python - 为什么 numpy.dot 会这样？

Question

我试图理解为什么 numpy 的 dot 函数会这样运行:

M = np.ones((9, 9))
V1 = np.ones((9,))
V2 = np.ones((9, 5))
V3 = np.ones((2, 9, 5))
V4 = np.ones((3, 2, 9, 5))

现在 np.dot(M, V1) 和 np.dot(M, V2) 表现为预期的。但是对于 V3 和 V4 结果出乎意料我:

>>> np.dot(M, V3).shape
(9, 2, 5)
>>> np.dot(M, V4).shape
(9, 3, 2, 5)

我预计 (2, 9, 5) 和 (3, 2, 9, 5) 分别。另一方面，np.matmul做我所期望的:矩阵乘法被广播在第二个参数的第一个 N - 2 维和结果具有相同的形状:

>>> np.matmul(M, V3).shape
(2, 9, 5)
>>> np.matmul(M, V4).shape
(3, 2, 9, 5)

所以我的问题是:这样做的理由是什么np.dot 表现如何？它是否用于某些特定目的，还是应用某些一般规则的结果？

Answer 1

来自 the docs for np.dot :

For 2-D arrays it is equivalent to matrix multiplication, and for 1-D arrays to inner product of vectors (without complex conjugation). For N dimensions it is a sum product over the last axis of a and the second-to-last of b:

dot(a, b)[i,j,k,m] = sum(a[i,j,:] * b[k,:,m])

对于np.dot(M, V3),

(9, 9), (2, 9, 5) --> (9, 2, 5)

For np.dot(M, V4),

(9, 9), (3, 2, 9, 5) --> (9, 3, 2, 5)

The strike-through represents dimensions that are summed over, and are therefore not present in the result.

---

In contrast, np.matmul treats N-dimensional arrays as 'stacks' of 2D matrices:

The behavior depends on the arguments in the following way.

• If both arguments are 2-D they are multiplied like conventional matrices.
• 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.

The same reductions are performed in both cases, but the order of the axes is different. np.matmul essentially does the equivalent of:

for ii in range(V3.shape[0]):
    out1[ii, :, :] = np.dot(M[:, :], V3[ii, :, :])

和

for ii in range(V4.shape[0]):
    for jj in range(V4.shape[1]):
        out2[ii, jj, :, :] = np.dot(M[:, :], V4[ii, jj, :, :])