Python:numpy数组的外部乘积的所有置换的总和

时间:2018-09-16 02:29:23

标签: python numpy numpy-broadcasting numpy-ndarray numpy-einsum

我有一个数组Ai的numpy数组,我希望将每个外部乘积(np.outer(Ai [i],Ai [j]))与比例乘子相加,以产生H。然后用比例因子矩阵对它们进行张量。我认为事情可以大大简化,但是还没有找到一种通用/高效的方法来进行ND。如何更轻松地生成Arr2D和H?注意:Arr2D可以是64个2D阵列,而不是8x8 2D阵列。

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]))

1 个答案:

答案 0 :(得分:1)

好的设置可以利用einsum

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

时间-

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+ 在那里加速!