theano-使用张量点计算两个张量的点积 [英] theano - use tensordot compute dot product of two tensor
问题描述
我想使用张量点来计算两个张量的特定暗点的点积.喜欢:
I want to use tensordot to compute the dot product of a specific dim of two tensors. Like:
A是张量,其形状为(3,4,5) B是张量,其形状为(3,5)
A is a tensor, whose shape is (3, 4, 5) B is a tensor, whose shape is (3, 5)
我想用A的第三个暗点和B的第二个暗点做一个点,并得到暗值为(3,4)的输出
I want to do a dot use A's third dim and B's second dim, and get a output whose dims is (3, 4)
像下面这样:
for i in range(3):
C[i] = dot(A[i], B[i])
如何通过张量点做到这一点?
How to do it by tensordot?
推荐答案
好吧,您是否要在numpy或Theano中使用它? 在这种情况下,如您所言,您希望使A的轴3相对于B的轴2收缩,两者都很简单:
Well, do you want this in numpy or in Theano? In the case, where, as you state, you would like to contract axis 3 of A against axis 2 of B, both are straightforward:
import numpy as np
a = np.arange(3 * 4 * 5).reshape(3, 4, 5).astype('float32')
b = np.arange(3 * 5).reshape(3, 5).astype('float32')
result = a.dot(b.T)
在Theano中写为
import theano.tensor as T
A = T.ftensor3()
B = T.fmatrix()
out = A.dot(B.T)
out.eval({A: a, B: b})
但是,输出的形状为(3,4,3).由于您似乎想要输出形状(3,4),因此numpy替代方法使用einsum,就像这样
however, the output then is of shape (3, 4, 3). Since you seem to want an output of shape (3, 4), the numpy alternative uses einsum, like so
einsum_out = np.einsum('ijk, ik -> ij', a, b)
但是,Thein中不存在einsum.因此,这里的具体情况可以如下模拟
However, einsum does not exist in Theano. So the specific case here can be emulated as follows
out = (a * b[:, np.newaxis]).sum(2)
也可以用Theano编写
which can also be written in Theano
out = (A * B.dimshuffle(0, 'x', 1)).sum(2)
out.eval({A: a, B: b})
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