斯坦福cs231n计算机视觉课训练营之矩阵类

ゞ浴缸里的玫瑰 2022-04-22 07:50 185阅读 0赞

根据python和numpy的教程，写一个矩阵的类。

M a t r i x . p y Matrix.py Matrix.py

class Matrix:
        def __init__(self, lst2d):
            self._values = [row[:] for row in lst2d]
    
        def __repr__(self):
            return "Matrix({})".format(self._values)
    
        __str__ = __repr__
    	
        # 获取矩阵的形状
        def shape(self):
            return len(self._values), len(self._values[0])
    	# 获取矩阵的行数
        def row_num(self):
            return self.shape()[0]
    
        __len__ = row_num
        
    	# 获取矩阵的列数
        def col_num(self):
            return self.shape()[1]
    	# 获取矩阵的大小
        def size(self):
            r, c = self.shape()
            return r * c
    	
        # 获取矩阵的元素
        def __getitem__(self, item):
            r, c = item
            return self._values[r][c]
    	
        # 获取矩阵指定行
        def row_vector(self, index):
            return list(self._values[index])
    	
        # 获取矩阵指定列
        def col_vector(self, index):
            return list([row[index] for row in self._values])
    	
        # 获取矩阵加法
        def __add__(self, other):
            assert self.shape() == other.shape(), \
                "Error in adding. Shape of matrix must be same."
            return Matrix([[a + b for a, b in zip(self.row_vector(i), other.row_vector(i))]
                           for i in range(self.row_num())])
    	
        # 获取矩阵减法
        def __sub__(self, other):
            assert self.shape() == other.shape(), \
                "Error in subtracting. Shape of matrix must be same."
            return Matrix([[a - b for a, b in zip(self.row_vector(i), other.row_vector(i))]
                           for i in range(self.row_num())])
    	# 获取矩阵乘法
        def __mul__(self, k):
            return Matrix([[e * k for e in self.row_vector(i)]
                           for i in range(self.row_num())])
    	# 获取矩阵右乘
        def __rmul__(self, k):
            return self * k
    	
        # 获取矩阵除法
        def __truediv__(self, other):
            return (1 / other) * self
    
        # 矩阵取正
        def __pos__(self):
            return 1 * self
    	
        # 矩阵取负
        def __neg__(self):
            return -1 * self
    	
        # 构造零矩阵
        @classmethod
        def zero(cls, r, c):
            return cls([[0] * c for _ in range(r)])
    	
        # 获取矩阵点乘
        def dot(self, other):
            if isinstance(other, Vector):
                # 矩阵和向量的乘法
                assert self.col_num() == len(other), \
                    "Error in Matrix-Vector Multiplication."
                return Vector([self.row_vector(i).dot(other) for i in range(self.row_num())])
    
            if isinstance(other, Matrix):
                # 矩阵和矩阵的乘法
                assert self.col_num() == other.row_num(), \
                    "Error in Matrix-Matrix Multiplication."
                return Matrix([[self.row_vector(i).dot(other.col_vector(j)) for j in range(other.col_num())]
                               for i in range(self.row_num())])
    	
        # 获取矩阵转置
        def T(self):
            return Matrix([[e for e in self.col_vector(i)]
                           for i in range(self.col_num())])

m a i n . p y main.py main.py

from playLA.Matrix import Matrix
    
    if __name__ == '__main__':
        matrix = Matrix([[1, 2], [3, 4]])
    
        print(matrix)
    
        print(matrix.shape())
        print(len(matrix))
    
        print(matrix[0, 0])
    
        print(matrix.col_vector(1))
    
        matrix2 = Matrix([[5, 6], [7, 8]])
    
        print(matrix + matrix2)
        print(matrix - matrix2)
    
        print(matrix * 2)
        print(2 * matrix)
    
        print(Matrix.zero(3, 2))
    
        matrix3 = Matrix([[1.5, 0], [0, 2]])
        matrix4 = list([5, 3])
    
        print("A:{}, B:{} \nA.dot(B)={}".format(matrix3, matrix4, matrix3.dot(matrix4)))
        print("A:{}, B:{} \nA.dot(B)={}".format(matrix, matrix2, matrix.dot(matrix2)))
    
        matrix5 = Matrix([[1, 2, 3], [4, 5, 6]])
        print("matrix5.T = {}".format(matrix5.T()))

##### output ###
    Matrix([[1, 2], [3, 4]])
    (2, 2)
    2
    1
    (2, 4)
    Matrix([[6, 8], [10, 12]])
    Matrix([[-4, -4], [-4, -4]])
    Matrix([[2, 4], [6, 8]])
    Matrix([[2, 4], [6, 8]])
    Matrix([[0, 0], [0, 0], [0, 0]])
    A:Matrix([[1.5, 0], [0, 2]]), B:(5, 3) 
    A.dot(B)=(7.5, 6)
    A:Matrix([[1, 2], [3, 4]]), B:Matrix([[5, 6], [7, 8]]) 
    A.dot(B)=Matrix([[19, 22], [43, 50]])
    matrix5.T = Matrix([[1, 4], [2, 5], [3, 6]])