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]]

matrix = [[1,2,3],
          [1,2,3],
          [1,2,3],
          [1,2,3],
          [1,2,3],
          [1,2,3],
          [1,2,3]]
    
rows = len(matrix)
cols = len(matrix[0])

transposed = []
while len(transposed) < cols:
    transposed.append([])
    while len(transposed[-1]) < rows:
        transposed[-1].append(0)

for i in range(rows):
    for j in range(cols):
        transposed[j][i] = matrix[i][j]

for i in transposed:
    print(i)

只是为了好玩，有效的矩形假设m[0]存在

>>> m = [[1,2,3],[4,5,6],[7,8,9]]
>>> [[row[i] for row in m] for i in range(len(m[0]))]
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

    #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

import numpy as np
r = list(map(list, np.transpose(l)))

也许不是最优雅的解决方案，但这里有一个使用嵌套while循环的解决方案:

def transpose(lst):
    newlist = []
    i = 0
    while i < len(lst):
        j = 0
        colvec = []
        while j < len(lst):
            colvec.append(lst[j][i])
            j = j + 1
        newlist.append(colvec)
        i = i + 1
    return newlist