在python中的实现矩阵的乘法 、矩阵的转置、学习numpy中的矩阵运算方法

创建 矩阵类文件 Matrix.py
创建 需要使用到的 矩阵函数功能

from.Vector import Vector
class Matrix:

    def __init__(self,list2d):
        self._values = [row[:]for row in list2d]

    def row_vector(self,index):
        '返回矩阵的第index个行向量'
        return Vector(self._values[index])

    def col_vector(self,index):
        '返回矩阵的第index个列向量'
        return Vector([row[index]] for row in self._values)

    def __getitem__(self, pos):
        '返回矩阵中pos位置的元素'
        r,c = pos
        return self._values[r][c]

    def size(self):
        '返回矩阵中元素的个数'
        r,c = self.shape()
        return r * c

    def row_num(self):
        '返回矩阵的行数'
        return self.shape()[0]

    __len__ = row_num

    def col_num(self):
        '返回矩阵的列数'
        return  self.shape()[1]

    def shape(self):
        '返回矩阵的形状：（行数，列数）'
        return len(self._values), len(self._values[0])  # 返回第0个元素即知道了有几列

    def __repr__(self):
        return 'Matrix({})'.format(self._values)

    __str__ = __repr__ # 返回同样的字符串，不再进行区分 

与Vector相同，建立main_matrix.py文件去验证矩阵函数功能能否实现。

from playLA.Matrix import Matrix

if __name__ == '__main__':
    matrix = Matrix([[1,2],[3,4]])
    print(matrix)
    print('matrix.shape = {}'.format(matrix.shape()))
    print('matrix.size = {}'.format(matrix.size()))
    print('len(matrix) = {}'.format(len(matrix)))
    print('matrix[0][0] = {}'.format(matrix[0,0]))

# 输出结果为
Matrix([[1, 2], [3, 4]])
matrix.shape = (2, 2)
matrix.size = 4
len(matrix) = 2
matrix[0][0] = 1 

加入矩阵基本运算函数

@classmethod
    def zero(cls,r,c):
        '返回一个r行c列的零矩阵'
        return cls([[0] * c for _ in range (r)])

    def __add__(self, another):
        '返回两个矩阵的加法结果'
        assert self.shape() == another.shape(), \
            'Error in adding. Shape of matrix must be same'
        return Matrix([[a + b for a,b in zip(self.row_vector(i),another.row_vector(i))]
                       for i in range(self.row_num())])

    def __sub__(self, another):
        '返回两个矩阵的减法结果'
        assert self.shape() == another.shape(), \
            'Error in adding. Shape of matrix must be same'
        return Matrix([[a - b for a,b in zip(self.row_vector(i),another.row_vector(i))]
                       for i in range(self.row_num())])

    def __mul__(self, k):
        '返回矩阵的数量乘法结果: self * k'
        return Matrix([[e * k for e in self.row_vector(i)]
                       for i in range(self.row_num())])

    def __rmul__(self,k):
        '返回矩阵的数量乘法结果：k * self'
        return self * k

    def __truediv__(self, k):
        '返回矩阵的数量除法结果：self/k'
        return (1/k) * self

    def __pos__(self):
        '返回矩阵取正结果'
        return 1 * self

    def __neg__(self):
        '返回矩阵取负结果'
        return -1 * self 

验证函数功能

 matrix2 = Matrix([[5,6],[7,8]])
    print('add: {}'.format(matrix + matrix2))
    print('subtract: {}'.format(matrix - matrix2))
    print('scalar-mul: {}'.format(matrix * 2))
    print('scalar-mul: {}'.format(2 * matrix))
    print('zero_2_3: {}'.format(Matrix.zero(2,3)))

# 输出结果为
add: Matrix([[6, 8], [10, 12]])
subtract: Matrix([[-4, -4], [-4, -4]])
scalar-mul: Matrix([[2, 4], [6, 8]])
scalar-mul: Matrix([[2, 4], [6, 8]])
zero_2_3: Matrix([[0, 0, 0], [0, 0, 0]]) 

添加矩阵的乘法运算函数

 def dot(self, another):
        """返回矩阵乘法的结果"""
        if isinstance(another, Vector):
            # 矩阵和向量的乘法
            assert self.col_num() == len(another), \
                "Error in Matrix-Vector Multiplication."
            return Vector([self.row_vector(i).dot(another) for i in range(self.row_num())])

        if isinstance(another, Matrix):
            # 矩阵和矩阵的乘法
            assert self.col_num() == another.row_num(), \
                "Error in Matrix-Matrix Multiplication."
            return Matrix([[self.row_vector(i).dot(another.col_vector(j)) for j in range(another.col_num())]
                           for i in range(self.row_num())]) 

验证函数

 T = Matrix([[1.5, 0], [0, 2]])
    p = Vector([5, 3])
    print("T.dot(p) = {}".format(T.dot(p)))

    P = Matrix([[0, 4, 5], [0, 0, 3]])
    # print("T.dot(P) = {}".format(T.dot(P)))

    # print("A.dot(B) = {}".format(matrix.dot(matrix2)))
    # print("B.dot(A) = {}".format(matrix2.dot(matrix))) 

添加转置功能

 def T(self):
        '返回矩阵的转置矩阵'
        return Matrix([[e for e in self.col_vector(i)]
                       for i in range(self.col_num())]) 

验证转置

print('P.T = {}'.format(P.T()))

# 输出结果
P.T = Matrix([[[0], [0]], [[4], [0]], [[5], [3]]]) 

学习使用numpy中的矩阵方法

import numpy as np

if __name__ == '__main__':

    # 矩阵的创建
    A = np.array([ [1,2], [3,4]])
    print(A)

    # 矩阵的属性
    print(A.shape)
    print(A.T)

    # 获取矩阵的元素
    print(A[1,1])
    print(A[0])  # 行向量
    print(A[:,0])  # 取列向量 ； 只有一个‘:’表示从头到尾
    print(A[1,:]) # 取第二行

    # 矩阵的基本运算
    B = np.array([[5,6] , [7,8]])
    print(A + B)
    print(A - B)
    print(10 * A)
    print(A * 10)
    print(A * B) # 对应元素相乘，不适于线性代数中
    print(A.dot(B)) # 真正的矩阵相乘结果

    p = np.array([10,100])
    print(A + p) # 不适于线性代数中
    print(A + 1) # 广播，高维数组可以与低维数组相加，不适于线性代数中

    print(A.dot(p))

# 输出结果
[[1 2]
 [3 4]]
 
(2, 2)

[[1 3]
 [2 4]]
 
4

[1 2]
[1 3]
[3 4]

[[ 6  8]
 [10 12]]
 
[[-4 -4]
 [-4 -4]]
 
[[10 20]
 [30 40]]
 
[[10 20]
 [30 40]]
 
[[ 5 12]
 [21 32]]
 
[[19 22]
 [43 50]]
 
[[ 11 102]
 [ 13 104]]
 
[[2 3]
 [4 5]]
 
[210 430] 

本文地址：https://blog.csdn.net/IAN27/article/details/108240490