Python实现矩阵转置的几种方法详解

import numpy as np
A = np.mat([[1,2,3],[4,5,6],[7,8,9]])
print(A.T)
print(A.swapaxes(0, 1))

import numpy as np
A = [[1,2,3],[4,5,6],[7,8,9]]
print(np.transpose(A))

>>> a = [1,2,3]
>>> b = [4,5,6]
>>> c = [4,5,6,7,8]
>>> A = zip(a,b)     # 打包为元组的列表
[(1, 4), (2, 5), (3, 6)]
>>> zip(a,c)              # 元素个数与最短的列表一致
[(1, 4), (2, 5), (3, 6)]
>>> zip(*A)          # 与 zip 相反，*A 可理解为解压，返回二维矩阵式
[(1, 2, 3), (4, 5, 6)]

A = [[1,2,3],[4,5,6],[7,8,9]]
print(*A) #[1, 2, 3] [4, 5, 6] [7, 8, 9]
#zip()返回的是一个对象。如需展示列表，需手动 list() 转换。
#print(zip(*A)) #<zip object at 0x000001CD7733A2C8>
print(list(zip(*A)))

A = [[1,2,3],[4,5,6],[7,8,9]]
#print(len(A)) #矩阵行数
#print(len(A[0])) #矩阵列数
B = [[A[j][i] for j in range(len(A))] for i in range(len(A[0]))]
print(B)

B = [[A[j][i] for j in range(len(A))] for i in range(len(A[0]))]

A = [[1,2,3],[4,5,6],[7,8,9]]
#print(len(A)) #矩阵行数
#print(len(A[0])) #矩阵列数
for i in range(len(A[0])):#len(A[0])矩阵列数
    for j in range(i,len(A)):#len(A)矩阵行数
        #转置就是A[i][j]和A[j][i]互换
        A[j][i], A[i][j] = A[i][j], A[j][i]
print(A)

A = [[1,2,3],[4,5,6],[7,8,9]]
B=[]
for i in range(len(A[0])):#len(A[0])矩阵列数
    temp = []
    for j in range(len(A)):#len(A)矩阵行数
        temp.append(A[j][i])
    B.append(temp)
print(B)