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

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]]

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])

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]

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