沿指定轴的两个3D矩阵之间的np.dot积 [英] np.dot product between two 3D matrices along specified axis

问题描述

我有两个3D矩阵:

a = np.random.normal(size=[3,2,5])
b = np.random.normal(size=[5,2,3])

我想要每个切片分别沿2和0轴的点积:

I want the dot product of each slice along 2 and 0 axes respectively:

c = np.zeros([3,3,5]) # c.size is 45
c[:,:,0] = a[:,:,0].dot(b[0,:,:])
c[:,:,1] = a[:,:,1].dot(b[1,:,:])
...

我想使用np.tensordot做到这一点(为了提高效率和速度) 我已经尝试过:

I would like to do that using np.tensordot (for efficiency and speed) I have tried:

c = np.tensordot(a, b, axes=[2,0])

但是我得到了一个包含36个元素(而不是45个)的4D数组. c.shape，c.size =((3L，2L，2L，3L)，36).我在这里发现了类似的问题( Numpy张量:张量的张量张量)，但这并不是我想要的，而且我无法推断该解决方案来解决我的问题. 总而言之，我可以使用np.tensordot计算上面显示的c数组吗?

but I get a 4D array with 36 elements (instead of 45). c.shape, c.size = ((3L, 2L, 2L, 3L), 36). I have found a similar question here (Numpy tensor: Tensordot over frontal slices of tensor) but it's not exactly what I want, and I was unable to extrapolate that solution to my problem. To summarise, can I use np.tensordot to compute c array show above?

更新#1

@hpaulj的答案是我想要的，但是在我的系统(python 2.7和np 1.13.3)中，这些方法相当缓慢:

The answer by @hpaulj is what I wanted, however in my system (python 2.7 and np 1.13.3) those aproaches are pretty slow:

n = 3000
a = np.random.normal(size=[n, 20, 5])
b = np.random.normal(size=[5, 20, n])


t = time.clock()
c_slice = a[:,:,0].dot(b[0,:,:])
print('one slice_x_5: {:.3f} seconds'.format( (time.clock()-t)*5 ))

t = time.clock()
c = np.zeros([n, n, 5])
for i in range(5):
     c[:,:,i] = a[:,:,i].dot(b[i,:,:])
print('for loop: {:.3f} seconds'.format(time.clock()-t))

t = time.clock()
d = np.einsum('abi,ibd->adi', a, b)
print('einsum: {:.3f} seconds'.format(time.clock()-t))

t = time.clock()
e = np.tensordot(a,b,[1,1])
e1 = e.transpose(0,3,1,2)[:,:,np.arange(5),np.arange(5)]
print('tensordot: {:.3f} seconds'.format(time.clock()-t))

a = a.transpose(2,0,1)
t = time.clock()
f = np.matmul(a,b)
print('matmul: {:.3f} seconds'.format(time.clock()-t))

推荐答案

与tensordot相比，使用einsum更容易.因此，让我们从这里开始:

It's easier to work with einsum than tensordot. So let's start there:

In [469]: a = np.random.normal(size=[3,2,5])
     ...: b = np.random.normal(size=[5,2,3])
     ...: 
In [470]: c = np.zeros([3,3,5]) # c.size is 45
In [471]: for i in range(5):
     ...:     c[:,:,i] = a[:,:,i].dot(b[i,:,:])
     ...:     

In [472]: d = np.einsum('abi,ibd->iad', a, b)
In [473]: d.shape
Out[473]: (5, 3, 3)
In [474]: d = np.einsum('abi,ibd->adi', a, b)
In [475]: d.shape
Out[475]: (3, 3, 5)
In [476]: np.allclose(c,d)
Out[476]: True

我不得不考虑一下尺寸.它有助于将a[:,:,i]集中为2d，而对于b[i,:,:]同样.因此，dot总和位于两个数组的中间维度(大小2)上.

I had to think a bit about to match up the dimensions. It helped to focus on a[:,:,i] as 2d, and similarly for b[i,:,:]. So the dot sum is over the middle dimension of both arrays (size 2).

在测试思路中，如果c的前两个维度不同，可能会有所帮助.混合它们的机会会更少.

In testing ideas it might help if the first 2 dimensions of c were different. There'd be less chance of mixing them up.

在tensordot中指定dot求和轴(轴)很容易，但是更难于约束其他尺寸的处理.这就是为什么要获得4D阵列的原因.

It's easy to specify the dot summation axis (axes) in tensordot, but harder to constrain the handling of the other dimensions. That's why you get a 4d array.

我可以使它与转置一起工作，然后取对角线:

I can get it to work with a transpose, followed by taking the diagonal:

In [477]: e = np.tensordot(a,b,[1,1])
In [478]: e.shape
Out[478]: (3, 5, 5, 3)
In [479]: e1 = e.transpose(0,3,1,2)[:,:,np.arange(5),np.arange(5)]
In [480]: e1.shape
Out[480]: (3, 3, 5)
In [481]: np.allclose(c,e1)
Out[481]: True

我计算出的值比需要的多得多，并且将其中的大部分扔掉了.

I've calculated a lot more values than needed, and thrown most of them away.

matmul进行一些移调可能会更好.

matmul with some transposing might work better.

In [482]: f = a.transpose(2,0,1)@b
In [483]: f.shape
Out[483]: (5, 3, 3)
In [484]: np.allclose(c, f.transpose(1,2,0))
Out[484]: True

我认为5维度是不断前进".这就是循环的作用.在einsum中，i的所有部分都相同.

I think of the 5 dimension as 'going-along-for-ride'. That's what your loop does. In einsum the i is the same in all parts.

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