numpy 3D点产品

Question

我有两个3dim numpy矩阵，我想在不使用循环的情况下根据一个轴做点积：

a=[ [[ 0, 0, 1, 1, 0,  0,  0,  0,  0,  0,  1,  0,  0,  1,  0],
  [ 1,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  1,  0,  1,  0],
  [ 0,  1,  0,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  0,  1],
  [ 0,  1,  0,  0,  0,  0,  1,  0,  0,  0,  1,  0,  1,  0,  0]],
    [[ 0,  0,  1,  1,  0,  0,  0,  0,  0,  0,  1,  0,  0,  1,  0],
  [ 1,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  1,  0,  1,  0],
  [ 0,  1,  0,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  0,  1],
  [ 0,  1,  0,  0,  0,  0,  1,  0,  0,  0,  1,  0,  1,  0,  0]],
 [ [ 0,  0,  1,  1,  0,  0,  0,  0,  0,  0,  1,  0,  0,  1,  0],
  [ 1,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  1,  0,  1,  0],
  [ 0,  1,  0,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  0,  1],
  [ 0,  1,  0,  0,  0,  0,  1,  0,  0,  0,  1,  0,  1,  0,  0]],
 [ [ 0,  0,  1,  1,  0,  0,  0,  0,  0,  0,  1,  0,  0,  1,  0],
  [ 1,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  1,  0,  1,  0],
  [ 0,  1,  0,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  0,  1],
  [ 0,  1,  0,  0,  0,  0,  1,  0,  0,  0,  1,  0,  1,  0,  0]],
 [[ 0,  0,  1,  1,  0,  0,  0,  0,  0,  0,  1,  0,  0,  1,  0],
  [ 1,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  1,  0,  1,  0],
  [ 0,  1,  0,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  0,  1],
  [ 0,  1,  0,  0,  0,  0,  1,  0,  0,  0,  1,  0,  1,  0,  0]],
 [[ 0,  0,  1,  1,  0,  0,  0,  0,  0,  0,  1,  0,  0,  1,  0],
  [ 1,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  1,  0,  1,  0],
  [ 0,  1,  0,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  0,  1],
  [ 0,  1,  0,  0,  0,  0,  1,  0,  0,  0,  1,  0,  1,  0,  0.]],
 [[ 0,  0,  1,  1,  0,  0,  0,  0,  0,  0,  1,  0,  0,  1,  0],
  [ 1,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  1,  0,  1,  0],
  [ 0,  1,  0,  0,  0,  1,  0,  0,  0,  0,  0,  0,  0,  0,  1],
  [ 0,  1,  0,  0,  0,  0,  1,  0,  0,  0,  1,  0,  1,  0,  0]]]

b=[[[ 0,  0,  1,  0,  0.],
  [ 1,  0,  0,  0,  0.],
  [ 0,  0,  0,  0,  0.],
  [ 0,  1,  0,  0,  0.]],
 [[ 0,  0,  1,  0,  0.],
  [ 1,  0,  0,  0,  0.],
  [ 0,  0,  0,  0,  0.],
  [ 0,  1,  0,  0,  0.]],
 [[ 0,  0,  1,  0,  0.],
  [ 1,  0,  0,  0,  0.],
  [ 0,  0,  0,  0,  0.],
  [ 0,  1,  0,  0,  0.]],
 [[ 0,  0,  1,  0,  0.],
  [ 1,  0,  0,  0,  0.],
  [ 0,  0,  0,  0,  0.],
  [ 0,  1,  0,  0,  0.]],
 [[ 0,  0,  1,  0,  0.],
  [ 1,  0,  0,  0,  0.],
  [ 0,  0,  0,  0,  0.],
  [ 0,  1,  0,  0,  0.]],
 [[ 0,  0,  1,  0,  0.],
  [ 1,  0,  0,  0,  0.],
  [ 0,  0,  0,  0,  0.],
  [ 0,  1,  0,  0,  0.]],
 [[ 0,  0,  1,  0,  0.],
  [ 1,  0,  0,  0,  0.],
  [ 0,  0,  0,  0,  0.],
  [ 0,  1,  0,  0,  0.]]]
dt = np.dtype(np.float32)
a=np.asarray(a,dtype=dt)
b=np.asarray(b,dtype=dt)
print(a.shape)
print(b.shape)

a的形状是（7，4，15），b的形状是（7，4，5）。 我要c=美国运输部（a，b）尺寸为（7,5,15），如下所示：

^{pr2}$

但我正在寻找一个没有for循环的解决方案。比如：

c = np.tensordot(a.reshape(4,7,5),b.reshape(7,4,15),axes=([1,0],[0,1]))

但这个并没有如预期的那样工作。在

我也试过了：

newaxes_a=[2,0,1]
newaxes_b=[1,0,2]

newshape_a=(-1,28)
newshape_b=(28,-1)
a_t = a.transpose(newaxes_a).reshape(newshape_a)
b_t = b.transpose(newaxes_b).reshape(newshape_b)
c = np.dot(a_t, b_t)

但并没有如预期的那样起作用。在

有什么想法吗？在

Answer 1

您可以使用^{}-

#to match the given example
c2 = np.einsum('ijk,ijl->ikl',a,b)
print np.allclose(c, c2)

另一个使用^{}-

^{pr2}$