沿第三轴的点积 [英] Dot product along third axis

问题描述

我正在尝试使用 tensordot 在 numpy 中获取张量点积，但我不确定我应该如何重塑我的数组以实现我的计算.(总的来说，我对张量数学还是个新手.)

I'm trying to take a tensor dot product in numpy using tensordot, but I'm not sure how I should reshape my arrays to achieve my computation. (I'm still new to the mathematics of tensors, in general.)

我有

arr = np.array([[[1, 1, 1],
                [0, 0, 0],
                [2, 2, 2]],

               [[0, 0, 0],
                [4, 4, 4],
                [0, 0, 0]]])

w = [1, 1, 1]

我想沿 axis=2 取点积，这样我就有了矩阵

And I want to take a dot product along axis=2, such that I have the matrix

array([[3, 0, 6],
       [0, 12, 0]])

什么是正确的 numpy 语法?np.tensordot(arr, [1, 1, 1], axes=2) 似乎引发了 ValueError.

What's the proper numpy syntax for this? np.tensordot(arr, [1, 1, 1], axes=2) seems to raise a ValueError.

推荐答案

对于 arr 和 axis=0 减少沿 axis=2> 为 w.因此，使用 np.tensordot，解决方案是-

The reduction is along axis=2 for arr and axis=0 for w. Thus, with np.tensordot, the solution would be -

np.tensordot(arr,w,axes=([2],[0]))

或者，也可以使用 np.einsum -

Alternatively, one can also use np.einsum -

np.einsum('ijk,k->ij',arr,w)

np.matmul 也有效

np.matmul(arr, w)

运行时测试 -

In [52]: arr = np.random.rand(200,300,300)

In [53]: w = np.random.rand(300)

In [54]: %timeit np.tensordot(arr,w,axes=([2],[0]))
100 loops, best of 3: 8.75 ms per loop

In [55]: %timeit np.einsum('ijk,k->ij',arr,w)
100 loops, best of 3: 9.78 ms per loop

In [56]: %timeit np.matmul(arr, w)
100 loops, best of 3: 9.72 ms per loop

hlin117 在 Macbook Pro OS X El Capitan 上测试，numpy 版本 1.10.4.

hlin117 tested on Macbook Pro OS X El Capitan, numpy version 1.10.4.

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