数组（python3）

Dear 丶 2022-05-22 11:56 49阅读 0赞

# 数组--------------------
    # 定义一个[n,m]的二维数组
    n, m = 3, 4
    array_2 = [[None for i in range(n)] for i in range(m)]
    # 定义一个[x,y,x]的三维数组
    x, y, z = 2, 3, 4
    array_3 = [[[None for i in range(x)] for i in range(y)] for i in range(z)]
    
    
    # 求[m,n]两矩阵和
    def add_matrix_2(a, b, m, n):  # a和b必须矩阵的行列数量相同
        sum = [[None for i in range(m)] for j in range(n)]
        for i in range(0, m):
            for j in range(0, n):
                sum[m][n] = a[m][n] + b[m][n]
        return sum
    
    
    # 二维矩阵相乘
    def multiply_matrix_2(a, b, a_line, a_column, b_line, b_column):  # a*b矩阵
        if a_line != b_column or a_column != b_line:
            print('不满足条件\n')
            return None
        else:
            multiply_matrix = [[None for i in range(b_column)] for j in range(a_line)]
            term = 0
            for x in range(a_line):
                for y in range(a_column):
                    term += a[a_line][a_column] * b[a_column][a_line]
                multiply_matrix[a_line][a_column] = term
            return multiply_matrix
    
    
    # 二维矩阵转置
    def transposition_matrix_2(a, m, n):
        a_t = [[None for i in range(m)] for j in range(n)]
        for i in range(m):
            for j in range(n):
                a_t[m][n] = a[n][m]
        return a_t
    
    
    # 稀疏矩阵压缩
    def compress_matrix_2(a, m, n):
        rar_matrix[0][0] = m  # 行数
        rar_matrix[0][1] = n  # 列数
        temp = 1
        flag = 0  # 为0的个数
        for i in range(m):
            for j in range(n):
                if a[m][n] != 0:
                    rar_matrix[temp][0] = i
                    rar_matrix[temp][1] = j
                    rar_matrix[temp][2] = a[m][n]
                    temp += 1
        rar_matrix[0][2] = temp
        return rar_matrix
    
    
    # 上三角矩阵转列表
    def change_matrix_2_to_list(a, m, n):
        list_1 = []
        for i in range(m):
            for j in range(n):
                if a[m][n] != 0:
                    list_1.append(a[m][n])
        return list_1