如何使用循环将结果分配给矩阵乘法中的新矩阵

Question

我编写了分别乘以2x3和3x2两个矩阵的算法。该算法表现完美。问题是如何将结果保存到2x2的结果矩阵中？

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)

Answer 1

我们可以使用相同的算法进行矩阵乘法，使用列表和Numpy数组

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]]

并且，对于检查，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包含在矩阵类中，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)

为什么我们有这个错误？尺寸和以前一样，为什么我们有不同的形状？

这就是原因，当我们查看m1的元素时，我们没有一维数组或非嵌套列表，我们还有另一个矩阵

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

这个故事的道德？

矩阵类具有固有的粘性，导致这种特殊的意外行为以及其他不同的意外行为。

如果不完全知道与使用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，如果一个人确切知道与使用np.matrix()相关的所有问题，通常他们不想使用np.matrix。

Answer 2

from numpy import *

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

m1_Row_Col = m1.shape
m2_Row_Col = m2.shape

m3 = [[0 for row in range(m2_Row_Col[1])] for col in range(m1_Row_Col[0])]

for k in range(len(m1)):
    for i in range(m2_Row_Col[1]):
        out = 0
        for j in range(len(m2)):
            out = out + m1[k, j] * m2[j, i]
        m3[k][i] = out
print(m3)