python-3.x - python : In place transpose of a matrix

Question

我尝试编写一个就地转置函数只是为了练习。谁能告诉我这个算法的时间和空间复杂度是多少？

from copy import *

def transpose(matrix):
    reference=deepcopy(matrix)
    col_num=len(reference[0])
    row_num=len(reference)
    matrix.clear()

    new=[list(map(lambda x: x[i],reference)) for i in range(col_num)]
    for i in new:
        matrix.append(new)
    return matrix

    x=[[ 1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]
    y=transpose(x)

编辑:使我的就地转置代码更加简洁

Answer 1

选项 1:如果是方阵 NxN，则:

def transpose(matrix):
    # Transpose O(N*N)
    size = len(matrix)
    for i in range(size):
        for j in range(i+1, size):
            matrix[j][i],matrix[i][j] = matrix[i][j],matrix[j][i]

选项 2:通用的“pytonic”方式是这样做的:

mat = [[1,2,3],[4,5,6],[7,8,9]]
transpose = list(zip(*mat))
transpose
>>> [(1, 4, 7), (2, 5, 8), (3, 6, 9)]

zip function reference