Python：数组的numpy数组外积的所有排列之和

Question

我有一个数组 Ai 的 numpy 数组，我希望每个外积 (np.outer(Ai[i],Ai[j])) 与缩放乘数相加以产生 H。我可以逐步完成并制作它们，然后使用比例因子矩阵张量它们。我认为事情可以大大简化，但还没有想出 general/efficient 方法来为 ND 做到这一点。 Arr2D和H如何更容易产生？注意：Arr2D 可以是 64 个二维数组而不是 8x8 个二维数组。

Ai = np.random.random((8,101))
Arr2D = np.zeros((Ai.shape[0], Ai.shape[0], Ai.shape[1], Ai.shape[1]))
Arr2D[:,:,:,:] = np.asarray([ np.outer(Ai[i], Ai[j]) for i in range(Ai.shape[0]) 
    for j in range(Ai.shape[0]) ]).reshape(Ai.shape[0],Ai.shape[0],Ai[0].size,Ai[0].size)
arr = np.random.random( (Ai.shape[0] * Ai.shape[0]) )
arr2D = arr.reshape(Ai.shape[0], Ai.shape[0])
H = np.tensordot(Arr2D, arr2D, axes=([0,1],[0,1]))

Answer 1

利用 einsum 的良好设置！

np.einsum('ij,kl,ik->jl',Ai,Ai,arr2D,optimize=True)

计时 -

In [71]: # Setup inputs
    ...: Ai = np.random.random((8,101))
    ...: arr = np.random.random( (Ai.shape[0] * Ai.shape[0]) )
    ...: arr2D = arr.reshape(Ai.shape[0], Ai.shape[0])

In [74]: %%timeit # Original soln
    ...: Arr2D = np.zeros((Ai.shape[0], Ai.shape[0], Ai.shape[1], Ai.shape[1]))
    ...: Arr2D[:,:,:,:] = np.asarray([ np.outer(Ai[i], Ai[j]) for i in range(Ai.shape[0]) 
    ...:     for j in range(Ai.shape[0]) ]).reshape(Ai.shape[0],Ai.shape[0],Ai[0].size,Ai[0].size)
    ...: H = np.tensordot(Arr2D, arr2D, axes=([0,1],[0,1]))
100 loops, best of 3: 4.5 ms per loop

In [75]: %timeit np.einsum('ij,kl,ik->jl',Ai,Ai,arr2D,optimize=True)
10000 loops, best of 3: 146 µs per loop

30x+ 那里加速了！

Python：数组的numpy数组外积的所有排列之和

Python: Sum of all permutations of outer products of numpy arrays of arrays

python

numpy

numpy-einsum

numpy-ndarray

array-broadcasting