numpy (n, m) 和 (n, k) 到 (n, m, k)

Question

让x是 np.array形状(n, m) .

让y是 np.array形状(n, k) .

计算张量的正确方法是什么 z形状(n, m, k)这样

for all i in [0, n - 1]
z[i] = np.dot(x[i][:, np.newaxis], y[i][np.newaxis, :])

？

换句话说，每对行(x_i, y_i)给出一个形状为 (m, k) 的矩阵.

我查看了np.tensordot但经过多次试验，我找不到其 axes 的正确值争论。我不确定它是否适合这项工作。

Answer 1

您可以像这样使用np.einsum():

z = np.einsum('ij,ik->ijk', x, y)

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根据快速测试，这也比基于 np.matmul() 的方法更快(除了非常小的输入):

import numpy as np


x = np.random.randint(1, 100, (2, 3))
y = np.random.randint(1, 100, (2, 4))

%timeit np.einsum('ij,ik->ijk', x, y)
# 100000 loops, best of 3: 3.14 µs per loop
%timeit np.matmul(x[:, :, None], y[:, None, :])
# 100000 loops, best of 3: 2,07 µs per loop


x = np.random.randint(1, 100, (20, 30))
y = np.random.randint(1, 100, (20, 40))

%timeit np.einsum('ij,ik->ijk', x, y)
# 10000 loops, best of 3: 32.1 µs per loop
%timeit np.matmul(x[:, :, None], y[:, None, :])
# 10000 loops, best of 3: 76.8 µs per loop


x = np.random.randint(1, 100, (200, 300))
y = np.random.randint(1, 100, (200, 400))

%timeit np.einsum('ij,ik->ijk', x, y)
# 10 loops, best of 3: 48.7 ms per loop
%timeit np.matmul(x[:, :, None], y[:, None, :])
# 10 loops, best of 3: 68.2 ms per loop

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将np.dot()应用于可广播 View ，如下所示:

np.dot(x[:, :, None], y[:, None, :])

不起作用(它甚至不会达到正确的形状)。

(已编辑)