python中数组和矩阵乘法及使用总结

朴灿烈づ我的快乐病毒、 2022-05-27 14:16 472阅读 0赞

Matrix是Array的一个小的分支，包含于Array。所以matrix 拥有array的所有特性。

但在数组乘和矩阵乘时,两者各有不同,如果a和b是两个matrices，那么a\*b，就是矩阵积

如果a,b是数组的话,则a\*b是数组的运算

1.对数组的操作

>>> import numpy as np

>>> a=np.array([[1,2,3],[4,5,6],[7,8,9]])
    >>> a
    array([[1, 2, 3],
           [4, 5, 6],
           [7, 8, 9]])
    >>> b=a.copy()
    >>> b
    array([[1, 2, 3],
           [4, 5, 6],
           [7, 8, 9]])
    >>> a+b#多维数组的加减,按对应位置操作
    array([[ 2,  4,  6],
           [ 8, 10, 12],
           [14, 16, 18]])
    >>> a*3#多维数组乘常数,则对数组中每一个元素乘该常数
    array([[ 3,  6,  9],
           [12, 15, 18],
           [21, 24, 27]])
    >>> np.dot(a,b)#数组的点乘运算通过np.dot(a,b)来实现,相当于矩阵乘
    array([[ 30,  36,  42],
           [ 66,  81,  96],
           [102, 126, 150]])
    >>> c=np.array([1,2,3])#构造一行三列的数组
    >>> c
    array([1, 2, 3])
    >>> c*a#c为一行三列,放于数组a之前,则对数组a中每行对应位置相乘
    array([[ 1,  4,  9],
           [ 4, 10, 18],
           [ 7, 16, 27]])
    >>> a*c#c为一行三列,放于数组a之后,依旧是对数组a中每行对应位置相乘
    array([[ 1,  4,  9],
           [ 4, 10, 18],
           [ 7, 16, 27]])
    >>> #如果想要矩阵运算,则需要np.dot()函数
    >>> np.dot(c,a)#c为一行三列,放于数组a之前,按正常矩阵方式运算
    array([30, 36, 42])
    >>> np.dot(a,c)#c为一行三列,放于数组a之后,相当于矩阵a乘以3行一列的c矩阵,返回结果值不变,格式为1行3列
    array([14, 32, 50])
    >>> #将c改为多行一列的形式
    >>> d=c.reshape(3,1)
    >>> d
    array([[1],
           [2],
           [3]])
    >>> #
    >>> np.dot(a,d)#值与np.dot(a,c)一致,但格式以改变为3行1列
    array([[14],
           [32],
           [50]])
    
    >>> a*a#数组使用*的运算其结果属于数组运算,对应位置元素之间的运算
    array([[ 1,  4,  9],
           [16, 25, 36],
           [49, 64, 81]])
    >>> #但是不能更改a,d点乘的位置,不符合矩阵运算格式
    >>> np.dot(d,a)
    Traceback (most recent call last):
      File "<pyshell#28>", line 1, in <module>
        np.dot(d,a)
    ValueError: shapes (3,1) and (3,3) not aligned: 1 (dim 1) != 3 (dim 0)

对于数组的转置,求逆,求迹运算请参考上篇文章

2.对矩阵的操作

>>> a=np.array([[1,2,3],[4,5,6],[7,8,9]])
    >>> a=np.mat(a)
    >>> a
    matrix([[1, 2, 3],
            [4, 5, 6],
            [7, 8, 9]])
    >>> b=a
    >>> b
    matrix([[1, 2, 3],
            [4, 5, 6],
            [7, 8, 9]])
    >>> a+b#矩阵的加减运算和数组运算一致
    matrix([[ 2,  4,  6],
            [ 8, 10, 12],
            [14, 16, 18]])
    >>> a-b
    matrix([[0, 0, 0],
            [0, 0, 0],
            [0, 0, 0]])
    >>> a*b#矩阵的乘用*即可表示
    matrix([[ 30,  36,  42],
            [ 66,  81,  96],
            [102, 126, 150]])
    >>> np.dot(a,b)#与*一致
    matrix([[ 30,  36,  42],
            [ 66,  81,  96],
            [102, 126, 150]])
    >>> b*a
    matrix([[ 30,  36,  42],
            [ 66,  81,  96],
            [102, 126, 150]])
    >>> np.dot(b,a)
    matrix([[ 30,  36,  42],
            [ 66,  81,  96],
            [102, 126, 150]])
    >>> c=np.array([1,2,3])#构造一行三列数组
    >>> c
    array([1, 2, 3])
    >>> c*a#矩阵运算
    matrix([[30, 36, 42]])
    >>> a*c#不合矩阵规则
    Traceback (most recent call last):
      File "<pyshell#63>", line 1, in <module>
        a*c
      File "F:\python3\anzhuang\lib\site-packages\numpy\matrixlib\defmatrix.py", line 309, in __mul__
        return N.dot(self, asmatrix(other))
    ValueError: shapes (3,3) and (1,3) not aligned: 3 (dim 1) != 1 (dim 0)
    >>> np.dot(c,a)#和矩阵运算一致
    matrix([[30, 36, 42]])
    >>> np.dot(a,c)#自动将a转换成3行1列参与运算,返回结果格式已经变为1行3列而非3行一列的矩阵
    matrix([[14, 32, 50]])
    >>> c=c.reshape(3,1)
    >>> c
    array([[1],
           [2],
           [3]])
    >>> a*c#和矩阵运算一致
    matrix([[14],
            [32],
            [50]])
    >>> c*a#不合矩阵运算格式
    Traceback (most recent call last):
      File "<pyshell#71>", line 1, in <module>
        c*a  
    ValueError: shapes (3,1) and (3,3) not aligned: 1 (dim 1) != 3 (dim 0)

矩阵运算的另一个好处就是方便于求转置,求逆,求迹

>>> a
    matrix([[1, 2, 3],
            [4, 5, 6],
            [7, 8, 9]])
    >>> a.T
    matrix([[1, 4, 7],
            [2, 5, 8],
            [3, 6, 9]])
    >>> a.H#共轭转置
    matrix([[1, 4, 7],
            [2, 5, 8],
            [3, 6, 9]])
    >>> b=np.eye(3)*3
    >>> b
    array([[3., 0., 0.],
           [0., 3., 0.],
           [0., 0., 3.]])
    >>> b=np.mat(b)
    >>> b.I#求逆运算
    matrix([[0.33333333, 0.        , 0.        ],
            [0.        , 0.33333333, 0.        ],
            [0.        , 0.        , 0.33333333]])
    >>> np.trace(b)#求迹运算
    9.0