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

Python 3:

# short circuits at shortest nested list if table is jagged:
list(map(list, zip(*l)))

# discards no data if jagged and fills short nested lists with None
list(map(list, itertools.zip_longest(*l, fillvalue=None)))

Python 2:

map(list, zip(*l))

[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

解释:

要了解发生了什么，我们需要知道两件事:

The signature of zip: zip(*iterables) This means zip expects an arbitrary number of arguments each of which must be iterable. E.g. zip([1, 2], [3, 4], [5, 6]). Unpacked argument lists: Given a sequence of arguments args, f(*args) will call f such that each element in args is a separate positional argument of f. itertools.zip_longest does not discard any data if the number of elements of the nested lists are not the same (homogenous), and instead fills in the shorter nested lists then zips them up.

回到问题l =[[1,2,3]，[4,5,6]，[7,8,9]]的输入，zip(*l)将等价于zip([1,2,3]，[4,5,6]，[7,8,9])。剩下的就是确保结果是列表的列表，而不是元组的列表。

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])

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]]

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，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]]