python - 当矩阵a是二维矩阵，b是三维矩阵时，如何理解matmul函数？

Question

a=np.arange(8).reshape(2,2,2)
b=np.arange(4).reshape(2,2)
print(np.matmul(a,b))

结果是:[[[ 2 3] [6 11]]

[[10 19] [14 27]]]如何理解结果？它是如何产生的

Answer 1

简短回答:它将第二个 2d 矩阵“广播”到 3d 矩阵，然后执行“映射”，以便将元素子矩阵映射到结果中的新子矩阵。

作为documentation on np.matmul [numpy-doc]说:

numpy.matmul(a, b, out=None)

Matrix product of two arrays.

The behavior depends on the arguments in the following way.

1. If both arguments are 2-D they are multiplied like conventional matrices.
2. If either argument is N-D, N > 2, it is treated as a stack of matrices residing in the last two indexes and broadcast accordingly.
3. If the first argument is 1-D, it is promoted to a matrix by prepending a 1 to its dimensions. After matrix multiplication the prepended 1 is removed.
4. If the second argument is 1-D, it is promoted to a matrix by appending a 1 to its dimensions. After matrix multiplication the appended 1 is removed.

所以这里第二项适用。因此，首先第二个矩阵也被“广播”到 3d 变体，这意味着我们要乘以:

array([[[0, 1],
        [2, 3]],

       [[4, 5],
        [6, 7]]])

与:

array([[[0, 1],
        [2, 3]],

       [[0, 1],
        [2, 3]]])

我们将它们视为堆叠矩阵。首先我们乘以:

array([[0, 1],      array([[0, 1],
       [2, 3]])  x        [2, 3]])

这给了我们:

array([[ 2,  3],
       [ 6, 11]])

然后是元素第二个子矩阵:

array([[4, 5],      array([[0, 1],
       [6, 7]])  x        [2, 3]])

这给了我们:

array([[10, 19],
       [14, 27]])

因此，我们将它们叠加到结果中，并获得:

>>> np.matmul(a, b)
array([[[ 2,  3],
        [ 6, 11]],

       [[10, 19],
        [14, 27]]])

尽管该行为已被完美定义，但最好谨慎使用此功能，因为对于 3d 矩阵与 2d 矩阵的“矩阵乘积”可能是什么样子，还有其他“有意义的”定义，因此这些是此处未使用。