]
    second_elements = [i[1] for i in ll]
    
    new_list = []
    new_list.append(tuple(first_elements))
    new_list.append(tuple(second_elements))
    return tuple(new_list)

ll = [["a1","a2"],["b1","b2"],["c1","c2"]]
list_of_list_to_tuple_of_tuple(ll)

现在，问题是：

1. 有没有其他方法可以轻松完成我所做的事情？

2. 如果我们有一个包含3 个内部列表且每个内部列表包含 n 个元素的列表，是否有任何方法可以轻松推广该算法？例如：

输入： ll =[["a1","a2","a3",..."an"],["b1","b2","b3",..."bn"],["c1","c2","c3",..."cn"]]

期望的输出： tt =(("a1","b1","c1"),("a2","b2","c2"),("a3","b3","c3"),...,("an","bn","cn"))

Answer 1

试试这个单行词——

tuple(zip(*l))

实施例1

l = [["a1","a2"],
     ["b1","b2"],
     ["c1","c2"]]

tuple(zip(*l))

(('a1', 'b1', 'c1'), 
 ('a2', 'b2', 'c2'))

实施例2

l2 = [["a1","a2","a3","an"],
      ["b1","b2","b3","bn"],
      ["c1","c2","c3","cn"]]

tuple(zip(*l2))

(('a1', 'b1', 'c1'),
 ('a2', 'b2', 'c2'),
 ('a3', 'b3', 'c3'),
 ('an', 'bn', 'cn'))

解释

1. 解包运算符允许您将列表解包到子列表中，并将它们作为单独的参数传递给zip，正如它所期望的那样。
2. 将zip每个子列表的第一个、第二个、第三个...第n个各自的元素组合成n个元组对象

3. 元组转换此 zip 对象将整个 zip 对象转换为元组。

奖金

直观上，此操作类似于矩阵的转置。如果将列表列表转换为 numpy 数组然后进行转置，就可以很容易地看出这一点。

import numpy as np

l = [["a1","a2"],["b1","b2"],["c1","c2"]]
arr = np.array(l)
transpose = arr.T
transpose

array([['a1', 'b1', 'c1'],
       ['a2', 'b2', 'c2']], dtype='<U2')