numpy 3D dot产品

时间:2015-11-30 16:42:53

标签: python numpy vectorization dot-product numpy-broadcasting

我有两个3dim numpy矩阵,我想根据一个轴做一个点积而不使用循环:

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具有(7,4,15)的形状,b具有(7,4,5)的形状。 我希望c = np.dot(a,b)的大小为(7,5,15),如下所示:

c = np.zeros((7,15,5))
for i in range(7):
   c[i,:,:] = np.dot(a[i,:,:].T , b[i,:,:])

但我正在寻找没有for-loop的解决方案。类似的东西:

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

但是这个没有按预期工作。

我也试过这个:

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)

没有按预期工作。

有什么想法吗?

1 个答案:

答案 0 :(得分:4)

您可以使用np.einsum -

c = (a[:,:,None,:]*b[...,None]).sum(1)

另一个使用broadcasting -

on('line')