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]]

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]]

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}]

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]]

下面是一个不一定是平方的列表的转置的解决方案:

maxCol = len(l[0])
for row in l:
    rowLength = len(row)
    if rowLength > maxCol:
        maxCol = rowLength
lTrans = []
for colIndex in range(maxCol):
    lTrans.append([])
    for row in l:
        if colIndex < len(row):
            lTrans[colIndex].append(row[colIndex])

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