让我们来:

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

我想要的结果是

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

而不是

r = [(1, 4, 7), (2, 5, 8), (3, 6, 9)]

当前回答

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]

其他回答

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

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]]

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

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])