有三种选择:

1. Zip地图

solution1 = map(list, zip(*l))

2. 列表理解

solution2 = [list(i) for i in zip(*l)]

3.For循环附加

solution3 = []
for i in zip(*l):
    solution3.append((list(i)))

查看结果:

print(*solution1)
print(*solution2)
print(*solution3)

# [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}]

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

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]

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]]