#Import functions from library
    from numpy import size, array
    #Transpose a 2D list
    def transpose_list_2d(list_in_mat):
        list_out_mat = []
        array_in_mat = array(list_in_mat)
        array_out_mat = array_in_mat.T
        nb_lines = size(array_out_mat, 0)
        for i_line_out in range(0, nb_lines):
            array_out_line = array_out_mat[i_line_out]
            list_out_line = list(array_out_line)
            list_out_mat.append(list_out_line)
        return list_out_mat

只是为了好玩:如果你想把它们都做成字典的话。

In [1]: l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
   ...: fruits = ["Apple", "Pear", "Peach",]
   ...: [dict(zip(fruits, j)) for j in [list(i) for i in zip(*l)]]
Out[1]:
[{'Apple': 1, 'Pear': 4, 'Peach': 7},
 {'Apple': 2, 'Pear': 5, 'Peach': 8},
 {'Apple': 3, 'Pear': 6, 'Peach': 9}]

More_itertools.unzip()很容易阅读，它也可以用于生成器。

import more_itertools
l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
r = more_itertools.unzip(l) # a tuple of generators.
r = list(map(list, r))      # a list of lists

或者同样的

import more_itertools
l = more_itertools.chunked(range(1,10), 3)
r = more_itertools.unzip(l) # a tuple of generators.
r = list(map(list, r))      # a list of lists

方阵的另一种方法。不使用numpy和itertools，使用(有效的)就地元素交换。

def transpose(m):
    for i in range(1, len(m)):
        for j in range(i):
            m[i][j], m[j][i] = m[j][i], m[i][j]

等价于耶拿的解决方案:

>>> l=[[1,2,3],[4,5,6],[7,8,9]]
>>> [list(i) for i in zip(*l)]
... [[1, 4, 7], [2, 5, 8], [3, 6, 9]]

一种方法是用NumPy转置。如需列出清单，请填写:

>>> import numpy as np
>>> np.array(l).T.tolist()
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

或者另一个没有zip (python < 3)的:

>>> map(list, map(None, *l))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

或者对于python >= 3:

>>> list(map(lambda *x: list(x), *l))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]