了解张量点

Question

在我学会了如何使用einsum 之后，我现在正在尝试了解np.tensordot 的工作原理。

但是，我有点迷茫，尤其是对于参数axes 的各种可能性。

为了理解它，由于我从未练习过张量演算，我使用以下示例：

A = np.random.randint(2, size=(2, 3, 5))
B = np.random.randint(2, size=(3, 2, 4))

在这种情况下，np.tensordot 可能有哪些不同？您将如何手动计算它？

Answer 1

上面的答案很好，对我理解tensordot有很大帮助。但它们没有显示操作背后的实际数学。这就是为什么我在 TF 2 中为自己做了类似的操作并决定在这里分享它们：

a = tf.constant([1,2.])
b = tf.constant([2,3.])
print(f"{tf.tensordot(a, b, 0)}\t tf.einsum('i,j', a, b)\t\t- ((the last 0 axes of a), (the first 0 axes of b))")
print(f"{tf.tensordot(a, b, ((),()))}\t tf.einsum('i,j', a, b)\t\t- ((() axis of a), (() axis of b))")
print(f"{tf.tensordot(b, a, 0)}\t tf.einsum('i,j->ji', a, b)\t- ((the last 0 axes of b), (the first 0 axes of a))")
print(f"{tf.tensordot(a, b, 1)}\t\t tf.einsum('i,i', a, b)\t\t- ((the last 1 axes of a), (the first 1 axes of b))")
print(f"{tf.tensordot(a, b, ((0,), (0,)))}\t\t tf.einsum('i,i', a, b)\t\t- ((0th axis of a), (0th axis of b))")
print(f"{tf.tensordot(a, b, (0,0))}\t\t tf.einsum('i,i', a, b)\t\t- ((0th axis of a), (0th axis of b))")

[[2. 3.]
 [4. 6.]]    tf.einsum('i,j', a, b)     - ((the last 0 axes of a), (the first 0 axes of b))
[[2. 3.]
 [4. 6.]]    tf.einsum('i,j', a, b)     - ((() axis of a), (() axis of b))
[[2. 4.]
 [3. 6.]]    tf.einsum('i,j->ji', a, b) - ((the last 0 axes of b), (the first 0 axes of a))
8.0          tf.einsum('i,i', a, b)     - ((the last 1 axes of a), (the first 1 axes of b))
8.0          tf.einsum('i,i', a, b)     - ((0th axis of a), (0th axis of b))
8.0          tf.einsum('i,i', a, b)     - ((0th axis of a), (0th axis of b))

对于(2,2) 形状：

a = tf.constant([[1,2],
                 [-2,3.]])

b = tf.constant([[-2,3],
                 [0,4.]])
print(f"{tf.tensordot(a, b, 0)}\t tf.einsum('ij,kl', a, b)\t- ((the last 0 axes of a), (the first 0 axes of b))")
print(f"{tf.tensordot(a, b, (0,0))}\t tf.einsum('ij,ik', a, b)\t- ((0th axis of a), (0th axis of b))")
print(f"{tf.tensordot(a, b, (0,1))}\t tf.einsum('ij,ki', a, b)\t- ((0th axis of a), (1st axis of b))")
print(f"{tf.tensordot(a, b, 1)}\t tf.matmul(a, b)\t\t- ((the last 1 axes of a), (the first 1 axes of b))")
print(f"{tf.tensordot(a, b, ((1,), (0,)))}\t tf.einsum('ij,jk', a, b)\t- ((1st axis of a), (0th axis of b))")
print(f"{tf.tensordot(a, b, (1, 0))}\t tf.matmul(a, b)\t\t- ((1st axis of a), (0th axis of b))")
print(f"{tf.tensordot(a, b, 2)}\t tf.reduce_sum(tf.multiply(a, b))\t- ((the last 2 axes of a), (the first 2 axes of b))")
print(f"{tf.tensordot(a, b, ((0,1), (0,1)))}\t tf.einsum('ij,ij->', a, b)\t\t- ((0th axis of a, 1st axis of a), (0th axis of b, 1st axis of b))")
[[[[-2.  3.]
   [ 0.  4.]]
  [[-4.  6.]
   [ 0.  8.]]]

