python_列表推导式_矩阵运算

带条件的列表推导式

even_number = [i for i in range(10) if i % 2 == 0]
even_number # output [0, 2, 4, 6, 8]

[0, 2, 4, 6, 8]

列表推导式的嵌套

matrix = [[i * j for i in range(1, 4)]for j in range(1, 4)]
matrix # output = [[1, 2, 3], [2, 4, 6], [3, 6, 9]]

[[1, 2, 3], [2, 4, 6], [3, 6, 9]]

矩阵的转置

A = [[1, 2], 
     [2, 4], 
     [3, 6]]
B = [[A[j][i] 
      for j in range(len(A))] 
          for i in range(len(A[0])) ]
B

[[1, 2, 3], [2, 4, 6]]

矩阵的逐项积

def harmard_prod(V1, V2):
    if(len(V1) != len(V2) or len(V1[0]) != len(V2[0])):
        raise ValueError("Matrices must have the same shape")
    result = [[V1[i][j] * V2[i][j]
               for j in range(len(V1[0]))] 
                    for i in range(len(V1))]
    return result

A = [[1, 2], 
     [2, 4]]
B = [[1, 3],
     [2, 6]]
harmard_prod(A,B)

[[1, 6], [4, 24]]

笛卡尔积

A = ['a', 'b', 'c']
B = [1, 2]

cartesian_product = [y+str(x) for x in B for y in A]
cartesian_product

['a1', 'b1', 'c1', 'a2', 'b2', 'c2']

矩阵的乘法

import numpy as np
A = [[1, 2, 3],
     [4, 5, 6]]
B = [[1, 2],
     [4, 5],
     [6, 7]]
C = np.zeros((len(A), len(A)))
for i in range(len(A)): # 3
    for j in range(len(B[0])): # 2 
        for k in range(len(A[0])):
            C[i][j] += A[i][k] * B[k][j]
C

array([[27., 33.],
       [60., 75.]])