问题

在Tensorflow 2中，您可以使用以下代码计算x' * A * y的点积：

import tensorflow as tf

# 假设您已经定义了矩阵 A, 向量 x, 和向量 y

result = tf.linalg.matmul(tf.transpose(x), tf.linalg.matmul(A, y))

这将给您一个标量结果，即x' * A * y的点积。

答案1

得分: 5

Use tf.linalg.tensordot(). See the documentation

As you have mentioned in the question that you are trying to find dot product. In this case tf.matmul() will not work, as it is only for cross product of matrices.

Demo code snippet:

import tensorflow as tf

A = tf.constant([[1,4,6],[2,1,5],[3,2,4]])
x = tf.constant([3,2,7])
result = tf.linalg.tensordot(tf.transpose(x), A, axes=1)
result = tf.linalg.tensordot(result, x, axes=1)
print(result)

And the result will be

>>>tf.Tensor(532, shape=(), dtype=int32)

Few points I want to mention here:

1. Don't forget the axes argument inside tf.linalg.tensordot()

2. When you create tf.zeros(5) it will create a list of shape 5 and it will be like [0,0,0,0,0], when you transpose this it will give you the same list. But if you create it like tf.zeros((5,1)), it would be a vector of shape (5,1) and the result will be

   [
   [0],[0],[0],[0],[0]
   ]
   

Now you can transpose this, and the result will be different, but I recommend you do the code snippet I have mentioned. In case of dot product, you don't have to bother much about this.

If you are still facing issues, will be very happy to help you.

import tensorflow as tf

A = tf.constant([[1,4,6],[2,1,5],[3,2,4]])
x = tf.constant([3,2,7])
result = tf.linalg.tensordot(tf.transpose(x), A, axes=1)
result = tf.linalg.tensordot(result, x, axes=1)
print(result)

>>>tf.Tensor(532, shape=(), dtype=int32)

答案2

得分: 1

仅执行以下操作，

```python
import tensorflow as tf

x = tf.constant([1,2])
a = tf.constant([[2,3],[3,4]])
y = tf.constant([2,3])


z = tf.reshape(tf.matmul(tf.matmul(x[tf.newaxis,:], a), y[:, tf.newaxis]),[])
print(z.numpy())

返回

>>> 49

<details>
<summary>英文:</summary>

Just do the following,

import tensorflow as tf

x = tf.constant([1,2])
a = tf.constant([[2,3],[3,4]])
y = tf.constant([2,3])

z = tf.reshape(tf.matmul(tf.matmul(x[tf.newaxis,:], a), y[:, tf.newaxis]),[])
print(z.numpy())

Returns 

>>> 49

</details>



# 答案3
**得分**: 0

只需使用`tf.transpose`和乘法运算符，如下所示：
`tf.transpose(x)* A * y`。

<details>
<summary>英文:</summary>

Just use `tf.transpose` and multiplication operator like this:
 `tf.transpose(x)* A * y` .

</details>



# 答案4
**得分**: 0

基于您的示例：

```py
x = tf.zeros(5)
A = tf.zeros((5,5))

怎么样

x = tf.expand_dims(x, -1)
tf.matmul(tf.matmul(x, A, transpose_a=True), x)

x = tf.zeros(5)
A = tf.zeros((5,5))

x = tf.expand_dims(x, -1)
tf.matmul(tf.matmul(x, A, transpose_a=True), x)