 [[[ 4. -6.]
   [-0. -8.]]
  [[-6.  9.]
   [ 0. 12.]]]]  tf.einsum('ij,kl', a, b)   - ((the last 0 axes of a), (the first 0 axes of b))
[[-2. -5.]
 [-4. 18.]]      tf.einsum('ij,ik', a, b)   - ((0th axis of a), (0th axis of b))
[[-8. -8.]
 [ 5. 12.]]      tf.einsum('ij,ki', a, b)   - ((0th axis of a), (1st axis of b))
[[-2. 11.]
 [ 4.  6.]]      tf.matmul(a, b)            - ((the last 1 axes of a), (the first 1 axes of b))
[[-2. 11.]
 [ 4.  6.]]      tf.einsum('ij,jk', a, b)   - ((1st axis of a), (0th axis of b))
[[-2. 11.]
 [ 4.  6.]]      tf.matmul(a, b)            - ((1st axis of a), (0th axis of b))
16.0    tf.reduce_sum(tf.multiply(a, b))    - ((the last 2 axes of a), (the first 2 axes of b))
16.0    tf.einsum('ij,ij->', a, b)          - ((0th axis of a, 1st axis of a), (0th axis of b, 1st axis of b))

Answer 2

tensordot 的想法非常简单 - 我们输入数组和相应的轴，沿这些轴进行求和。参与减和的轴在输出中被删除，输入数组中的所有剩余轴都展开作为输出中的不同轴，保持输入数组的顺序喂。

让我们看几个具有一个和两个轴减和的示例案例，并交换输入位置，看看如何在输出中保持顺序。

我。一个轴减和

输入：

 In [7]: A = np.random.randint(2, size=(2, 6, 5))
   ...:  B = np.random.randint(2, size=(3, 2, 4))
   ...: 

案例#1：

In [9]: np.tensordot(A, B, axes=((0),(1))).shape
Out[9]: (6, 5, 3, 4)

A : (2, 6, 5) -> reduction of axis=0
B : (3, 2, 4) -> reduction of axis=1

Output : `(2, 6, 5)`, `(3, 2, 4)` ===(2 gone)==> `(6,5)` + `(3,4)` => `(6,5,3,4)`

案例 #2（与案例 #1 相同，但输入是交换的）：

In [8]: np.tensordot(B, A, axes=((1),(0))).shape
Out[8]: (3, 4, 6, 5)

B : (3, 2, 4) -> reduction of axis=1
A : (2, 6, 5) -> reduction of axis=0

Output : `(3, 2, 4)`, `(2, 6, 5)` ===(2 gone)==> `(3,4)` + `(6,5)` => `(3,4,6,5)`.

二。减和的两个轴

输入：

In [11]: A = np.random.randint(2, size=(2, 3, 5))
    ...: B = np.random.randint(2, size=(3, 2, 4))
    ...: 

案例#1：

In [12]: np.tensordot(A, B, axes=((0,1),(1,0))).shape
Out[12]: (5, 4)

A : (2, 3, 5) -> reduction of axis=(0,1)
B : (3, 2, 4) -> reduction of axis=(1,0)

Output : `(2, 3, 5)`, `(3, 2, 4)` ===(2,3 gone)==> `(5)` + `(4)` => `(5,4)`

案例#2：

In [14]: np.tensordot(B, A, axes=((1,0),(0,1))).shape
Out[14]: (4, 5)

B : (3, 2, 4) -> reduction of axis=(1,0)
A : (2, 3, 5) -> reduction of axis=(0,1)

Output : `(3, 2, 4)`, `(2, 3, 5)` ===(2,3 gone)==> `(4)` + `(5)` => `(4,5)`

我们可以将其扩展到尽可能多的轴。

Answer 3

tensordot 交换轴并重塑输入，以便它可以将np.dot 应用于 2 个二维数组。然后它交换并重塑回目标。实验可能比解释更容易。没有特殊的张量数学，只是扩展dot 以在更高维度上工作。 tensor 仅表示大于 2d 的数组。如果您已经对einsum 感到满意，那么将结果与之比较是最简单的。

一个样本测试，在一对轴上求和

In [823]: np.tensordot(A,B,[0,1]).shape
Out[823]: (3, 5, 3, 4)
In [824]: np.einsum('ijk,lim',A,B).shape
Out[824]: (3, 5, 3, 4)
In [825]: np.allclose(np.einsum('ijk,lim',A,B),np.tensordot(A,B,[0,1]))
Out[825]: True

另一个，对两个求和。

In [826]: np.tensordot(A,B,[(0,1),(1,0)]).shape
Out[826]: (5, 4)
In [827]: np.einsum('ijk,jim',A,B).shape
Out[827]: (5, 4)
In [828]: np.allclose(np.einsum('ijk,jim',A,B),np.tensordot(A,B,[(0,1),(1,0)]))
Out[828]: True

我们可以对(1,0) 做同样的事情。考虑到维度的混合，我认为没有其他组合。