Python 3.x 在循环矩阵乘法中如何将结果赋给新矩阵

我编写了分别乘以2x3和3x2的两个矩阵的算法。算法运行良好。问题是如何将结果保存到2x2的合成矩阵中

下面是代码

from numpy import *

m1 = matrix('1 2 3; 4 5 6')
m2 = matrix('7 8; 9 10; 11 12')
m4 = matrix('')

for k in range(len(m1)):
    for i in range(len(m1)):
        m3 = 0
        for j in range(len(m2)):
            m3 = m3 + m1[k, j] * m2[j, i]
        print(m3)

---

对于使用列表列表和Numpy数组的矩阵乘法，我们可以使用相同的算法

In [30]: l1 = [[1,2,3],[4,5,6]] ; l2 = [[7,8],[9,10],[11,12]]                                                                         
In [31]: l3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*l2)] for row in l1]                    
In [32]: a1 = np.array(l1) ; a2 = np.array(l2)                                                               
In [33]: a3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*a2)] for row in a1]                    
In [34]: a3                                                                                                  
Out[34]: [[58, 64], [139, 154]]
In [35]: l3                                                                                                  
Out[35]: [[58, 64], [139, 154]]

为了进行检查，让我们使用内置矩阵乘法运算符

@

In [36]: a1@a2                                                                                               
Out[36]: 
array([[ 58,  64],
       [139, 154]])

现在，让我们看看如果我们的数据被包装在一个矩阵类中，我们是否可以使用相同的矩阵乘法算法

In [37]: m1 = np.matrix(l1) ; m2 = np.matrix(l2)                                                             
In [38]: m3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*m2)] for row in m1]                    
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-38-2cd8454aa248> in <module>
----> 1 m3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*m2)] for row in m1]
<ipython-input-38-2cd8454aa248> in <listcomp>(.0)
----> 1 m3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*m2)] for row in m1]
<ipython-input-38-2cd8454aa248> in <listcomp>(.0)
----> 1 m3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*m2)] for row in m1]
<ipython-input-38-2cd8454aa248> in <genexpr>(.0)
----> 1 m3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*m2)] for row in m1]
~/lib/miniconda3/lib/python3.7/site-packages/numpy/matrixlib/defmatrix.py in __mul__(self, other)
    218         if isinstance(other, (N.ndarray, list, tuple)) :
    219             # This promotes 1-D vectors to row vectors
--> 220             return N.dot(self, asmatrix(other))
    221         if isscalar(other) or not hasattr(other, '__rmul__') :
    222             return N.dot(self, other)

ValueError: shapes (1,3) and (1,2) not aligned: 3 (dim 1) != 1 (dim 0)

这个故事的寓意是什么

matrix类具有固有的粘性，这会导致这种特殊的意外行为以及其他不同的意外行为

如果不能确切了解与使用

np.matrix（）

相关的所有问题，最好的选择是使用

np.array

实例化的2D数组

另一方面，如果一个人确切地知道与使用

np.matrix（）

相关的所有问题，通常他们不想使用

np.matrix

，上面说“不再建议使用这个类，即使是线性代数。改为使用正则数组。这个类将来可能会被删除。”-said“

In [37]: m1 = np.matrix(l1) ; m2 = np.matrix(l2)                                                             
In [38]: m3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*m2)] for row in m1]                    
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-38-2cd8454aa248> in <module>
----> 1 m3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*m2)] for row in m1]
<ipython-input-38-2cd8454aa248> in <listcomp>(.0)
----> 1 m3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*m2)] for row in m1]
<ipython-input-38-2cd8454aa248> in <listcomp>(.0)
----> 1 m3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*m2)] for row in m1]
<ipython-input-38-2cd8454aa248> in <genexpr>(.0)
----> 1 m3 = [[sum(e1*e2 for e1, e2 in zip(row,col)) for col in zip(*m2)] for row in m1]
~/lib/miniconda3/lib/python3.7/site-packages/numpy/matrixlib/defmatrix.py in __mul__(self, other)
    218         if isinstance(other, (N.ndarray, list, tuple)) :
    219             # This promotes 1-D vectors to row vectors
--> 220             return N.dot(self, asmatrix(other))
    221         if isscalar(other) or not hasattr(other, '__rmul__') :
    222             return N.dot(self, other)

ValueError: shapes (1,3) and (1,2) not aligned: 3 (dim 1) != 1 (dim 0)

In [39]: m1[0]                                                                                               
Out[39]: matrix([[1, 2, 3]])