Python矩阵乘法; numpy的阵列

Question

我有一些问题，矩阵乘法：

我想多样信息例如a和b：

a=array([1,3])      # a is random and is array!!! (I have no impact on that) 
            # there is a just for example what I want to do... 

b=[[[1], [2]],      #b is also random but always size(b)= even 
    [[3], [2]], 
    [[4], [6]], 
    [[2], [3]]] 

所以，我想是这样

[1,3]*[1;2]=7 
[1,3]*[3;2]=9 
[1,3]*[4;6]=22 
[1,3]*[2;3]=11 

到多样信息所以结果我需要看看：

x1=[7,9] 
x2=[22,8] 

我知道，是非常复杂的，但是我试着2小时实现这一点，但没有成功:(

Answer 1

如何：

In [16]: a 
Out[16]: array([1, 3]) 

In [17]: b 
Out[17]: 
array([[1, 2], 
     [3, 2], 
     [4, 6], 
     [2, 3]]) 

In [18]: np.array([np.dot(a,row) for row in b]).reshape(-1,2) 
Out[18]: 
array([[ 7, 9], 
     [22, 11]]) 

Answer 2

result = \ 
[[sum(reduce(lambda x,y:x[0]*y[0]+x[1]*y[1],(a,[b1 for b1 in row]))) \ 
    for row in b][i:i+2] \ 
    for i in range(0, len(b),2)] 

Answer 3

你b似乎有必要维度。

有了适当的b你可以使用dot(.)，如：

In []: a 
Out[]: array([1, 3]) 
In []: b 
Out[]: 
array([[1, 2], 
     [3, 2], 
     [4, 6], 
     [2, 3]]) 
In []: dot(b, a).reshape((2, -1)) 
Out[]: 
array([[ 7, 9], 
     [22, 11]])